**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Snakes
time_limit_ms: 3000
memory_limit_mb: 512
Description: Suppose you play a game where the game field looks like a strip of $$$1 \times 10^9$$$ square cells, numbered from $$$1$$$ to $$$10^9$$$.

You have $$$n$$$ snakes (numbered from $$$1$$$ to $$$n$$$) you need to place into some cells. Initially, each snake occupies exactly one cell, and you can't place more than one snake into one cell. After that, the game starts.

The game lasts for $$$q$$$ seconds. There are two types of events that may happen each second:

- snake $$$s_i$$$ enlarges: if snake $$$s_i$$$ occupied cells $$$[l, r]$$$, it enlarges to a segment $$$[l, r + 1]$$$;
- snake $$$s_i$$$ shrinks: if snake $$$s_i$$$ occupied cells $$$[l, r]$$$, it shrinks to a segment $$$[l + 1, r]$$$.

Each second, exactly one of the events happens.

If at any moment of time, any snake runs into some obstacle (either another snake or the end of the strip), you lose. Otherwise, you win with the score equal to the maximum cell occupied by any snake so far.

What is the minimum possible score you can achieve?

Input: The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n \le 20$$$; $$$1 \le q \le 2 \cdot 10^5$$$) — the number of snakes and the number of events. Next $$$q$$$ lines contain the description of events — one per line.

The $$$i$$$-th line contains

- either "$$$s_i$$$ +" ($$$1 \le s_i \le n$$$) meaning that the $$$s_i$$$-th snake enlarges
- or "$$$s_i$$$ -" ($$$1 \le s_i \le n$$$) meaning that the $$$s_i$$$-th snake shrinks.

Additional constraint on the input: the given sequence of events is valid, i. e. a snake of length $$$1$$$ never shrinks.

Output: Print one integer — the minimum possible score.

Examples:
input:
3 6
1 +
1 -
3 +
3 -
2 +
2 -
output:
4

input:
5 13
5 +
3 +
5 -
2 +
4 +
3 +
5 +
5 -
2 +
3 -
3 +
3 -
2 +
output:
11

Note: In the first test, the optimal strategy is to place the second snake at cell $$$1$$$, the third snake — at $$$2$$$, and the first one — at $$$3$$$. The maximum occupied cell is cell $$$4$$$, and it's the minimum possible score.

In the second test, one of the optimal strategies is to place:

- snake $$$2$$$ at position $$$1$$$;
- snake $$$3$$$ at position $$$4$$$;
- snake $$$5$$$ at position $$$6$$$;
- snake $$$1$$$ at position $$$9$$$;
- snake $$$4$$$ at position $$$10$$$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;

#define int long long
#define all(v)                ((v).begin()), ((v).end())
#define allr(v)                ((v).rbegin()), ((v).rend())
#define sz(v)                ((int)((v).size()))
#define clr(v, d)            memset(v, d, sizeof(v))
#define rep(i, v)        for(int i=0;i<sz(v);++i)
#define lp(i, n)        for(int i=0;i<(int)(n);++i)
#define lpi(i, j, n)    for(int i=(j);i<(int)(n);++i)
#define lpd(i, j, n)    for(int i=(j);i>=(int)(n);--i)
#define ceils(x, y) (x + y - 1) / y

//for pairs
#define pii pair<int, int>
#define vii vector<pii>

const double EPS = (1e-7);
int dcmp(double x, double y) {    return fabs(x - y) <= EPS ? 0 : x < y ? -1 : 1;    }
int fix(int a, int n) {
    return (a % n + n) % n;
}

#define pb                    push_back
#define MP                    make_pair
#define P(x)                cout<<#x<<" = { "<<x<<" }\n"
typedef long double         ld;
typedef vector<int>       vi;
typedef vector<double>    vd;
typedef vector< vi >      vvi;
typedef vector< vd >      vvd;
typedef vector<string>    vs;

#define sumv(v) accumulate(all(v), 0ll)
#define maxv(v) *max_element(all(v))
#define minv(v) *min_element(all(v))

// Handling vector, vector of pairs
template <typename T1, typename T2> // cin >> pair<T1, T2>
istream& operator>>(istream& istream, pair<T1, T2>& p) {return (istream >> p.first >> p.second);}
template <typename T1, typename T2> // cout << pair<T1, T2>
ostream& operator<<(ostream& ostream, const pair<T1, T2>& p) {return (ostream << p.first << " " << p.second);}
template<typename T> // cin >> vector<T>
istream& operator>>(istream &in, vector<T> &v) {for (auto &it : v) cin >> it; return in;}
template<typename T> //cout << vector<T>
ostream& operator<<(ostream &out, const vector<T> &c) {for (auto &it : c) cout << it << " "; return out;}

int binpow(int a, int b) {int res = 1; while (b > 0) {if (b & 1)res = res * a; a = a * a; b >>= 1;} return res;}

int gcd (int a, int b) { while (b) {a %= b; swap(a, b);} return a;}
int lcm (int a, int b) {return a / gcd(a, b) * b;}
int int_sqrt (long long x) {int ans = 0; for (int k = 1LL << 30; k != 0; k /= 2) {if ((ans + k) * (ans + k) <= x) {ans += k;}} return ans;}
bool isOverflow(int x, int y) {return (x > LLONG_MAX / y or y > LLONG_MAX / x);}

const int mod = 1e9 + 7;

int inv(int i) {if (i == 1) return 1; return (mod - ((mod / i) * inv(mod % i)) % mod) % mod;}
int mod_mul(int a, int b) {a = a % mod; b = b % mod; return (((a * b) % mod) + mod) % mod;}
int mod_add(int a, int b) {a = a % mod; b = b % mod; return (((a + b) % mod) + mod) % mod;}

const int inf = (int)1e18;
const int tp=21;
vii vec;
int mp[tp];
int n, q;
int final;
int dis[tp][tp];
const int N=(1ll<<tp);

   int Dis(int i, int j)
    {
        int ans = 1;
        int r = 0;
        int l = 1;

        for (auto it : vec)
        {
            if (it.first == i)
            {
                if (it.second == 1)
                {
                    if (r + 1 == l)
                    {
                        l++;
                        ans++;
                    }
                    r++;
                }
            }

            if (it.first == j)
            {
                if (it.second == -1)
                    l++;
            }

        }
        return ans;
    }

    void precomputeDistances() {
    lp(i, n) {
        lp(j, n) {
            if (i == j) continue;
            int ans = 1;
            int r = 0, l = 1;
            for (auto it : vec) {
                if (it.first == i && it.second == 1) {
                    if (r + 1 == l) {
                        l++;
                        ans++;
                    }
                    r++;
                }
                if (it.first == j && it.second == -1) {
                    l++;
                }
            }
            dis[i][j] = ans;
        }
    }
}

int fun(int last,int mask,vvi &dp)
{
    if(mask==final)
    {
        return mp[last];
    }

int ret=dp[last][mask];
if(ret!=-1)
return ret;

   int ans=inf;

    lp(i,n)
    {
        if(i==last)continue;
        if((mask &(1ll<<i))==0)
            ans=min(ans,dis[last][i]+fun(i,mask|(1ll<<i),dp));

    }
    return ret=ans;

}

int solve() {

    cin >> n >> q;

final=(1ll<<n)-1ll;

vvi dp(n+1,vi(final+1,-1ll));
    lp(i, q)
    {
        int x;
        cin >> x;
        x--;
        char ch;
        cin >> ch;
        if (ch == '+')
        {
            mp[x]++;
            vec.pb({x, 1});
        } else {
            vec.pb({x, -1});
        }

    }

precomputeDistances();

    // lp(i, n)
    // {
    //     lp(j, n)
    //     {
    //         if (i == j)continue;
    //         dis[i][j] = Dis(i, j);
    //         //cout<<dis[i][j]<<" ";
    //     }
    //     //cout<<"\n";
    // }

// lp(i,n)
// {
//     lp(j,n)
//     {

//         cout<<dis[i][j]<<" ";
//     }
//     cout<<"\n";
// }

int ans=inf;
lp(i,n)
{
    int mask=(1ll<<i);
    int curr=fun(i,mask,dp)+1;
    ans=min(ans,curr);
}

cout<<ans;

    return 0;
}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);

    int t = 1;
    // cin >> t;
    while (t--)
    {
        solve();
        cout << "\n";
    }

    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can cause the algorithm to fail. The key aspects to consider are:

1. **Small-scale input**: The input should be small enough to be manageable but still challenging.
2. **Edge cases**: Consider cases where snakes are placed in such a way that they can easily collide or cause the algorithm to miscalculate the minimum possible score.
3. **Complex event sequences**: Include a mix of enlargement and shrinkage events that can test the algorithm's ability to handle dynamic changes in snake positions.

Given these considerations, let's create a test input:

### Test Input
```plaintext
3 6
1 +
2 +
3 +
1 -
2 -
3 -
```

### Explanation
- **Number of snakes (n = 3)**: A small number to keep the input manageable.
- **Number of events (q = 6)**: A reasonable number of events to test the algorithm's handling of dynamic changes.
- **Event sequence**:
  - Snake 1 enlarges: Snake 1 moves to the right.
  - Snake 2 enlarges: Snake 2 moves to the right.
  - Snake 3 enlarges: Snake 3 moves to the right.
  - Snake 1 shrinks: Snake 1 moves to the left.
  - Snake 2 shrinks: Snake 2 moves to the left.
  - Snake 3 shrinks: Snake 3 moves to the left.

This sequence ensures that the algorithm needs to handle both enlargement and shrinkage events, and it can test the algorithm's ability to correctly compute the minimum possible score.

### Expected Output
The expected output for this test input is:
```plaintext
4
```

### Reasoning
- Initially, the snakes can be placed at positions 1, 2, and 3.
- After the first three events, the snakes will occupy positions 2, 3, and 4.
- After the next three events, the snakes will move back to positions 1, 2, and 3.
- The maximum cell occupied by any snake during the entire sequence is 4, which is the minimum possible score.

This test input should expose any issues in the algorithm's handling of snake movements and collision detection.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Snakes
time_limit_ms: 3000
memory_limit_mb: 512
Description: Suppose you play a game where the game field looks like a strip of $$$1 \times 10^9$$$ square cells, numbered from $$$1$$$ to $$$10^9$$$.

You have $$$n$$$ snakes (numbered from $$$1$$$ to $$$n$$$) you need to place into some cells. Initially, each snake occupies exactly one cell, and you can't place more than one snake into one cell. After that, the game starts.

The game lasts for $$$q$$$ seconds. There are two types of events that may happen each second:

- snake $$$s_i$$$ enlarges: if snake $$$s_i$$$ occupied cells $$$[l, r]$$$, it enlarges to a segment $$$[l, r + 1]$$$;
- snake $$$s_i$$$ shrinks: if snake $$$s_i$$$ occupied cells $$$[l, r]$$$, it shrinks to a segment $$$[l + 1, r]$$$.

Each second, exactly one of the events happens.

If at any moment of time, any snake runs into some obstacle (either another snake or the end of the strip), you lose. Otherwise, you win with the score equal to the maximum cell occupied by any snake so far.

What is the minimum possible score you can achieve?

Input: The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n \le 20$$$; $$$1 \le q \le 2 \cdot 10^5$$$) — the number of snakes and the number of events. Next $$$q$$$ lines contain the description of events — one per line.

The $$$i$$$-th line contains

- either "$$$s_i$$$ +" ($$$1 \le s_i \le n$$$) meaning that the $$$s_i$$$-th snake enlarges
- or "$$$s_i$$$ -" ($$$1 \le s_i \le n$$$) meaning that the $$$s_i$$$-th snake shrinks.

Additional constraint on the input: the given sequence of events is valid, i. e. a snake of length $$$1$$$ never shrinks.

Output: Print one integer — the minimum possible score.

Examples:
input:
3 6
1 +
1 -
3 +
3 -
2 +
2 -
output:
4

input:
5 13
5 +
3 +
5 -
2 +
4 +
3 +
5 +
5 -
2 +
3 -
3 +
3 -
2 +
output:
11

Note: In the first test, the optimal strategy is to place the second snake at cell $$$1$$$, the third snake — at $$$2$$$, and the first one — at $$$3$$$. The maximum occupied cell is cell $$$4$$$, and it's the minimum possible score.

In the second test, one of the optimal strategies is to place:

- snake $$$2$$$ at position $$$1$$$;
- snake $$$3$$$ at position $$$4$$$;
- snake $$$5$$$ at position $$$6$$$;
- snake $$$1$$$ at position $$$9$$$;
- snake $$$4$$$ at position $$$10$$$.

Buggy Code:
// जय श्री राम

// #define case3_local

#include<bits/stdc++.h>
#ifndef ONLINE_JUDGE
#include<algo/debug.cpp>
#else
#define debug(...)
#define debugArr(...)
#endif
using namespace std;

#define int long long
void solve() {
    int n, q;
    cin >> n >> q;
    n++;
    vector<pair<int,char>> moves(q);
    for(auto& val : moves)
        cin >> val.first >> val.second;
    vector<vector<int>> cost(n+1, vector<int>(n+1, 0));
    auto getCost = [&](int s1, int s2) -> int {
        if(s1 == s2)
            return 0;
        if(s1 == n)
            return 0;
        int dist = 1;
        int l1 = 0, l2 = 1, r1 = 0, r2 = 1;
        for(auto& val : moves) {
            if(val.first == s1) {
                if(val.second == '+')
                    r1++;
                else
                    l1++;
            }
            if(val.first == s2) {
                if(val.second == '+')
                    r2++;
                else
                    l2++;
            }
            if(r1 >= l2) {
                l2++;
                r2++;
                ++dist;
            }
        }
        return dist;
    };
    for(int i=1;i<=n;i++) {
        for(int j=1;j<=n;j++) {
            cost[i][j] = getCost(i,j);
        }
    }
    debug(cost);

    map<pair<int,int>, int> dp;
    int endState = ((1<<(n+1)) -1);

    auto tsp = [&](auto self, int cnode, int state) -> int {
        if(state == endState)
            return cost[cnode][1];

        if(dp.find({cnode,state}) != dp.end())
            return dp[{cnode,state}];

        int ans = -1;
        for(int i=1;i<=n;i++) {
            int curr = (1<<i);
            if(!(state&curr)) {
                int cans = cost[cnode][i] + self(self, i, (state|curr));
                debug(i, cnode, cans);
                if(ans == -1)
                    ans = cans;
                ans = min(ans, cans);
            }
        }

        return dp[{cnode,state}] = ans;

    };

    cout << tsp(tsp, 1, 3) << "\n";

}
#undef int

int main() {
    #ifdef case3_local
        auto begin = std::chrono::high_resolution_clock::now();
    #endif
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t=1;
    // cin >> t;
    while(t--){
        solve();
    }
    #ifdef case3_local
        auto end = std::chrono::high_resolution_clock::now();
        cerr << setprecision(4) << fixed;
        cerr << "Execution time: " << std::chrono::duration_cast<std::chrono::duration<double>>(end - begin).count() << " seconds" << endl;
    #endif
    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that can maximize the failure probability. Here are the steps to think through this:

1. **Initial Placement of Snakes**: We need to place snakes in a way that their movements can lead to a conflict or an optimal score that is hard to achieve.
2. **Sequence of Events**: The sequence of events should be such that it forces the snakes to move in a way that can expose the bug in the code.
3. **Edge Cases**: Consider edge cases like snakes at the boundaries of the strip, snakes that grow and shrink in a specific pattern, and snakes that interact with each other.

Given the constraints and the nature of the problem, let's create a small-scale test input that is complete and valid:

### Test Input
```plaintext
2 5
1 +
2 +
1 +
2 +
1 +
```

### Explanation
- **Initial Placement**: We have 2 snakes and 5 events.
- **Sequence of Events**:
  - Snake 1 enlarges: Moves from its initial cell to the next cell.
  - Snake 2 enlarges: Moves from its initial cell to the next cell.
  - Snake 1 enlarges again: Moves further.
  - Snake 2 enlarges again: Moves further.
  - Snake 1 enlarges again: Moves even further.

### Expected Output
The expected output for this test input is:
```plaintext
4
```

### Reasoning
- If we place Snake 1 at cell 1 and Snake 2 at cell 2, the sequence of events will result in:
  - After the first event, Snake 1 occupies cells [1, 2].
  - After the second event, Snake 2 occupies cells [2, 3].
  - After the third event, Snake 1 occupies cells [1, 3].
  - After the fourth event, Snake 2 occupies cells [2, 4].
  - After the fifth event, Snake 1 occupies cells [1, 4].

- The maximum cell occupied by any snake is cell 4, which is the minimum possible score.

This test input is designed to expose any issues in the code related to handling the movements of snakes and ensuring that the score is calculated correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Snakes
time_limit_ms: 3000
memory_limit_mb: 512
Description: Suppose you play a game where the game field looks like a strip of $$$1 \times 10^9$$$ square cells, numbered from $$$1$$$ to $$$10^9$$$.

You have $$$n$$$ snakes (numbered from $$$1$$$ to $$$n$$$) you need to place into some cells. Initially, each snake occupies exactly one cell, and you can't place more than one snake into one cell. After that, the game starts.

The game lasts for $$$q$$$ seconds. There are two types of events that may happen each second:

- snake $$$s_i$$$ enlarges: if snake $$$s_i$$$ occupied cells $$$[l, r]$$$, it enlarges to a segment $$$[l, r + 1]$$$;
- snake $$$s_i$$$ shrinks: if snake $$$s_i$$$ occupied cells $$$[l, r]$$$, it shrinks to a segment $$$[l + 1, r]$$$.

Each second, exactly one of the events happens.

If at any moment of time, any snake runs into some obstacle (either another snake or the end of the strip), you lose. Otherwise, you win with the score equal to the maximum cell occupied by any snake so far.

What is the minimum possible score you can achieve?

Input: The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n \le 20$$$; $$$1 \le q \le 2 \cdot 10^5$$$) — the number of snakes and the number of events. Next $$$q$$$ lines contain the description of events — one per line.

The $$$i$$$-th line contains

- either "$$$s_i$$$ +" ($$$1 \le s_i \le n$$$) meaning that the $$$s_i$$$-th snake enlarges
- or "$$$s_i$$$ -" ($$$1 \le s_i \le n$$$) meaning that the $$$s_i$$$-th snake shrinks.

Additional constraint on the input: the given sequence of events is valid, i. e. a snake of length $$$1$$$ never shrinks.

Output: Print one integer — the minimum possible score.

Examples:
input:
3 6
1 +
1 -
3 +
3 -
2 +
2 -
output:
4

input:
5 13
5 +
3 +
5 -
2 +
4 +
3 +
5 +
5 -
2 +
3 -
3 +
3 -
2 +
output:
11

Note: In the first test, the optimal strategy is to place the second snake at cell $$$1$$$, the third snake — at $$$2$$$, and the first one — at $$$3$$$. The maximum occupied cell is cell $$$4$$$, and it's the minimum possible score.

In the second test, one of the optimal strategies is to place:

- snake $$$2$$$ at position $$$1$$$;
- snake $$$3$$$ at position $$$4$$$;
- snake $$$5$$$ at position $$$6$$$;
- snake $$$1$$$ at position $$$9$$$;
- snake $$$4$$$ at position $$$10$$$.

Buggy Code:
import sys
z=sys.stdin.read().split();n,q=map(int,z[:2]);p=2
E=[0]*n;S=[0]*n
ex=[[0]*(q+1)for _ in range(n)]
sh=[[0]*(q+1)for _ in range(n)]
for t_ in range(1,q+1):
 i_=int(z[p])-1;op=z[p+1];p+=2
 if op=='+':E[i_]+=1
 else:S[i_]+=1
 for r_ in range(n):
  ex[r_][t_]=ex[r_][t_-1]
  sh[r_][t_]=sh[r_][t_-1]
 if op=='+':ex[i_][t_]+=1
 else:sh[i_][t_]+=1
M=[max(ex[i_][t]for t in range(q+1))for i_ in range(n)]
G=[[0]*n for _ in range(n)]
for i_ in range(n):
 for j_ in range(n):
  if i_==j_:G[i_][j_]=0
  else:
   mx=-10**15
   for t_ in range(q+1):
    tmp=ex[i_][t_]-sh[j_][t_]
    if tmp>mx:mx=tmp
   G[i_][j_]=mx+1
N=1<<n
dp=[[10**15]*n for _ in range(N)]
ct=[[10**15]*n for _ in range(N)]
for i_ in range(n):
 dp[1<<i_][i_]=0
 ct[1<<i_][i_]=1+M[i_]
for mask in range(N):
 for l_ in range(n):
  if mask&(1<<l_):
   sg=dp[mask][l_]
   sc=ct[mask][l_]
   for j_ in range(n):
    if not(mask&(1<<j_)):
     nm=mask|(1<<j_)
     ng=sg+G[l_][j_]
     nc=max(sc,(1+ng)+M[j_])
     if ng<dp[nm][j_]or nc<ct[nm][j_]:
      if ng<dp[nm][j_]:
       dp[nm][j_]=ng;ct[nm][j_]=nc
      elif ng==dp[nm][j_]and nc<ct[nm][j_]:
       ct[nm][j_]=nc
fm=(1<<n)-1;ans=10**15
for i_ in range(n):
 ans=min(ans,ct[fm][i_])
print(ans)

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
2 4
1 +
1 -
2 +
2 -
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Snakes
time_limit_ms: 3000
memory_limit_mb: 512
Description: Suppose you play a game where the game field looks like a strip of $$$1 \times 10^9$$$ square cells, numbered from $$$1$$$ to $$$10^9$$$.

You have $$$n$$$ snakes (numbered from $$$1$$$ to $$$n$$$) you need to place into some cells. Initially, each snake occupies exactly one cell, and you can't place more than one snake into one cell. After that, the game starts.

The game lasts for $$$q$$$ seconds. There are two types of events that may happen each second:

- snake $$$s_i$$$ enlarges: if snake $$$s_i$$$ occupied cells $$$[l, r]$$$, it enlarges to a segment $$$[l, r + 1]$$$;
- snake $$$s_i$$$ shrinks: if snake $$$s_i$$$ occupied cells $$$[l, r]$$$, it shrinks to a segment $$$[l + 1, r]$$$.

Each second, exactly one of the events happens.

If at any moment of time, any snake runs into some obstacle (either another snake or the end of the strip), you lose. Otherwise, you win with the score equal to the maximum cell occupied by any snake so far.

What is the minimum possible score you can achieve?

Input: The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n \le 20$$$; $$$1 \le q \le 2 \cdot 10^5$$$) — the number of snakes and the number of events. Next $$$q$$$ lines contain the description of events — one per line.

The $$$i$$$-th line contains

- either "$$$s_i$$$ +" ($$$1 \le s_i \le n$$$) meaning that the $$$s_i$$$-th snake enlarges
- or "$$$s_i$$$ -" ($$$1 \le s_i \le n$$$) meaning that the $$$s_i$$$-th snake shrinks.

Additional constraint on the input: the given sequence of events is valid, i. e. a snake of length $$$1$$$ never shrinks.

Output: Print one integer — the minimum possible score.

Examples:
input:
3 6
1 +
1 -
3 +
3 -
2 +
2 -
output:
4

input:
5 13
5 +
3 +
5 -
2 +
4 +
3 +
5 +
5 -
2 +
3 -
3 +
3 -
2 +
output:
11

Note: In the first test, the optimal strategy is to place the second snake at cell $$$1$$$, the third snake — at $$$2$$$, and the first one — at $$$3$$$. The maximum occupied cell is cell $$$4$$$, and it's the minimum possible score.

In the second test, one of the optimal strategies is to place:

- snake $$$2$$$ at position $$$1$$$;
- snake $$$3$$$ at position $$$4$$$;
- snake $$$5$$$ at position $$$6$$$;
- snake $$$1$$$ at position $$$9$$$;
- snake $$$4$$$ at position $$$10$$$.

Buggy Code:
import sys
input = sys.stdin.readline

n,q=map(int,input().split())

T=[input().split() for i in range(q)]

ANS=n
for x,com in T:
    if com=="+":
        ANS+=1

LIST=[[] for i in range(n)]

for i in range(q):
    x,com=T[i]
    LIST[int(x)-1].append((i,com))

DIS=[[-1]*n for i in range(n)]

for i in range(n):
    for j in range(n):
        if i==j:
            continue

        X=LIST[i]
        Y=LIST[j]

        score=0
        xminus=0

        indx=0
        indy=0

        while indx<len(X) or indy<len(Y):
            if indx==len(X):
                if Y[indy][1]=="+":
                    if xminus>0:
                        xminus-=1
                        score+=1
                indy+=1
                continue

            if indy==len(Y):
                if X[indx][1]=="-":
                        xminus+=1

                indx+=1
                continue

            if X[indx][0]<Y[indy][0]:
                if X[indx][1]=="-":
                        xminus+=1

                indx+=1
                continue
            else:
                if Y[indy][1]=="+":
                    if xminus>0:
                        xminus-=1
                        score+=1
                indy+=1
                continue
        DIS[i][j]=score

DP=[[-1000000000]*n for i in range(1<<n)]

for i in range(n):
    DP[1<<i][i]=0

for i in range(1<<n):
    for j in range(n):
        if DP[i][j]==-1000000000:
            continue

        for to in range(n):
            if ((1<<to) & i)!=0:
                continue
            DP[i|(1<<to)][to]=max(DP[i|(1<<to)][to],DIS[j][to]+DP[i][j])

ANS2=0

for i in range(n):
    ANS2=max(ANS2,DP[-1][i])

print(ANS-ANS2)

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases that can cause the algorithm to fail. The key issues to consider are:

1. **Overlap and Collision**: Ensure that the snakes can potentially collide with each other or with the end of the strip.
2. **Sequence of Events**: Create a sequence of events that can lead to complex interactions between snakes.
3. **Initial Placement**: Place the snakes in such a way that the sequence of events can lead to a high score if not handled correctly.

Here's a small-scale test input that should expose the bug:

```plaintext
3 6
1 +
2 +
3 +
1 -
2 -
3 -
```

### Explanation:

1. **Initial Placement**:
   - Place snake 1 at cell 1.
   - Place snake 2 at cell 2.
   - Place snake 3 at cell 3.

2. **Sequence of Events**:
   - In the first second, snake 1 enlarges, moving from cell 1 to cell 2.
   - In the second second, snake 2 enlarges, moving from cell 2 to cell 3.
   - In the third second, snake 3 enlarges, moving from cell 3 to cell 4.
   - In the fourth second, snake 1 shrinks, moving back from cell 2 to cell 1.
   - In the fifth second, snake 2 shrinks, moving back from cell 3 to cell 2.
   - In the sixth second, snake 3 shrinks, moving back from cell 4 to cell 3.

3. **Expected Output**:
   - The maximum cell occupied by any snake is cell 4.

This test case is designed to check if the algorithm correctly handles the sequence of enlarging and shrinking events, ensuring that the maximum cell occupied by any snake is correctly calculated. If the buggy code fails to handle these events correctly, it will produce an incorrect score.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Snakes
time_limit_ms: 3000
memory_limit_mb: 512
Description: Suppose you play a game where the game field looks like a strip of $$$1 \times 10^9$$$ square cells, numbered from $$$1$$$ to $$$10^9$$$.

You have $$$n$$$ snakes (numbered from $$$1$$$ to $$$n$$$) you need to place into some cells. Initially, each snake occupies exactly one cell, and you can't place more than one snake into one cell. After that, the game starts.

The game lasts for $$$q$$$ seconds. There are two types of events that may happen each second:

- snake $$$s_i$$$ enlarges: if snake $$$s_i$$$ occupied cells $$$[l, r]$$$, it enlarges to a segment $$$[l, r + 1]$$$;
- snake $$$s_i$$$ shrinks: if snake $$$s_i$$$ occupied cells $$$[l, r]$$$, it shrinks to a segment $$$[l + 1, r]$$$.

Each second, exactly one of the events happens.

If at any moment of time, any snake runs into some obstacle (either another snake or the end of the strip), you lose. Otherwise, you win with the score equal to the maximum cell occupied by any snake so far.

What is the minimum possible score you can achieve?

Input: The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n \le 20$$$; $$$1 \le q \le 2 \cdot 10^5$$$) — the number of snakes and the number of events. Next $$$q$$$ lines contain the description of events — one per line.

The $$$i$$$-th line contains

- either "$$$s_i$$$ +" ($$$1 \le s_i \le n$$$) meaning that the $$$s_i$$$-th snake enlarges
- or "$$$s_i$$$ -" ($$$1 \le s_i \le n$$$) meaning that the $$$s_i$$$-th snake shrinks.

Additional constraint on the input: the given sequence of events is valid, i. e. a snake of length $$$1$$$ never shrinks.

Output: Print one integer — the minimum possible score.

Examples:
input:
3 6
1 +
1 -
3 +
3 -
2 +
2 -
output:
4

input:
5 13
5 +
3 +
5 -
2 +
4 +
3 +
5 +
5 -
2 +
3 -
3 +
3 -
2 +
output:
11

Note: In the first test, the optimal strategy is to place the second snake at cell $$$1$$$, the third snake — at $$$2$$$, and the first one — at $$$3$$$. The maximum occupied cell is cell $$$4$$$, and it's the minimum possible score.

In the second test, one of the optimal strategies is to place:

- snake $$$2$$$ at position $$$1$$$;
- snake $$$3$$$ at position $$$4$$$;
- snake $$$5$$$ at position $$$6$$$;
- snake $$$1$$$ at position $$$9$$$;
- snake $$$4$$$ at position $$$10$$$.

Buggy Code:
// package q7;

// import java.io.*;
// import java.util.*;

// public class Main {
//           static int MAX_MASK_VAL;
//           public static long fun1(int idx, int n, long[][] dp, int curr_mask, int[] arr, long[][] minDist) {
//                     curr_mask = (curr_mask | (1 << idx));
//                     if(dp[curr_mask][idx] != -1) return dp[curr_mask][idx];
//                     int flag = 0;
//                     long ans = Long.MAX_VALUE;
//                     for(int i = 0; i < n; i++){
//                               int myMask = (1 << i);
//                               if((curr_mask & myMask) == 0){
//                                         long x = fun1(i, n, dp, curr_mask, arr, minDist);
//                                         ans = Math.min(ans, x + minDist[idx][i]);
//                                         flag = 1;
//                               }
//                     }
//                     curr_mask = (curr_mask ^ (1 << idx));
//                     if(flag == 0) {
//                               return dp[curr_mask][idx] = arr[idx];
//                     }
//                     return dp[curr_mask][idx] = ans;
//           }

//           public static long fun(int n, int q, int[][] queries) {
//                     long[][] minDist = new long[n][n]; // minDist[i][j] means minDist between snake i -> j
//                     int[] arr = new int[n];
//                     for(int i = 0; i < q; i++){
//                               if(queries[i][1] == 1) arr[queries[i][0]]++;
//                     }
//                     for (int i = 0; i < n; i++) { // my snake
//                               for (int j = 0; j < n; j++) { // jiske aage mai rehna chata hu
//                                         int myMinus = 0, yourAdd = 0, gap = 0;
//                                         for (int[] query : queries) {
//                                                   if (query[0] == i && query[1] == -1)
//                                                             myMinus++;
//                                                   else if (query[0] == j && query[1] == 1) {
//                                                             yourAdd++;
//                                                             gap = Math.max(yourAdd - myMinus, gap);
//                                                   }
//                                         }

//                                         minDist[j][i] = gap + 1;
//                               }
//                     }
//                     MAX_MASK_VAL = (1 << n) - 1;
//                     long[][] dp = new long[MAX_MASK_VAL + 1][n];          // [set_of_indices_selected_so_far][curr_idx_to_include]
//                     for(int i = 0; i <= MAX_MASK_VAL; i++) Arrays.fill(dp[i], -1);
//                     for(int i = 0; i < n; i++){
//                               fun1(i, n, dp, 0, arr, minDist);
//                     }
//                     long ans = Long.MAX_VALUE;
//                     for(int i = 0; i < n; i++) ans = Math.min(ans, 1 + dp[0][i]);

//                     return ans;
//           }

//           public static void main(String[] args) throws IOException {
//                     BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
//                     BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
//                     StringTokenizer st = new StringTokenizer(br.readLine());
//                     // int t = Integer.parseInt(st.nextToken());
//                     // while(t-- > 0){
//                     // st = new StringTokenizer(br.readLine());
//                     // int n = Integer.parseInt(st.nextToken());
//                     // int q = Integer.parseInt(st.nextToken());
//                     // int[][] queries = new int[q][2];
//                     // for(int i = 0; i < q; i++){
//                     // st = new StringTokenizer(br.readLine());
//                     // queries[i][0] = Integer.parseInt(st.nextToken()) - 1;
//                     // char ch = st.nextToken().charAt(0);
//                     // if(ch == '+') queries[i][1] = 1;
//                     // else queries[i][1] = -1;
//                     // }

//                     // bw.write(fun(n, q, queries) + "\n");
//                     // }

//                     int n = Integer.parseInt(st.nextToken());
//                     int q = Integer.parseInt(st.nextToken());
//                     int[][] queries = new int[q][2];
//                     for (int i = 0; i < q; i++) {
//                               st = new StringTokenizer(br.readLine());
//                               queries[i][0] = Integer.parseInt(st.nextToken()) - 1;
//                               char ch = st.nextToken().charAt(0);
//                               if (ch == '+')
//                                         queries[i][1] = 1;
//                               else
//                                         queries[i][1] = -1;
//                     }

//                     bw.write(fun(n, q, queries) + "\n");

//                     br.close();
//                     bw.flush();
//                     bw.close();
//           }
// }

// class Pair {
//           int node;
//           long dist;

//           Pair(int n, long d) {
//                     node = n;
//                     dist = d;
//           }
// }

import java.io.*;
import java.util.*;

public class Main {
          static int MAX_MASK_VAL;

          public static long fun1(int idx, int n, long[][] dp, int curr_mask, int[] arr, long[][] cost) {
                    curr_mask = (curr_mask | (1 << idx));
                    if (dp[curr_mask][idx] != -1)
                              return dp[curr_mask][idx];

                    long ans = Long.MAX_VALUE;
                    boolean hasNext = false;

                    // Directly access precomputed costs for valid transitions
                    for (int next = 0; next < n; next++) {
                              if ((curr_mask & (1 << next)) == 0) { // Valid transition
                                        hasNext = true;
                                        ans = Math.min(ans, fun1(next, n, dp, curr_mask, arr, cost) + cost[idx][next]);
                              }
                    }

                    curr_mask = (curr_mask ^ (1 << idx));
                    if (!hasNext) // Base case: if no valid transitions, return the cost of `arr[idx]`
                              return dp[curr_mask][idx] = arr[idx];
                    return dp[curr_mask][idx] = ans;
          }

          public static long fun(int n, int q, int[][] queries) {
                    long[][] minDist = new long[n][n]; // minDist[i][j] means minDist between snake i -> j
                    int[] arr = new int[n];

                    // Step 1: Populate `arr` with the count of '+1' operations for each node
                    for (int i = 0; i < q; i++) {
                              if (queries[i][1] == 1)
                                        arr[queries[i][0]]++;
                    }

                    // Step 2: Compute `minDist[i][j]` for all pairs (i, j)
                    for (int i = 0; i < n; i++) { // My snake
                              for (int j = 0; j < n; j++) { // The snake I want to stay ahead of
                                        int myMinus = 0, yourAdd = 0, gap = 0;
                                        for (int[] query : queries) {
                                                  if (query[0] == i && query[1] == -1)
                                                            myMinus++;
                                                  else if (query[0] == j && query[1] == 1) {
                                                            yourAdd++;
                                                            gap = Math.max(yourAdd - myMinus, gap);
                                                  }
                                        }

                                        minDist[j][i] = gap + 1;
                              }
                    }

                    // Step 3: Precompute the cost array for all transitions
                    long[][] cost = new long[n][n];
                    for (int i = 0; i < n; i++) {
                              for (int j = 0; j < n; j++) {
                                        cost[i][j] = minDist[i][j];
                              }
                    }

                    // Step 4: Initialize DP table
                    MAX_MASK_VAL = (1 << n) - 1;
                    long[][] dp = new long[MAX_MASK_VAL + 1][n]; // [set_of_indices_selected_so_far][current_index]
                    for (int i = 0; i <= MAX_MASK_VAL; i++)
                              Arrays.fill(dp[i], -1);

                    // Step 5: Compute DP for all starting indices
                    for (int i = 0; i < n; i++) {
                              fun1(i, n, dp, 0, arr, cost);
                    }

                    // Step 6: Find the minimum value over all starting indices
                    long ans = Long.MAX_VALUE;
                    for (int i = 0; i < n; i++)
                              ans = Math.min(ans, 1 + dp[0][i]);

                    return ans;
          }

          public static void main(String[] args) throws IOException {
                    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
                    BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
                    StringTokenizer st = new StringTokenizer(br.readLine());

                    int n = Integer.parseInt(st.nextToken()); // Number of nodes
                    int q = Integer.parseInt(st.nextToken()); // Number of queries
                    int[][] queries = new int[q][2];
                    for (int i = 0; i < q; i++) {
                              st = new StringTokenizer(br.readLine());
                              queries[i][0] = Integer.parseInt(st.nextToken()) - 1; // Convert to 0-based index
                              char ch = st.nextToken().charAt(0);
                              if (ch == '+')
                                        queries[i][1] = 1;
                              else
                                        queries[i][1] = -1;
                    }

                    bw.write(fun(n, q, queries) + "\n");

                    br.close();
                    bw.flush();
                    bw.close();
          }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. Here are the steps to think through this:

1. **Understand the Bug**: The buggy code has a potential issue with how it handles the dynamic programming (DP) transitions and the computation of the minimum distance (`minDist`). Specifically, it might not correctly handle the cases where snakes are placed in such a way that they interfere with each other's movements.

2. **Identify Edge Cases**:
   - **Single Snake**: Test with a single snake to ensure the basic logic works.
   - **Multiple Snakes**: Test with multiple snakes to check how the code handles interactions.
   - **Edge of the Strip**: Test scenarios where snakes are placed near the edge of the strip (cell 1 or cell \(10^9\)).
   - **Sequential Enlarging and Shrinking**: Test sequences where snakes enlarge and shrink in a way that could cause interference.

3. **Generate the Test Input**:
   - Start with a small number of snakes and a reasonable number of events to keep the input manageable.
   - Ensure the test input includes a mix of enlarging and shrinking events.
   - Place snakes in positions that are likely to cause interference when they move.

Here is a small-scale test input that should expose potential bugs in the code:

```plaintext
4 8
1 +
2 +
3 +
4 +
1 -
2 -
3 -
4 -
```

### Explanation:
- **4 Snakes**: We use 4 snakes to ensure there are enough interactions.
- **8 Events**: We use 8 events to include a sequence of enlarging and shrinking.
- **Initial Enlarging**: The first four events enlarge each snake, causing them to potentially overlap if not placed correctly.
- **Subsequent Shrinking**: The next four events shrink each snake, which should reverse the enlarging effects.

This test input should help identify issues with how the code handles the dynamic placement and movement of snakes. If the code correctly manages the placement and movement, it should produce the correct minimum score. If not, it will likely fail to handle the interactions properly, leading to an incorrect score.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Snakes
time_limit_ms: 3000
memory_limit_mb: 512
Description: Suppose you play a game where the game field looks like a strip of $$$1 \times 10^9$$$ square cells, numbered from $$$1$$$ to $$$10^9$$$.

You have $$$n$$$ snakes (numbered from $$$1$$$ to $$$n$$$) you need to place into some cells. Initially, each snake occupies exactly one cell, and you can't place more than one snake into one cell. After that, the game starts.

The game lasts for $$$q$$$ seconds. There are two types of events that may happen each second:

- snake $$$s_i$$$ enlarges: if snake $$$s_i$$$ occupied cells $$$[l, r]$$$, it enlarges to a segment $$$[l, r + 1]$$$;
- snake $$$s_i$$$ shrinks: if snake $$$s_i$$$ occupied cells $$$[l, r]$$$, it shrinks to a segment $$$[l + 1, r]$$$.

Each second, exactly one of the events happens.

If at any moment of time, any snake runs into some obstacle (either another snake or the end of the strip), you lose. Otherwise, you win with the score equal to the maximum cell occupied by any snake so far.

What is the minimum possible score you can achieve?

Input: The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n \le 20$$$; $$$1 \le q \le 2 \cdot 10^5$$$) — the number of snakes and the number of events. Next $$$q$$$ lines contain the description of events — one per line.

The $$$i$$$-th line contains

- either "$$$s_i$$$ +" ($$$1 \le s_i \le n$$$) meaning that the $$$s_i$$$-th snake enlarges
- or "$$$s_i$$$ -" ($$$1 \le s_i \le n$$$) meaning that the $$$s_i$$$-th snake shrinks.

Additional constraint on the input: the given sequence of events is valid, i. e. a snake of length $$$1$$$ never shrinks.

Output: Print one integer — the minimum possible score.

Examples:
input:
3 6
1 +
1 -
3 +
3 -
2 +
2 -
output:
4

input:
5 13
5 +
3 +
5 -
2 +
4 +
3 +
5 +
5 -
2 +
3 -
3 +
3 -
2 +
output:
11

Note: In the first test, the optimal strategy is to place the second snake at cell $$$1$$$, the third snake — at $$$2$$$, and the first one — at $$$3$$$. The maximum occupied cell is cell $$$4$$$, and it's the minimum possible score.

In the second test, one of the optimal strategies is to place:

- snake $$$2$$$ at position $$$1$$$;
- snake $$$3$$$ at position $$$4$$$;
- snake $$$5$$$ at position $$$6$$$;
- snake $$$1$$$ at position $$$9$$$;
- snake $$$4$$$ at position $$$10$$$.

Buggy Code:
public class Main {
    private static final int MAX_N = 20;
    private static final int INF = 200000000;

    static class Pair {
        int x;
        char y;

        Pair(int x, char y) {
            this.x = x;
            this.y = y;
        }
    }

    private static int max(int a, int b) {
        return Math.max(a, b);
    }

    private static int min(int a, int b) {
        return Math.min(a, b);
    }

    public static void main(String[] args) {
        java.util.Scanner scanner = new java.util.Scanner(System.in);
        int n = scanner.nextInt();
        int q = scanner.nextInt();

        int[] inc = new int[n + 1];
        Pair[] a = new Pair[q];

        // Read input pairs
        for (int i = 0; i < q; i++) {
            int x = scanner.nextInt();
            char y = scanner.next().charAt(0);
            a[i] = new Pair(x, y);
            if (a[i].y == '+') {
                inc[a[i].x]++;
            }
        }

        // Create 2D cost array
        int[][] cost = new int[n + 1][n + 1];
        for (int i = 0; i <= n; i++) {
            java.util.Arrays.fill(cost[i], 1);
        }

        // Calculate costs
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                int x = 0, y = 0;
                for (int k = 0; k < q; k++) {
                    if (a[k].x == i) {
                        x += (a[k].y == '+' ? 1 : 0);
                    }
                    if (a[k].x == j) {
                        y += (a[k].y == '-' ? 1 : 0);
                    }
                    cost[i-1][j-1] = max(cost[i-1][j-1], x - y + 1);
                }
            }
        }

        // Create dp array
        int size = (1 << n);
        int[][] dp = new int[size][n + 1];
        for (int i = 0; i < size; i++) {
            java.util.Arrays.fill(dp[i], INF);
        }

        // Initialize dp array
        for (int i = 0; i < n; i++) {
            dp[1 << i][i] = 1;
        }

        // Dynamic programming
        for (int i = 1; i < (1 << n); i++) {
            for (int j = 0; j < n; j++) {
                if ((i & (1 << j)) == 0) {
                    continue;
                }
                for (int k = 0; k < n; k++) {
                    if ((i & (1 << k)) == 0) {
                        dp[i | (1 << k)][k] = min(dp[i | (1 << k)][k], dp[i][j] + cost[j][k]);
                    }
                }
            }
        }

        // Find minimum answer
        int ans = INF;
        for (int i = 0; i < n; i++) {
            ans = min(ans, dp[(1 << n) - 1][i] + inc[i + 1]);
        }

        System.out.print(ans);
        scanner.close();
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The key points to consider are:

1. **Initial Placement of Snakes**: The initial placement of snakes can significantly affect the outcome. We need to place snakes in positions that can lead to conflicts or maximize the score if not handled correctly.
2. **Sequence of Events**: The sequence of events (enlarging or shrinking) should be designed to test the logic of the code, especially the dynamic programming part and the cost calculation.

Given these considerations, let's create a test input that:

- Places snakes in positions that can cause conflicts if not managed properly.
- Includes a sequence of events that can lead to the snakes overlapping or reaching the end of the strip.

Here is a small-scale test input:

```plaintext
4 8
1 +
2 +
3 +
4 +
1 -
2 -
3 -
4 -
```

### Explanation:

- **Initial Placement**:
  - Snake 1 starts at position 1.
  - Snake 2 starts at position 2.
  - Snake 3 starts at position 3.
  - Snake 4 starts at position 4.

- **Sequence of Events**:
  - Snakes 1, 2, 3, and 4 each enlarge by 1 cell.
  - Snakes 1, 2, 3, and 4 each shrink by 1 cell.

### Expected Behavior:
- The snakes should not overlap or run into the end of the strip.
- The minimum possible score should be the maximum cell occupied by any snake after all the events.

### Why This Test Input is Challenging:
- **Initial Placement**: The snakes are placed consecutively, which can cause conflicts if the code does not handle the cost calculation correctly.
- **Sequence of Events**: The sequence of enlarging and shrinking events should test the dynamic programming logic and ensure that the cost array is updated correctly.

This test input should help expose any issues in the buggy code, particularly in the cost calculation and dynamic programming parts.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Beautiful Triple Pairs
time_limit_ms: 4000
memory_limit_mb: 256
Description: Polycarp was given an array $a$ of $n$ integers. He really likes triples of numbers, so for each $j$ ($1 \le j \le n - 2$) he wrote down a triple of elements $[a_j, a_{j + 1}, a_{j + 2}]$.

Polycarp considers a pair of triples $b$ and $c$ beautiful if they differ in exactly one position, that is, one of the following conditions is satisfied:

  * $b_1 \ne c_1$ and $b_2 = c_2$ and $b_3 = c_3$;    * $b_1 = c_1$ and $b_2 \ne c_2$ and $b_3 = c_3$;    * $b_1 = c_1$ and $b_2 = c_2$ and $b_3 \ne c_3$.

Find the number of beautiful pairs of triples among the written triples $[a_j, a_{j + 1}, a_{j + 2}]$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($3 \le n \le 2 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$) — the elements of the array.

It is guaranteed that the sum of the values of $n$ for all test cases in the test does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer — the number of beautiful pairs of triples among the pairs of the form $[a_j, a_{j + 1}, a_{j + 2}]$.

Note that the answer may not fit into 32-bit data types.

Examples:
input:
8
5
3 2 2 2 3
5
1 2 1 2 1
8
1 2 3 2 2 3 4 2
4
2 1 1 1
8
2 1 1 2 1 1 1 1
7
2 1 1 1 1 1 1
6
2 1 1 1 1 1
5
2 1 1 1 1
output:
2
0
3
1
8
4
3
2

Note: In the first example, $a = [3, 2, 2, 2, 3]$, Polycarp will write the following triples:

  1. $[3, 2, 2]$;    2. $[2, 2, 2]$;    3. $[2, 2, 3]$.

The beautiful pairs are triple $1$ with triple $2$ and triple $2$ with triple $3$.

In the third example, $a = [1, 2, 3, 2, 2, 3, 4, 2]$, Polycarp will write the following triples:

  1. $[1, 2, 3]$;    2. $[2, 3, 2]$;    3. $[3, 2, 2]$;    4. $[2, 2, 3]$;    5. $[2, 3, 4]$;    6. $[3, 4, 2]$;

The beautiful pairs are triple $1$ with triple $4$, triple $2$ with triple $5$, and triple $3$ with triple $6$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define Buildup() ios::sync_with_stdio(false),cin.tie(nullptr)

#define make_unique(x) (x).resize(unique(all((x)))-(x).begin())
#define all(x) (x).begin(),(x).end()
#define allr(x) (x).rbegin(),(x).rend()
#define ceil(x,y) (x+y-1)/(y)
#define sz(x) (int)(x).size()
#define PI 3.141592653589793238
#define ld long double
#define ull unsigned long long
#define ll long long
#define nl '\n'
#define ff first
#define ss second
#define IM INT_MAX
#define IN INT_MIN
#define LN LLONG_MIN
#define LM LLONG_MAX
#define pb push_back
#define pp pop_back
#define ins insert
#define lb lower_bound
#define ub upper_bound

#define Nl '\n'
#define itn int

// const int M = 998244353;
// const int M = 1e9+7;
// const int N = 2e5+5;
// const int N = 1e6+5;

#ifndef ONLINE_JUDGE
#define dbg(x...) cerr << #x << " : ["; _print(x)
#else
#define dbg(x...)
#endif

void __print(int x) {cerr << x;}
void __print(ll x) {cerr << x;}
void __print(ull x) {cerr << x;}
void __print(ld x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}

template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.ff); cerr << ','; __print(x.ss); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}

/* .....!! Start by doing what's necessary; then do what's possible;
                            and suddenly you are doing impossible !!..... */

int main(){
    #ifndef ONLINE_JUDGE
        freopen("errorf.in", "w", stderr);
    #endif

    Buildup();

        int __ = 1;
        cin >> __;

        while(__--){

            int n;
            cin>>n;

            vector<itn> v(n);
            for(auto &it : v) cin>>it;

            map<pair<int,int>,vector<int>> m1, m2, m3;
            ll ans = 0;
            for(int i=0; i<n-2; i++){
                m1[{v[i],v[i+1]}].pb(v[i+2]);
            }
            for(int i=0; i<n-2; i++){
                m2[{v[i+1],v[i+2]}].pb(v[i]);
            }
            for(int i=0; i<n-2; i++){
                m3[{v[i],v[i+2]}].pb(v[i+1]);
            }

            auto get = [&](map<pair<int,int>,vector<int>> m){
                ll ans = 0;
                for(auto &[x,y] : m){
                    ll t = 1, res = 1;
                    sort(all(y));
                    if(y[0]==y[sz(y)-1]){
                        continue;
                    }
                    for(int i=0; i<sz(y)-1; i++){
                        if(y[i]==y[i+1]) t++;
                        else{
                            res *= t;
                            t = 1;
                        }
                    }
                    res *= t;
                    ans += res;
                }
                return ans;
            };

            cout<<get(m1)+get(m2)+get(m3)<<nl;

        }

}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
6
1 2 1 2 1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Beautiful Triple Pairs
time_limit_ms: 4000
memory_limit_mb: 256
Description: Polycarp was given an array $a$ of $n$ integers. He really likes triples of numbers, so for each $j$ ($1 \le j \le n - 2$) he wrote down a triple of elements $[a_j, a_{j + 1}, a_{j + 2}]$.

Polycarp considers a pair of triples $b$ and $c$ beautiful if they differ in exactly one position, that is, one of the following conditions is satisfied:

  * $b_1 \ne c_1$ and $b_2 = c_2$ and $b_3 = c_3$;    * $b_1 = c_1$ and $b_2 \ne c_2$ and $b_3 = c_3$;    * $b_1 = c_1$ and $b_2 = c_2$ and $b_3 \ne c_3$.

Find the number of beautiful pairs of triples among the written triples $[a_j, a_{j + 1}, a_{j + 2}]$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($3 \le n \le 2 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$) — the elements of the array.

It is guaranteed that the sum of the values of $n$ for all test cases in the test does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer — the number of beautiful pairs of triples among the pairs of the form $[a_j, a_{j + 1}, a_{j + 2}]$.

Note that the answer may not fit into 32-bit data types.

Examples:
input:
8
5
3 2 2 2 3
5
1 2 1 2 1
8
1 2 3 2 2 3 4 2
4
2 1 1 1
8
2 1 1 2 1 1 1 1
7
2 1 1 1 1 1 1
6
2 1 1 1 1 1
5
2 1 1 1 1
output:
2
0
3
1
8
4
3
2

Note: In the first example, $a = [3, 2, 2, 2, 3]$, Polycarp will write the following triples:

  1. $[3, 2, 2]$;    2. $[2, 2, 2]$;    3. $[2, 2, 3]$.

The beautiful pairs are triple $1$ with triple $2$ and triple $2$ with triple $3$.

In the third example, $a = [1, 2, 3, 2, 2, 3, 4, 2]$, Polycarp will write the following triples:

  1. $[1, 2, 3]$;    2. $[2, 3, 2]$;    3. $[3, 2, 2]$;    4. $[2, 2, 3]$;    5. $[2, 3, 4]$;    6. $[3, 4, 2]$;

The beautiful pairs are triple $1$ with triple $4$, triple $2$ with triple $5$, and triple $3$ with triple $6$.

Buggy Code:
#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <cstring>
#include <set>
#include <queue>
#include <stack>
#include <map>
#include <cmath>
#define endl '\n'
using namespace std;
const int N = 2e5 + 10;
typedef long long ll;

void solve()
{
	int n ;
	ll ans = 0 ;
	cin >> n ;
	vector <int> v(n) ;
	map< pair<int,int> , ll > a , b , c , w ;
	map< tuple<int,int,int> , ll > t ;
	for(int i = 1 ; i <= n ; i ++){
		cin >> v[i] ;
	}

	for(int i = 1 ; i <= n-2 ; i ++){
		ans += a[make_pair(v[i] , v[i+1])] ;
		ans += b[make_pair(v[i] , v[i+2])] ;
		ans += c[make_pair(v[i+1] , v[i+2])] ;
		ans -= 3LL * t[make_tuple(v[i] , v[i+1] , v[i+2])] ;
		a[make_pair(v[i] , v[i+1])] ++ ;
		b[make_pair(v[i] , v[i+2])] ++ ;
		c[make_pair(v[i+1] , v[i+2])] ++ ;
		t[make_tuple(v[i] , v[i+1] , v[i+2])] ++ ;
	}
	cout << ans << '\n' ;
}

int main()
{
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);

	int t;
	cin >> t;
	while (t--)
		solve();

	return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
6
1 2 2 1 1 2
7
1 1 2 2 1 1 2
8
1 2 2 1 1 2 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Beautiful Triple Pairs
time_limit_ms: 4000
memory_limit_mb: 256
Description: Polycarp was given an array $a$ of $n$ integers. He really likes triples of numbers, so for each $j$ ($1 \le j \le n - 2$) he wrote down a triple of elements $[a_j, a_{j + 1}, a_{j + 2}]$.

Polycarp considers a pair of triples $b$ and $c$ beautiful if they differ in exactly one position, that is, one of the following conditions is satisfied:

  * $b_1 \ne c_1$ and $b_2 = c_2$ and $b_3 = c_3$;    * $b_1 = c_1$ and $b_2 \ne c_2$ and $b_3 = c_3$;    * $b_1 = c_1$ and $b_2 = c_2$ and $b_3 \ne c_3$.

Find the number of beautiful pairs of triples among the written triples $[a_j, a_{j + 1}, a_{j + 2}]$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($3 \le n \le 2 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$) — the elements of the array.

It is guaranteed that the sum of the values of $n$ for all test cases in the test does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer — the number of beautiful pairs of triples among the pairs of the form $[a_j, a_{j + 1}, a_{j + 2}]$.

Note that the answer may not fit into 32-bit data types.

Examples:
input:
8
5
3 2 2 2 3
5
1 2 1 2 1
8
1 2 3 2 2 3 4 2
4
2 1 1 1
8
2 1 1 2 1 1 1 1
7
2 1 1 1 1 1 1
6
2 1 1 1 1 1
5
2 1 1 1 1
output:
2
0
3
1
8
4
3
2

Note: In the first example, $a = [3, 2, 2, 2, 3]$, Polycarp will write the following triples:

  1. $[3, 2, 2]$;    2. $[2, 2, 2]$;    3. $[2, 2, 3]$.

The beautiful pairs are triple $1$ with triple $2$ and triple $2$ with triple $3$.

In the third example, $a = [1, 2, 3, 2, 2, 3, 4, 2]$, Polycarp will write the following triples:

  1. $[1, 2, 3]$;    2. $[2, 3, 2]$;    3. $[3, 2, 2]$;    4. $[2, 2, 3]$;    5. $[2, 3, 4]$;    6. $[3, 4, 2]$;

The beautiful pairs are triple $1$ with triple $4$, triple $2$ with triple $5$, and triple $3$ with triple $6$.

Buggy Code:
//******************************************************//
                        //*                AUTHOR: NAMAN OSTWAL                *//
                        //******************************************************//
//______________________________________________________________________________________________________//
#include <bits/stdc++.h>
using namespace std;
#define ll long long
long long int mod = 1e9+7;
const int N = 1e6+7;
vector<bool> isprime(N, true);

inline void fast_io() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
}

void print(vector<ll>&v) { for(auto it : v) cout<<it<<' '; }
void sort(vector < ll > &v) { sort(v.begin() , v.end()); }
void rsort(vector < ll > &v) { sort(v.begin() , v.end(), greater<int>()); }
ll mul(ll a, ll b) { return ((a % mod) * (b % mod)) % mod; }
ll add(ll a, ll b) { return ((a % mod) + (b % mod) + mod) % mod; }
ll sub(ll a, ll b) { return ((a % mod) - (b % mod) + mod) % mod; }
ll lcm(ll a, ll b) { return ((a * b) / (__gcd(a,b))); }

ll bin_expo(ll a, ll b) {
    a %= mod;
    ll ans = 1;
    while (b > 0) {
        if (b & 1) {
            ans = (ans * a) % mod;
        }
        a = (a * a) % mod;
        b >>= 1;
    }
    return ans % mod;
}

void sieve_of_eratosthenes() {
    isprime[0] = isprime[1] = false;
    for(int i = 2; i<=N; i++) {
        if(isprime[i] == true) {
            for(int j = 2*i; j<=N; j+=i) {
                isprime[j] = false;
            }
        }
    }
}

bool ispal(string s) {
    int n = s.size();
    for (int i = 0; i < n / 2; ++i) {
        if (s[i] != s[n - i - 1]) {
            return false;
        }
    }
    return true;
}

int main() {

    fast_io();

    int t;
    cin >> t;
    while (t--)
    {
        ll n;
        cin>>n;
        vector<ll>v(n);
        for(int i = 0; i<n; i++)
        {
            cin>>v[i];
        }
        map<pair<ll, ll>, map<ll, ll>>mp;

        ll ans = 0;

        for(int i = 0; i<n-2; i++)
        {
            mp[{v[i], v[i+1]}][v[i+2]]++;
        }

        for(auto it : mp)
        {
            ll len = it.second.size();
            len = len * (len - 1) / 2;

            for(auto i : it.second)
            {
                len *= i.second;
            }

            ans += len;
        }

        mp.clear();

        for(int i = 0; i<n-2; i++)
        {
            mp[{v[i], v[i+2]}][v[i+1]]++;
        }

        for(auto it : mp)
        {
            ll len = it.second.size();
            len = len * (len - 1) / 2;

            for(auto i : it.second)
            {
                len *= i.second;
            }

            ans += len;
        }

        mp.clear();

        for(int i = 0; i<n-2; i++)
        {
            mp[{v[i+2], v[i+1]}][v[i]]++;
        }

        for(auto it : mp)
        {
            ll len = it.second.size();
            len = len * (len - 1) / 2;

            for(auto i : it.second)
            {
                len *= i.second;
            }

            ans += len;
        }

        cout<<ans<<endl;
    }

    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
3
1 1 1
4
1 1 1 2
5
1 1 1 2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Beautiful Triple Pairs
time_limit_ms: 4000
memory_limit_mb: 256
Description: Polycarp was given an array $a$ of $n$ integers. He really likes triples of numbers, so for each $j$ ($1 \le j \le n - 2$) he wrote down a triple of elements $[a_j, a_{j + 1}, a_{j + 2}]$.

Polycarp considers a pair of triples $b$ and $c$ beautiful if they differ in exactly one position, that is, one of the following conditions is satisfied:

  * $b_1 \ne c_1$ and $b_2 = c_2$ and $b_3 = c_3$;    * $b_1 = c_1$ and $b_2 \ne c_2$ and $b_3 = c_3$;    * $b_1 = c_1$ and $b_2 = c_2$ and $b_3 \ne c_3$.

Find the number of beautiful pairs of triples among the written triples $[a_j, a_{j + 1}, a_{j + 2}]$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($3 \le n \le 2 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$) — the elements of the array.

It is guaranteed that the sum of the values of $n$ for all test cases in the test does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer — the number of beautiful pairs of triples among the pairs of the form $[a_j, a_{j + 1}, a_{j + 2}]$.

Note that the answer may not fit into 32-bit data types.

Examples:
input:
8
5
3 2 2 2 3
5
1 2 1 2 1
8
1 2 3 2 2 3 4 2
4
2 1 1 1
8
2 1 1 2 1 1 1 1
7
2 1 1 1 1 1 1
6
2 1 1 1 1 1
5
2 1 1 1 1
output:
2
0
3
1
8
4
3
2

Note: In the first example, $a = [3, 2, 2, 2, 3]$, Polycarp will write the following triples:

  1. $[3, 2, 2]$;    2. $[2, 2, 2]$;    3. $[2, 2, 3]$.

The beautiful pairs are triple $1$ with triple $2$ and triple $2$ with triple $3$.

In the third example, $a = [1, 2, 3, 2, 2, 3, 4, 2]$, Polycarp will write the following triples:

  1. $[1, 2, 3]$;    2. $[2, 3, 2]$;    3. $[3, 2, 2]$;    4. $[2, 2, 3]$;    5. $[2, 3, 4]$;    6. $[3, 4, 2]$;

The beautiful pairs are triple $1$ with triple $4$, triple $2$ with triple $5$, and triple $3$ with triple $6$.

Buggy Code:
def count_beautiful_pairs(t, test_cases):
    results = []

    for i in range(t):
        n, array = test_cases[i]
        triples = [tuple(array[j:j+3]) for j in range(n-2)]
        count = 0

        for j in range(len(triples) - 1):
            for k in range(j + 1, len(triples)):
                b, c = triples[j], triples[k]
                differences = sum(b[l] != c[l] for l in range(3))
                if differences == 1:
                    count += 1

        results.append(count)

    return results

# Example usage
t = 8
test_cases = [
    (5, [3, 2, 2, 2, 3]),
    (5, [1, 2, 1, 2, 1]),
    (8, [1, 2, 3, 2, 2, 3, 4, 2]),
    (4, [2, 1, 1, 1]),
    (8, [2, 1, 1, 2, 1, 1, 1, 1]),
    (7, [2, 1, 1, 1, 1, 1, 1]),
    (6, [2, 1, 1, 1, 1, 1]),
    (5, [2, 1, 1, 1, 1])
]

results = count_beautiful_pairs(t, test_cases)
for result in results:
    print(result)

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
2
6
1 2 2 1 2 2
5
1 1 1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Beautiful Triple Pairs
time_limit_ms: 4000
memory_limit_mb: 256
Description: Polycarp was given an array $a$ of $n$ integers. He really likes triples of numbers, so for each $j$ ($1 \le j \le n - 2$) he wrote down a triple of elements $[a_j, a_{j + 1}, a_{j + 2}]$.

Polycarp considers a pair of triples $b$ and $c$ beautiful if they differ in exactly one position, that is, one of the following conditions is satisfied:

  * $b_1 \ne c_1$ and $b_2 = c_2$ and $b_3 = c_3$;    * $b_1 = c_1$ and $b_2 \ne c_2$ and $b_3 = c_3$;    * $b_1 = c_1$ and $b_2 = c_2$ and $b_3 \ne c_3$.

Find the number of beautiful pairs of triples among the written triples $[a_j, a_{j + 1}, a_{j + 2}]$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($3 \le n \le 2 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$) — the elements of the array.

It is guaranteed that the sum of the values of $n$ for all test cases in the test does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer — the number of beautiful pairs of triples among the pairs of the form $[a_j, a_{j + 1}, a_{j + 2}]$.

Note that the answer may not fit into 32-bit data types.

Examples:
input:
8
5
3 2 2 2 3
5
1 2 1 2 1
8
1 2 3 2 2 3 4 2
4
2 1 1 1
8
2 1 1 2 1 1 1 1
7
2 1 1 1 1 1 1
6
2 1 1 1 1 1
5
2 1 1 1 1
output:
2
0
3
1
8
4
3
2

Note: In the first example, $a = [3, 2, 2, 2, 3]$, Polycarp will write the following triples:

  1. $[3, 2, 2]$;    2. $[2, 2, 2]$;    3. $[2, 2, 3]$.

The beautiful pairs are triple $1$ with triple $2$ and triple $2$ with triple $3$.

In the third example, $a = [1, 2, 3, 2, 2, 3, 4, 2]$, Polycarp will write the following triples:

  1. $[1, 2, 3]$;    2. $[2, 3, 2]$;    3. $[3, 2, 2]$;    4. $[2, 2, 3]$;    5. $[2, 3, 4]$;    6. $[3, 4, 2]$;

The beautiful pairs are triple $1$ with triple $4$, triple $2$ with triple $5$, and triple $3$ with triple $6$.

Buggy Code:
test=int(input())

for ksdjf in range(test):
    n=int(input())

    a=list(map(int,input().split()))

    b=[]
    cnt=0
    for i in range(n-2):
        c=[a[i],a[i+1],a[i+2]]
        b.append(c)
        cnt+=1

    dinku={}
    ans=0
    jhanku={}

    rancho=0
    # print(b)
    for i in range(cnt-1):
        ara=b[i]
        dinku={}
        tinku=tuple(ara)
        rancho=0
        if(tinku in jhanku):
            ans+=jhanku[tinku]

        else:

            for j in range(i+1,cnt):

                mil=0
                cra=b[j]

                titu=tuple(cra)

                if( titu in dinku):
                    ans+=dinku[titu]
                    rancho+=dinku[titu]
                else:
                    for k in range(3):
                        if(ara[k]==cra[k]):
                            mil+=1

                    if(mil==2):
                        dinku[titu]=1
                        ans+=1
                        rancho+=1
                    else:
                        dinku[titu]=0
            tinku=tuple(ara)
            jhanku[tinku]=rancho

    # print(jhanku)
    print(ans)

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
4
5
1 2 3 2 1
6
1 1 2 2 3 3
7
2 2 2 1 1 1 1
8
1 2 3 3 2 2 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Beautiful Triple Pairs
time_limit_ms: 4000
memory_limit_mb: 256
Description: Polycarp was given an array $a$ of $n$ integers. He really likes triples of numbers, so for each $j$ ($1 \le j \le n - 2$) he wrote down a triple of elements $[a_j, a_{j + 1}, a_{j + 2}]$.

Polycarp considers a pair of triples $b$ and $c$ beautiful if they differ in exactly one position, that is, one of the following conditions is satisfied:

  * $b_1 \ne c_1$ and $b_2 = c_2$ and $b_3 = c_3$;    * $b_1 = c_1$ and $b_2 \ne c_2$ and $b_3 = c_3$;    * $b_1 = c_1$ and $b_2 = c_2$ and $b_3 \ne c_3$.

Find the number of beautiful pairs of triples among the written triples $[a_j, a_{j + 1}, a_{j + 2}]$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($3 \le n \le 2 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$) — the elements of the array.

It is guaranteed that the sum of the values of $n$ for all test cases in the test does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer — the number of beautiful pairs of triples among the pairs of the form $[a_j, a_{j + 1}, a_{j + 2}]$.

Note that the answer may not fit into 32-bit data types.

Examples:
input:
8
5
3 2 2 2 3
5
1 2 1 2 1
8
1 2 3 2 2 3 4 2
4
2 1 1 1
8
2 1 1 2 1 1 1 1
7
2 1 1 1 1 1 1
6
2 1 1 1 1 1
5
2 1 1 1 1
output:
2
0
3
1
8
4
3
2

Note: In the first example, $a = [3, 2, 2, 2, 3]$, Polycarp will write the following triples:

  1. $[3, 2, 2]$;    2. $[2, 2, 2]$;    3. $[2, 2, 3]$.

The beautiful pairs are triple $1$ with triple $2$ and triple $2$ with triple $3$.

In the third example, $a = [1, 2, 3, 2, 2, 3, 4, 2]$, Polycarp will write the following triples:

  1. $[1, 2, 3]$;    2. $[2, 3, 2]$;    3. $[3, 2, 2]$;    4. $[2, 2, 3]$;    5. $[2, 3, 4]$;    6. $[3, 4, 2]$;

The beautiful pairs are triple $1$ with triple $4$, triple $2$ with triple $5$, and triple $3$ with triple $6$.

Buggy Code:
def count_beautiful_pairs(triples):
    count = 0
    length = len(triples)

    triples.sort()

    for i in range(length):
        for j in range(i + 1, length):
            diff_count = 0
            for k in range(3):
                if triples[i][k] != triples[j][k]:
                    diff_count += 1
            if diff_count == 1:
                count += 1
            elif diff_count > 1:
                break

    return count

def process_test_case(n, array):
    triples = []
    for j in range(n - 2):
        triples.append((array[j], array[j + 1], array[j + 2]))
    return count_beautiful_pairs(triples)

t = int(input().strip())
results = []
for _ in range(t):
    n = int(input().strip())
    a = list(map(int, input().split()))
    print(process_test_case(n, a))

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
4
6
1 2 3 4 5 6
6
1 1 1 2 2 2
6
1 2 1 2 1 2
6
1 1 2 2 3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Beautiful Triple Pairs
time_limit_ms: 4000
memory_limit_mb: 256
Description: Polycarp was given an array $a$ of $n$ integers. He really likes triples of numbers, so for each $j$ ($1 \le j \le n - 2$) he wrote down a triple of elements $[a_j, a_{j + 1}, a_{j + 2}]$.

Polycarp considers a pair of triples $b$ and $c$ beautiful if they differ in exactly one position, that is, one of the following conditions is satisfied:

  * $b_1 \ne c_1$ and $b_2 = c_2$ and $b_3 = c_3$;    * $b_1 = c_1$ and $b_2 \ne c_2$ and $b_3 = c_3$;    * $b_1 = c_1$ and $b_2 = c_2$ and $b_3 \ne c_3$.

Find the number of beautiful pairs of triples among the written triples $[a_j, a_{j + 1}, a_{j + 2}]$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($3 \le n \le 2 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$) — the elements of the array.

It is guaranteed that the sum of the values of $n$ for all test cases in the test does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer — the number of beautiful pairs of triples among the pairs of the form $[a_j, a_{j + 1}, a_{j + 2}]$.

Note that the answer may not fit into 32-bit data types.

Examples:
input:
8
5
3 2 2 2 3
5
1 2 1 2 1
8
1 2 3 2 2 3 4 2
4
2 1 1 1
8
2 1 1 2 1 1 1 1
7
2 1 1 1 1 1 1
6
2 1 1 1 1 1
5
2 1 1 1 1
output:
2
0
3
1
8
4
3
2

Note: In the first example, $a = [3, 2, 2, 2, 3]$, Polycarp will write the following triples:

  1. $[3, 2, 2]$;    2. $[2, 2, 2]$;    3. $[2, 2, 3]$.

The beautiful pairs are triple $1$ with triple $2$ and triple $2$ with triple $3$.

In the third example, $a = [1, 2, 3, 2, 2, 3, 4, 2]$, Polycarp will write the following triples:

  1. $[1, 2, 3]$;    2. $[2, 3, 2]$;    3. $[3, 2, 2]$;    4. $[2, 2, 3]$;    5. $[2, 3, 4]$;    6. $[3, 4, 2]$;

The beautiful pairs are triple $1$ with triple $4$, triple $2$ with triple $5$, and triple $3$ with triple $6$.

Buggy Code:
import java.util.Scanner;
public class Cp2 {

    public static void main(String[] args) {
        Scanner scan=new Scanner(System.in);
        int times=scan.nextInt();
        while(times-->0)
        {
         int n=scan.nextInt();
         int [][]x=new int[n][2];
         for(int i=0;i<n;i++)
         x[i][1]=scan.nextInt();
         int a,b,c;int []ind=new int[n];int in=0;
         long an=0;
         for(int i=0;i<n-3;i++)
         {
          if(x[i][0]==1)
          continue;
          in=0;
          for(int j=i+1;j<n-2;j++)
          {

            if(x[j][1]==x[i][1]||x[j+1][1]==x[i+1][1]||x[j+2][1]==x[i+2][1])
            {
              ind[in++]=j;if(check(x,i,j))
              {
                x[j][0]=1;an++;

              }

            }
          }
          for(int j=0;j<in-1;j++)
          {
            for(int k=j+1;k<in;k++)
            {
              if(x[ind[j]][0]==1&&check(x,ind[j],ind[k]))
              an++;
            }
          }

         }
         System.out.println(an);

         } scan.close();
        }
        public static boolean check(int [][]x,int a,int b)
        {
          if(x[a][1]==x[b][1])
          {
            if(x[a+1][1]==x[b+1][1])
            {
              if(x[a+2][1]!=x[b+2][1])
              return true;
              return false;
            }
            else
            {if(x[a+2][1]==x[b+2][1])
            return true;
            return false;
            }
          }
          else if(x[a+1][1]==x[b+1][1]&&x[a+2][1]==x[b+2][1])
          return true;
          return false;

        }

      }

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
5
2 2 2 2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Beautiful Triple Pairs
time_limit_ms: 4000
memory_limit_mb: 256
Description: Polycarp was given an array $a$ of $n$ integers. He really likes triples of numbers, so for each $j$ ($1 \le j \le n - 2$) he wrote down a triple of elements $[a_j, a_{j + 1}, a_{j + 2}]$.

Polycarp considers a pair of triples $b$ and $c$ beautiful if they differ in exactly one position, that is, one of the following conditions is satisfied:

  * $b_1 \ne c_1$ and $b_2 = c_2$ and $b_3 = c_3$;    * $b_1 = c_1$ and $b_2 \ne c_2$ and $b_3 = c_3$;    * $b_1 = c_1$ and $b_2 = c_2$ and $b_3 \ne c_3$.

Find the number of beautiful pairs of triples among the written triples $[a_j, a_{j + 1}, a_{j + 2}]$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($3 \le n \le 2 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$) — the elements of the array.

It is guaranteed that the sum of the values of $n$ for all test cases in the test does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer — the number of beautiful pairs of triples among the pairs of the form $[a_j, a_{j + 1}, a_{j + 2}]$.

Note that the answer may not fit into 32-bit data types.

Examples:
input:
8
5
3 2 2 2 3
5
1 2 1 2 1
8
1 2 3 2 2 3 4 2
4
2 1 1 1
8
2 1 1 2 1 1 1 1
7
2 1 1 1 1 1 1
6
2 1 1 1 1 1
5
2 1 1 1 1
output:
2
0
3
1
8
4
3
2

Note: In the first example, $a = [3, 2, 2, 2, 3]$, Polycarp will write the following triples:

  1. $[3, 2, 2]$;    2. $[2, 2, 2]$;    3. $[2, 2, 3]$.

The beautiful pairs are triple $1$ with triple $2$ and triple $2$ with triple $3$.

In the third example, $a = [1, 2, 3, 2, 2, 3, 4, 2]$, Polycarp will write the following triples:

  1. $[1, 2, 3]$;    2. $[2, 3, 2]$;    3. $[3, 2, 2]$;    4. $[2, 2, 3]$;    5. $[2, 3, 4]$;    6. $[3, 4, 2]$;

The beautiful pairs are triple $1$ with triple $4$, triple $2$ with triple $5$, and triple $3$ with triple $6$.

Buggy Code:
import java.util.*;

public class Problem {

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int t = scan.nextInt();
        while (t --> 0) {
            Map<Integer, Map<Integer, Map<Integer, Integer>>> map = new HashMap<>();
            int n = scan.nextInt();
            int[] arr = new int[n];
            for (int i = 0; i < n; i++) {
                arr[i] = scan.nextInt();
            }
            for (int i = 0; i < n - 2; i++) {
                if (!map.containsKey(arr[i])) map.put(arr[i], new HashMap<>());
                if (!map.get(arr[i]).containsKey(arr[i + 1])) map.get(arr[i]).put(arr[i + 1], new HashMap<>());
                if (!map.get(arr[i]).get(arr[i + 1]).containsKey(arr[i + 2])) map.get(arr[i]).get(arr[i + 1]).put(arr[i + 2], 0);
                map.get(arr[i]).get(arr[i + 1]).put(arr[i + 2], map.get(arr[i]).get(arr[i + 1]).get(arr[i + 2]) + 1);
            }
            long ans = 0;
            for (int i1 : map.keySet()) {
                for (int i2 : map.get(i1).keySet()) {
                    if (map.get(i1).get(i2).size() <= 1) continue;
                    long temp = 1;
                    for (int i3 : map.get(i1).get(i2).keySet()) {
                        temp *= (long) map.get(i1).get(i2).get(i3);
                    }
                    ans += temp;
                }
            }
            map = new HashMap<>();
            for (int i = 0; i < n - 2; i++) {
                if (!map.containsKey(arr[i + 1])) map.put(arr[i + 1], new HashMap<>());
                if (!map.get(arr[i + 1]).containsKey(arr[i + 2])) map.get(arr[i + 1]).put(arr[i + 2], new HashMap<>());
                if (!map.get(arr[i + 1]).get(arr[i + 2]).containsKey(arr[i])) map.get(arr[i + 1]).get(arr[i + 2]).put(arr[i], 0);
                map.get(arr[i + 1]).get(arr[i + 2]).put(arr[i], map.get(arr[i + 1]).get(arr[i + 2]).get(arr[i]) + 1);
            }
            for (int i1 : map.keySet()) {
                for (int i2 : map.get(i1).keySet()) {
                    if (map.get(i1).get(i2).size() <= 1) continue;
                    long temp = 1;
                    for (int i3 : map.get(i1).get(i2).keySet()) {
                        temp *= (long) map.get(i1).get(i2).get(i3);
                    }
                    ans += temp;
                }
            }
            map = new HashMap<>();
            for (int i = 0; i < n - 2; i++) {
                if (!map.containsKey(arr[i])) map.put(arr[i], new HashMap<>());
                if (!map.get(arr[i]).containsKey(arr[i + 2])) map.get(arr[i]).put(arr[i + 2], new HashMap<>());
                if (!map.get(arr[i]).get(arr[i + 2]).containsKey(arr[i + 1])) map.get(arr[i]).get(arr[i + 2]).put(arr[i + 1], 0);
                map.get(arr[i]).get(arr[i + 2]).put(arr[i + 1], map.get(arr[i]).get(arr[i + 2]).get(arr[i + 1]) + 1);
            }
            for (int i1 : map.keySet()) {
                for (int i2 : map.get(i1).keySet()) {
                    if (map.get(i1).get(i2).size() <= 1) continue;
                    long temp = 1;
                    for (int i3 : map.get(i1).get(i2).keySet()) {
                        temp *= (long) map.get(i1).get(i2).get(i3);
                    }
                    ans += temp;
                }
            }
            System.out.println(ans);
        }
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
4
1 1 1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Beautiful Triple Pairs
time_limit_ms: 4000
memory_limit_mb: 256
Description: Polycarp was given an array $a$ of $n$ integers. He really likes triples of numbers, so for each $j$ ($1 \le j \le n - 2$) he wrote down a triple of elements $[a_j, a_{j + 1}, a_{j + 2}]$.

Polycarp considers a pair of triples $b$ and $c$ beautiful if they differ in exactly one position, that is, one of the following conditions is satisfied:

  * $b_1 \ne c_1$ and $b_2 = c_2$ and $b_3 = c_3$;    * $b_1 = c_1$ and $b_2 \ne c_2$ and $b_3 = c_3$;    * $b_1 = c_1$ and $b_2 = c_2$ and $b_3 \ne c_3$.

Find the number of beautiful pairs of triples among the written triples $[a_j, a_{j + 1}, a_{j + 2}]$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($3 \le n \le 2 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$) — the elements of the array.

It is guaranteed that the sum of the values of $n$ for all test cases in the test does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer — the number of beautiful pairs of triples among the pairs of the form $[a_j, a_{j + 1}, a_{j + 2}]$.

Note that the answer may not fit into 32-bit data types.

Examples:
input:
8
5
3 2 2 2 3
5
1 2 1 2 1
8
1 2 3 2 2 3 4 2
4
2 1 1 1
8
2 1 1 2 1 1 1 1
7
2 1 1 1 1 1 1
6
2 1 1 1 1 1
5
2 1 1 1 1
output:
2
0
3
1
8
4
3
2

Note: In the first example, $a = [3, 2, 2, 2, 3]$, Polycarp will write the following triples:

  1. $[3, 2, 2]$;    2. $[2, 2, 2]$;    3. $[2, 2, 3]$.

The beautiful pairs are triple $1$ with triple $2$ and triple $2$ with triple $3$.

In the third example, $a = [1, 2, 3, 2, 2, 3, 4, 2]$, Polycarp will write the following triples:

  1. $[1, 2, 3]$;    2. $[2, 3, 2]$;    3. $[3, 2, 2]$;    4. $[2, 2, 3]$;    5. $[2, 3, 4]$;    6. $[3, 4, 2]$;

The beautiful pairs are triple $1$ with triple $4$, triple $2$ with triple $5$, and triple $3$ with triple $6$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;

int main() {
    int t;
    cin >> t;
    while (t--) {
        int size;
        cin >> size;

        vector<int> arr(size);
        for (int i = 0; i < size; i++) cin >> arr[i];

        map<vector<int>, int> mpp;
        for (int i = 0; i <= size - 3; i++) {
            vector<int> temp = {arr[i], arr[i + 1], arr[i + 2]};
            mpp[temp]++;
        }

        long long ans = 0;
        for (auto p = mpp.begin(); p != mpp.end(); ++p) {
            vector<int> key1 = p->first;
            int count1 = p->second;

            for (auto q = next(p); q != mpp.end(); ++q) {
                vector<int> key2 = q->first;
                int count2 = q->second;

                int diffCount = 0;
                for (int k = 0; k < 3; k++) {
                    if (key1[k] != key2[k]) diffCount++;
                }

                if (diffCount == 1) {
                    ans += static_cast<long long>(count1) * count2;
                }
            }
        }

        cout << ans << endl;
    }
    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To create a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases or scenarios that can maximize the failure probability. The primary issue in the buggy code is that it only considers pairs of triples that are different in exactly one position, but it does not handle the case where a triple can form a beautiful pair with itself if it appears multiple times in the array.

Here are the steps to create a test input:

1. **Identify the Bug**: The code does not handle the case where a triple can form a beautiful pair with itself if it appears multiple times in the array. This means that if a triple `[a, b, c]` appears more than once, the code should count the number of pairs formed by these occurrences, but it does not.

2. **Create a Test Case**:
   - **Array with Repeated Triples**: We need an array where a specific triple appears multiple times. This will ensure that the code should count the pairs formed by these repeated triples.
   - **Small Scale**: The test input should be small enough to be manageable but large enough to demonstrate the bug clearly.

3. **Example Test Input**:
   - **Test Case 1**: An array with a single repeated triple.
   - **Test Case 2**: An array with multiple repeated triples.

Let's create the test input:

```plaintext
2
5
1 1 1 1 1
6
1 1 1 1 1 1
```

### Explanation:
- **Test Case 1**: The array is `[1, 1, 1, 1, 1]`.
  - The triples are `[1, 1, 1]`, `[1, 1, 1]`, and `[1, 1, 1]`.
  - The beautiful pairs are:
    - Triple 1 with Triple 2
    - Triple 1 with Triple 3
    - Triple 2 with Triple 3
  - The expected output is `3`.

- **Test Case 2**: The array is `[1, 1, 1, 1, 1, 1]`.
  - The triples are `[1, 1, 1]`, `[1, 1, 1]`, `[1, 1, 1]`, and `[1, 1, 1]`.
  - The beautiful pairs are:
    - Triple 1 with Triple 2
    - Triple 1 with Triple 3
    - Triple 1 with Triple 4
    - Triple 2 with Triple 3
    - Triple 2 with Triple 4
    - Triple 3 with Triple 4
  - The expected output is `6`.

This test input will help expose the bug in the buggy code by showing that it fails to count the pairs formed by repeated triples.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Beautiful Triple Pairs
time_limit_ms: 4000
memory_limit_mb: 256
Description: Polycarp was given an array $a$ of $n$ integers. He really likes triples of numbers, so for each $j$ ($1 \le j \le n - 2$) he wrote down a triple of elements $[a_j, a_{j + 1}, a_{j + 2}]$.

Polycarp considers a pair of triples $b$ and $c$ beautiful if they differ in exactly one position, that is, one of the following conditions is satisfied:

  * $b_1 \ne c_1$ and $b_2 = c_2$ and $b_3 = c_3$;    * $b_1 = c_1$ and $b_2 \ne c_2$ and $b_3 = c_3$;    * $b_1 = c_1$ and $b_2 = c_2$ and $b_3 \ne c_3$.

Find the number of beautiful pairs of triples among the written triples $[a_j, a_{j + 1}, a_{j + 2}]$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($3 \le n \le 2 \cdot 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$) — the elements of the array.

It is guaranteed that the sum of the values of $n$ for all test cases in the test does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer — the number of beautiful pairs of triples among the pairs of the form $[a_j, a_{j + 1}, a_{j + 2}]$.

Note that the answer may not fit into 32-bit data types.

Examples:
input:
8
5
3 2 2 2 3
5
1 2 1 2 1
8
1 2 3 2 2 3 4 2
4
2 1 1 1
8
2 1 1 2 1 1 1 1
7
2 1 1 1 1 1 1
6
2 1 1 1 1 1
5
2 1 1 1 1
output:
2
0
3
1
8
4
3
2

Note: In the first example, $a = [3, 2, 2, 2, 3]$, Polycarp will write the following triples:

  1. $[3, 2, 2]$;    2. $[2, 2, 2]$;    3. $[2, 2, 3]$.

The beautiful pairs are triple $1$ with triple $2$ and triple $2$ with triple $3$.

In the third example, $a = [1, 2, 3, 2, 2, 3, 4, 2]$, Polycarp will write the following triples:

  1. $[1, 2, 3]$;    2. $[2, 3, 2]$;    3. $[3, 2, 2]$;    4. $[2, 2, 3]$;    5. $[2, 3, 4]$;    6. $[3, 4, 2]$;

The beautiful pairs are triple $1$ with triple $4$, triple $2$ with triple $5$, and triple $3$ with triple $6$.

Buggy Code:
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#define aaaaa_iam_gonna_kill_myself ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL);
#define ll long long
#define ld long double
#define dd double
#define endl "\n"
#define rep(n) for(int i=0;i<n;i++)
using namespace std;
int N=1e5+7;/*
bool isprime(int n)
{
    for(int i=2; i*i<=n; i++)
    {
        if(n%i==0)
            return 0;
    }
    return n>1;
}
ll GCD(ll a,ll b)
{
    if(a%b==0)
    {
        return b;
    }
    else
    {
        ll x=a;
        a=b;
        b=x-((x/a)*a);
        return GCD(a,b);
    }
}
vector<ll>v;
void primes()
{
    for(int i=2; i<=N; i++)
    {
        if(isprime(i))
        {
            v.push_back(i);
        }
    }
}*/

int main()
{
    aaaaa_iam_gonna_kill_myself;
    int t=1;
    cin>>t;

    while(t--)
    {
        int n;
        cin>>n;
        int arr[n];
        rep(n)cin>>arr[i];
        ll res=0;
        vector<vector<int>>v(n-2);
        for(int i=0;i<n-2;i++)
        {
            v[i].push_back(arr[i]);
            v[i].push_back(arr[i+1]);
            v[i].push_back(arr[i+2]);
        }
        for(int i=0;i<v.size();i++)
        {
            for(int j=i+1;j<v.size();j++)
            {
               ll k=0;
               if(v[i][0]!=v[j][0])k++;
               if(v[i][1]!=v[j][1])k++;
               if(v[i][2]!=v[j][2])k++;
               if(k==1)
                  res++;

            }
        }
cout<<res<<endl;
    }
    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases or scenarios that maximize the probability of failure. The code provided uses a brute-force approach to check all pairs of triples, which can be inefficient and prone to errors, especially in edge cases.

### Step-by-Step Analysis:
1. **Edge Cases:**
   - **Small Arrays:** Arrays with the minimum length (3) to ensure the code handles the smallest possible input correctly.
   - **Large Arrays:** Arrays with the maximum length (200,000) to test the code's efficiency and correctness under large input sizes.
   - **Arrays with Repeated Elements:** Arrays where elements are repeated to test the logic for identifying beautiful pairs.
   - **Arrays with Unique Elements:** Arrays where all elements are unique to test the logic for identifying beautiful pairs.

2. **Efficiency:**
   - The current code has a time complexity of \(O(n^2)\), which can be inefficient for large inputs. We need to ensure that the test case is small enough to run within the time limit but challenging enough to expose any logical issues.

3. **Logical Issues:**
   - The code checks all pairs of triples, which can lead to incorrect results if the logic for identifying beautiful pairs is flawed.

### Chosen Test Input:
Let's create a small-scale test input that is complete and valid. We will use an array with repeated elements to ensure that the code's logic for identifying beautiful pairs is tested thoroughly.

```plaintext
1
6
2 1 1 1 1 1
```

### Explanation:
- **Test Case 1:**
  - **Array:** `2 1 1 1 1 1`
  - **Triples:**
    - `[2, 1, 1]`
    - `[1, 1, 1]`
    - `[1, 1, 1]`
    - `[1, 1, 1]`
    - `[1, 1, 1]`
  - **Beautiful Pairs:**
    - `[2, 1, 1]` and `[1, 1, 1]` (1st and 2nd triple)
    - `[2, 1, 1]` and `[1, 1, 1]` (1st and 3rd triple)
    - `[2, 1, 1]` and `[1, 1, 1]` (1st and 4th triple)
    - `[2, 1, 1]` and `[1, 1, 1]` (1st and 5th triple)
    - `[1, 1, 1]` and `[1, 1, 1]` (2nd and 3rd triple)
    - `[1, 1, 1]` and `[1, 1, 1]` (2nd and 4th triple)
    - `[1, 1, 1]` and `[1, 1, 1]` (2nd and 5th triple)
    - `[1, 1, 1]` and `[1, 1, 1]` (3rd and 4th triple)
    - `[1, 1, 1]` and `[1, 1, 1]` (3rd and 5th triple)
    - `[1, 1, 1]` and `[1, 1, 1]` (4th and 5th triple)

However, since the triples are identical except for the first element, the only beautiful pairs are those involving the first triple and any of the others. Therefore, the correct output should be 4.

### Expected Output:
```plaintext
4
```

This test input should expose any logical issues in the code's implementation of identifying beautiful pairs.