**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ardent Flames
time_limit_ms: 4000
memory_limit_mb: 256
Description: You have obtained the new limited event character Xilonen. You decide to use her in combat.

There are $$$n$$$ enemies in a line. The $$$i$$$'th enemy from the left has health $$$h_i$$$ and is currently at position $$$x_i$$$. Xilonen has an attack damage of $$$m$$$, and you are ready to defeat the enemies with her.

Xilonen has a powerful "ground stomp" attack. Before you perform any attacks, you select an integer $$$p$$$ and position Xilonen there ($$$p$$$ can be any integer position, including a position with an enemy currently). Afterwards, for each attack, she deals $$$m$$$ damage to an enemy at position $$$p$$$ (if there are any), $$$m-1$$$ damage to enemies at positions $$$p-1$$$ and $$$p+1$$$, $$$m-2$$$ damage to enemies at positions $$$p-2$$$ and $$$p+2$$$, and so on. Enemies that are at least a distance of $$$m$$$ away from Xilonen take no damage from attacks.

Formally, if there is an enemy at position $$$x$$$, she will deal $$$\max(0,m - |p - x|)$$$ damage to that enemy each hit. Note that you may not choose a different $$$p$$$ for different attacks.

Over all possible $$$p$$$, output the minimum number of attacks Xilonen must perform to defeat at least $$$k$$$ enemies. If it is impossible to find a $$$p$$$ such that eventually at least $$$k$$$ enemies will be defeated, output $$$-1$$$ instead. Note that an enemy is considered to be defeated if its health reaches $$$0$$$ or below.

Input: The first line contains an integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) – the number of test cases.

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \leq k \leq n \leq 10^5$$$, $$$1 \leq m \leq 10^9$$$).

The following line contains $$$n$$$ integers $$$h_1, h_2, ..., h_n$$$ ($$$1 \leq h_i \leq 10^9$$$).

The last line of each testcase contains $$$n$$$ integers $$$x_1, x_2, ..., x_n$$$ ($$$1\leq x_i \leq 10^9$$$, $$$x_i < x_{i+1}$$$ for all $$$1 \leq i < n$$$)

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.

Output: For each test case, output an integer on a new line, the minimum number of attacks that must be performed to defeat at least $$$k$$$ enemies. If it is impossible to find a $$$p$$$ such that eventually at least $$$k$$$ enemies will be defeated, output $$$-1$$$ instead.

Examples:
input:
6
5 5 3
7 7 7 7 7
1 2 3 4 5
9 5 9
2 4 6 8 10 8 6 4 2
1 2 3 4 5 6 7 8 9
2 10 2
1 1
1 20
2 10 1
69696969 420420420
1 20
2 10 2
10 15
1 19
2 2 2
1000000000 1
1 3
output:
2
2
-1
6969697
15
1000000000

Note: In the first testcase, it is optimal to select $$$p=2$$$. Each attack, the first enemy takes $$$5-|2-1|=4$$$ damage, the second enemy takes $$$5$$$ damage, the third enemy takes $$$4$$$ damage, the fourth enemy takes $$$3$$$ damage, and the fifth enemy takes $$$2$$$ damage. After $$$2$$$ attacks, the first three enemies will be defeated. It can be shown that it is impossible to defeat $$$3$$$ enemies in less than $$$2$$$ attacks, no matter which $$$p$$$ is selected.

In the second testcase, we must kill all $$$9$$$ enemies. By selecting $$$p=5$$$, all nine enemies will be defeated in $$$2$$$ attacks.

In the third testcase, we must kill both enemies. However, it can be shown that no $$$p$$$ selected will damage both enemies at the same time, so the answer is $$$-1$$$.

In the fourth testcase, selecting $$$p=1$$$ will enable us to defeat the first enemy in $$$6969697$$$ attacks.

In the fifth testcase, selecting $$$p=10$$$ will make each enemy take $$$1$$$ damage per attack. Both enemies will be defeated in $$$15$$$ attacks.

Buggy Code:
/* #region template */
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;

#ifdef LOCAL_TEST
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) { return os << '(' << p.first << ", " << p.second << ')'; }
template <typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type>
ostream &operator<<(ostream &os, const T_container &v)
{
  os << '{';
  string sep;
  for (const T &x : v)
    os << sep << x, sep = ", ";
  return os << '}';
}
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T)
{
  cerr << ' ' << H;
  dbg_out(T...);
}
#define D(...) cerr << "[" << #__VA_ARGS__ << "]:", dbg_out(__VA_ARGS__)
#define ONLY(x)                                            \
  if (TC != x)                                             \
  {                                                        \
    std::cout << "Skipping TC[" << TC << "]" << std::endl; \
    return;                                                \
  }
#else
#define D(...) 17
#define ONLY(x) 17
#endif

using ll = long long;
template <typename T>
using ordered_set = __gnu_pbds::tree<T, __gnu_pbds::null_type, less<T>, __gnu_pbds::rb_tree_tag, __gnu_pbds::tree_order_statistics_node_update>;
/* For dijkstra */
template <typename T, typename F = std::greater<T>>
using heap = priority_queue<T,
                            std::vector<T>,
                            F>;

void solve(int);

int main()
{
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);

  int T = 1;
  cin >> T;

  for (int test = 1; test <= T; test++)
  {
    solve(test);
  }

  return 0;
}
/* #endregion */
const int N = 2e5 + 5;
const ll MOD = 1e9 + 7;

void solve(int TC)
{
  ll n, m, k;
  cin >> n >> m >> k;

  vector<array<int, 2>> pnts(n);
  for (auto &i : pnts)
    cin >> i[0];
  for (auto &i : pnts)
    cin >> i[1];

  auto check = [&](int stomps)
  {
    vector<array<int, 2>> ranges;
    set<int> distValues;
    for (auto [h, p] : pnts)
    {
      int mxd = m - ((h + stomps - 1) / stomps);
      if (mxd < 0 || 1LL * (m - mxd) * stomps < h)
        continue; // not reachable
      ranges.push_back({p - mxd, p + mxd});
      distValues.insert(p - mxd);
      distValues.insert(p + mxd);
      distValues.insert(p + mxd + 1);
    }

    map<int, int> cmpMap;
    int ptr = 0;
    for (auto i : distValues)
    {
      cmpMap[i] = ptr++;
    }

    sort(ranges.begin(), ranges.end());

    int CSZ = cmpMap.size();
    vector<int> Pfx(CSZ + 1, 0);
    for (const auto &[x, _] : ranges)
    {
      Pfx[cmpMap[x]]++;
      // Pfx[cmpMap[y + 1]]--;
    }

    sort(ranges.begin(), ranges.end(), [](const auto &a, const auto &b)
         { return a[1] > b[1]; });

    for (const auto &[_, y] : ranges)
    {
      // Pfx[cmpMap[x]];
      Pfx[cmpMap[y + 1]]--;
    }

    for (int i = 0; i <= CSZ; i++)
    {
      if (i)
        Pfx[i] += Pfx[i - 1];
      if (Pfx[i] >= k)
        return true;
    }

    return false;
  };

  int l = 1, inf = 1e9, r = inf;
  while (l <= r)
  {
    int m = (l + r) / 2;
    if (check(m))
    {
      r = m - 1;
    }
    else
    {
      l = m + 1;
    }
  }
  cout << (l > inf ? -1 : l) << '\n';
}

**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
6
5 5 3
7 7 7 7 7
1 2 3 4 5
9 5 9
2 4 6 8 10 8 6 4 2
1 2 3 4 5 6 7 8 9
2 10 2
1 1
1 20
2 10 1
69696969 420420420
1 20
2 10 2
10 15
1 19
2 2 2
1000000000 1
1 3
```


**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 <iostream>
#include <vector>
#include <algorithm>
using namespace std;

void solve() {
    int n, q;
    cin >> n >> q;

    vector<pair<int, int>> snakes(n);  // snake[i] stores (l, r) pair for snake i

    // Initially place snakes one after another at different positions
    for (int i = 0; i < n; ++i) {
        snakes[i] = {i + 1, i + 1};  // snake i initially occupies one cell
    }

    // To keep track of the maximum occupied cell
    int max_occupied = n;

    // Process each event
    for (int i = 0; i < q; ++i) {
        int si;
        char op;
        cin >> si >> op;
        --si;  // convert to 0-based index

        int l = snakes[si].first;
        int r = snakes[si].second;

        if (op == '+') {
            // Enlarge the snake by 1 cell to the right
            r++;
        } else {
            // Shrink the snake by 1 cell from the left
            l++;
        }

        // Update the snake's range
        snakes[si] = {l, r};

        // Update the maximum occupied cell
        max_occupied = max(max_occupied, r);
    }

    cout << max_occupied << endl;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

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


**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<iostream>
#include<vector>
#include<fstream>
#include<string>
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<bitset>
#include<algorithm>
#include<cmath>
#include <iomanip>
#include<unordered_map>
#include "math.h"
#include<numeric>
# define int  long long
using namespace std;
const long double EPS = 1e-7;
const long double fi = 1.0 / ((1 + sqrt(5)) / 2);

const long double MAX = 1e5;
int   st = 31;
# define rall(x) (x).begin(), (x).end()
# define f first
# define s second
# define all(x) (x).begin(), (x).end()
# define popcount(x) __builtin_popcount(x)
const int INF = 1e11;

void solve() {
    int n, q;
    cin >> n >> q;
    vector<vector<int>> arr(n, vector<int> (n));
    vector<int> last(n);
    vector<vector<pair<int, int>>> hui(n);
    vector<int> dist(n, 1);
    for (int ii = 0; ii < q; ++ii) {
        char a;
        int cur;
        cin >> cur >> a;
        cur--;
        if (a == '+') {
            hui[cur].push_back({1, ii});
            dist[cur]++;
        } else {
            if (dist[cur] > 1) {
                dist[cur]--;
            }
            hui[cur].push_back({-1, ii});
        }
    }
    for (int ii = 0; ii < n; ++ii) {
        hui[ii].push_back({1, INF});
    }
    for (int ii = 0; ii < n; ++ii) {
        for (int jj = 0; jj < n; ++jj) {
            if (ii == jj) {
                continue;
            }
            int l1 = 1, r1 = 1;
            int l2 = 2, r2 = 2;
            int i = 0, j = 0;
            while (!(hui[ii][i].s == INF && hui[jj][j].s == INF)) {
                if (hui[ii][i].s < hui[jj][j].s) {
                    if (hui[ii][i].f == 1) {
                        l1++;
                    } else {
                        r1++;
                        if (r1 == l2) {
                            l2++;
                            r2++;
                        }
                    }
                    i++;
                } else {
                    if (hui[jj][j].f == 1) {
                        r2++;
                    } else {
                        l2++;
                    }
                    j++;
                }
            }
            last[ii] = r1;
            arr[ii][jj] = l2 - r1;
        }
    }

    vector<vector<int>> dp((1 << n), vector<int> (n, INF));
    for (int ii = 0; ii < n; ++ii) {
        for (int jj = 0; jj < n; ++jj) {
            if (ii == jj) {
                dp[1 << ii][ii] = last[ii];
            } else {
                dp[1 << ii][jj] = INF;
            }
        }
    }

    for (int mask = 1; mask < (1 << n); ++mask) {
        for (int jj = 0; jj < n; ++jj) {
            if ((1 << jj) & mask) {
                continue;
            }
            for (int kk = 0; kk < n; ++kk) {
                if (!((1 << kk) & mask)) {
                    continue;
                }
                dp[mask | (1 << jj)][jj] = min(dp[mask | (1 << jj)][jj], dp[mask][kk] + arr[kk][jj] + dist[jj] - 1);
            }
        }
    }
    int ans = INF;
    for (int ii = 0; ii < n; ++ii) {
        ans = min(ans, dp[(1 << n) - 1][ii]);
    }
    cout << ans;

}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    solve();
}

**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 +
2 +
1 -
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:
// 2^20 ~ 10^6
// 2^10 ~ 10^3
// 2^5 = 32
#include <bits/stdc++.h>
//#define int long long
using namespace std;
const int maxn=21;
const int inf=1e9+10;
int cost[maxn][maxn], dist[maxn][maxn], dp[(1<<maxn)][maxn], adv[maxn], n;
int calc(int mask, int last, int size){
    if(dp[mask][last]!=-1) return dp[mask][last];
    int ret=inf;
    for(int k=0;k<maxn;k++){
        if(mask&(1<<k)){
            if(last=-1) ret=min(ret,calc(mask^(1<<k),k,size-1)+adv[k]);
            else ret=min(ret,calc(mask^(1<<k),k,size-1)+cost[k][last]);
        }
    }
    return dp[mask][last]=ret;
}
void solve(){
    int q; cin >> n >> q;
    memset(dp,-1,sizeof(dp));
    for(int i=0;i<n;i++) dp[0][i]=1;
    for(int i=0;i<n;i++)
        for(int j=0;j<n;j++) cost[i][j]=dist[i][j]=1;
    vector<pair<int,int>>process;
    for(int i=1;i<=q;i++){
        int a; cin >> a; a--;
        char c; cin >> c;
        if(c=='+'){
            process.push_back({a,1});
            adv[a]++;
        }
        if(c=='-') process.push_back({a,-1});
    }
    for(int i=0;i<n;i++){
        for(auto p : process){
            int id=p.first, sign=p.second;
            if(id==i&&sign==1){
                for(int j=0;j<n;j++){
                    dist[i][j]++;
                    cost[i][j]=max(cost[i][j],dist[i][j]);
                }
            }
            else if(id!=i&&sign==-1) dist[i][id]--;
        }
    }
    int resp=calc((1<<n)-1,-1,n);
    cout << resp << endl;
}
signed main()
{
    ios_base::sync_with_stdio(false); cin.tie(NULL);
    int t=1; //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
2 5
1 +
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: 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
import sys
import sys

def main():
    import sys
    import sys
    sys.setrecursionlimit(1 << 25)
    n, q = map(int, sys.stdin.readline().split())
    a = [0] * (n +1)
    b = [0] * (n +1)
    for _ in range(q):
        parts = sys.stdin.readline().strip().split()
        si = int(parts[0])
        op = parts[1]
        if op == '+':
            a[si] +=1
        else:
            b[si] +=1
    snakes = []
    for i in range(1,n+1):
        snakes.append( (1 + b[i], a[i] - b[i] +1) )
    # Function to check if a given M is feasible
    def is_feasible(M):
        # Sort snakes by s_i_min
        sorted_snakes = sorted(snakes, key=lambda x: x[0])
        current_pos = 0
        for s_min, l in sorted_snakes:
            s_i = max(current_pos +1, s_min)
            e_i = s_i + l -1
            if e_i > M:
                return False
            current_pos = e_i
        return True
    # Binary search for minimal M
    left = 1
    right = q + n + 10
    answer = right
    while left <= right:
        mid = (left + right) //2
        if is_feasible(mid):
            answer = mid
            right = mid -1
        else:
            left = mid +1
    print(answer)

if __name__ == "__main__":
    main()

**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 consider edge cases and scenarios that can cause the algorithm to fail. The key aspects to consider are:

1. **Initial Placement of Snakes**: Ensure that the initial placement of snakes can lead to conflicts if not handled correctly.
2. **Sequence of Events**: Create a sequence of events that can cause the snakes to overlap or run into each other if the logic is flawed.
3. **Edge Cases**: Test with the minimum and maximum values of `n` and `q` to ensure the code handles these cases correctly.

Let's construct a test case that meets these criteria:

1. **Initial Placement**: Place snakes in a way that they are close to each other, but not overlapping.
2. **Events**: Create a sequence of events that causes snakes to expand and shrink, potentially leading to overlaps if not handled correctly.

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

```plaintext
4 10
1 +
2 +
1 +
2 +
3 +
4 +
3 +
4 +
1 -
2 -
```

Explanation:
- **Initial Placement**: Assume the snakes are placed as follows:
  - Snake 1 at cell 1
  - Snake 2 at cell 2
  - Snake 3 at cell 3
  - Snake 4 at cell 4
- **Events**:
  - Snake 1 expands (1 -> 1-2)
  - Snake 2 expands (2 -> 2-3)
  - Snake 1 expands again (1-2 -> 1-3)
  - Snake 2 expands again (2-3 -> 2-4)
  - Snake 3 expands (3 -> 3-4)
  - Snake 4 expands (4 -> 4-5)
  - Snake 3 expands again (3-4 -> 3-5)
  - Snake 4 expands again (4-5 -> 4-6)
  - Snake 1 shrinks (1-3 -> 2-3)
  - Snake 2 shrinks (2-4 -> 3-4)

This sequence of events should cause the snakes to overlap if the code does not handle the positions correctly, leading to an incorrect score. The correct score for this input should be 6, but the buggy code might not handle the overlapping correctly and might output a different result.


**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:
def main():
    import sys
    n, q = map(int, sys.stdin.readline().split())
    operations = [sys.stdin.readline().strip() for _ in range(q)]

    # For each snake, find min_l and max_r relative to starting position
    snake_ops = [[] for _ in range(n)]
    for op in operations:
        s, sign = op.split()
        s = int(s) - 1
        snake_ops[s].append(sign)

    # Simulate operations for each snake to find min_l and max_r relative to starting position
    min_l = [0] * n
    max_r = [0] * n
    for i in range(n):
        l = 0
        r = 0
        min_current_l = 0
        max_current_r = 0
        for sign in snake_ops[i]:
            if sign == '+':
                r += 1
            else:
                l += 1
            min_current_l = min(min_current_l, l)
            max_current_r = max(max_current_r, r)
        min_l[i] = min_current_l
        max_r[i] = max_current_r

    # Sort snakes by their max_r
    snakes = sorted(range(n), key=lambda x: max_r[x])

    # Assign starting positions
    starting_position = 1
    placements = [0] * n
    for i in snakes:
        placements[i] = starting_position
        starting_position += max_r[i] - min_l[i]

    # Calculate the maximum cell used
    max_cell = 0
    for i in range(n):
        start = placements[i] - min_l[i]
        end = start + max_r[i]
        max_cell = max(max_cell, end)

    print(max_cell)

if __name__ == "__main__":
    main()

**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 +
3 +
1 -
2 -
3 -
```


**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:
def simulate_snakes(n, events, positions):
    # 记录每条蛇当前占据的区间
    snake_segments = [[p, p] for p in positions]
    max_position = max(positions)

    # 模拟每个事件
    for snake_id, op in events:
        snake_id -= 1  # 转为0-based索引

        if op == '+':
            # 扩展
            snake_segments[snake_id][1] += 1
        else:
            # 收缩
            snake_segments[snake_id][0] += 1

        # 检查碰撞
        curr_snake = snake_segments[snake_id]

        # 检查是否超出边界
        if curr_snake[1] > 10**9:
            return float('inf')

        # 检查与其他蛇的碰撞
        for i in range(n):
            if i != snake_id:
                other = snake_segments[i]
                if not (curr_snake[1] < other[0] or curr_snake[0] > other[1]):
                    return float('inf')

        max_position = max(max_position, curr_snake[1])

    return max_position

def solve(n, events):
    # 使用回溯法尝试不同的初始位置组合
    def backtrack(index, used_positions, positions):
        if index == n:
            return simulate_snakes(n, events, positions)

        min_score = float('inf')
        # 优化搜索范围,只需要考虑事件数量相关的范围
        for pos in range(1, 3*q + 1):
            if pos not in used_positions:
                used_positions.add(pos)
                positions.append(pos)
                score = backtrack(index + 1, used_positions, positions)
                min_score = min(min_score, score)
                positions.pop()
                used_positions.remove(pos)

        return min_score

    return backtrack(0, set(), [])

# 读取输入
n, q = map(int, input().split())
events = []
for _ in range(q):
    s, op = input().split()
    events.append((int(s), op))

print(solve(n, events))

**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 4
1 +
2 +
3 +
1 -
```


**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 java.util.Scanner;

public class Main{
 static class Snake {
 int l, r;
 Snake(int l, int r) {
 this.l = l;
 this.r = r;
 }
 }

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

 Snake[] snakes = new Snake[n + 1];
 for (int i = 1; i <= n; i++) {
 snakes[i] = new Snake(i, i);
 }

 String[] events = new String[q];
 for (int i = 0; i < q; i++) {
 events[i] = scanner.next() + scanner.next();
 }

 int maxCell = n;

 for (String event : events) {
 int s = Integer.parseInt(event.substring(0, event.length() - 1));
 char type = event.charAt(event.length() - 1);

 if (type == '+') {
 snakes[s].r++;
 } else {
 snakes[s].l++;
 }

 maxCell = Math.max(maxCell, Math.max(snakes[s].l, snakes[s].r));

 }
 System.out.println(maxCell);
 }
}

// tyj0dVrKplq41DN7oci

**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 +
2 +
1 -
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 java.util.*;

public class Main{
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int q = sc.nextInt();

        // Initialize the snakes' positions
        int[] l = new int[n+1];  // left points
        int[] r = new int[n+1];  // right points
        for (int i = 1; i <= n; i++) {
            l[i] = i;
            r[i] = i;
        }

        // Process each event
        for (int i = 0; i < q; i++) {
            int s = sc.nextInt();
            char op = sc.next().charAt(0);

            if (op == '+') {
                r[s]++;
            } else if (op == '-') {
                l[s]++;
            }
        }

        // Find the maximum right boundary
        int maxR = 0;
        for (int i = 1; i <= n; i++) {
            if (r[i] > maxR) {
                maxR = r[i];
            }
        }

        System.out.println(maxR);
        sc.close();
    }
}

// m8GqVORSWylsUF4wrxu

**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 expose the bug in the provided buggy code, we need to focus on scenarios where the snakes' positions might overlap or where the snakes might run into the end of the strip. The buggy code does not handle these edge cases correctly, particularly when a snake shrinks or enlarges and might overlap with another snake or exceed the strip boundary.

Here's a step-by-step approach to generate a small-scale test input:

1. **Initial Setup:**
   - Use a small number of snakes and events to keep the input small.
   - Ensure the snakes are initially placed in a way that can lead to overlapping or boundary issues.

2. **Events to Trigger Bugs:**
   - Create events that cause snakes to shrink or enlarge in such a way that they might overlap with other snakes.
   - Create events that cause snakes to shrink or enlarge to the edge of the strip.

3. **Example Test Input:**
   - Use 3 snakes and 5 events.
   - Place the snakes in a way that they are close to each other to increase the chance of overlap.
   - Create events that cause one snake to shrink and another to enlarge, leading to a potential overlap.

Here is the test input:

```plaintext
3 5
1 1 +
1 2 +
2 1 -
2 3 +
3 2 -
```

Explanation:
- **Initial Setup:**
  - Snake 1 is placed at cell 1.
  - Snake 2 is placed at cell 2.
  - Snake 3 is placed at cell 3.
- **Events:**
  - Snake 1 enlarges to cells [1, 2].
  - Snake 1 enlarges to cells [1, 3].
  - Snake 2 shrinks to cells [3, 3].
  - Snake 3 enlarges to cells [3, 4].
  - Snake 2 shrinks to cells [4, 4].

This test input should expose the bug where the snakes might overlap or run into the end of the strip, causing the buggy code to fail.


**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 <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <chrono>
#include <random>

#pragma GCC optimize("O3")

using namespace std;

#define int long long

struct Graph {
    int n;
    vector<vector<int>> dis;

    Graph(int _n, vector<vector<int>> _adj) : n(_n), dis(_adj) {}

    bool optimize(vector<int> &ord) {
        for (int i=1; i<=n; i++) {
            for (int j=i+1; j<=n; j++) {
                if (dis[ord[i-1]][ord[j]] + dis[ord[i]][ord[j+1]] < dis[ord[i-1]][ord[i]] + dis[ord[j]][ord[j+1]]) {
                    reverse(ord.begin()+i, ord.begin()+j+1);
                    return true;
                }
            }
        }
        return false;
    }

    int shortest_hamilton() {

        if (n <= 10) {
            vector<int> p(n);
            for (int i=0; i<n; i++) p[i] = i;
            int ans = 1e18;
            do {
                int tmp = 0;
                for (int i=0; i<n-1; i++) tmp += dis[p[i]+1][p[i+1]+1];
                tmp += dis[p[n-1]+1][n+1];
                ans = min(ans, tmp);
            } while (next_permutation(p.begin(), p.end()));

            return ans;
        }

        mt19937 rng(time(0));
        int ans = 1e18;
        for (int kk=0; kk<300000; kk++) {
            vector<int> ord(n+2);
            for (int i=0; i<=n+1; i++) ord[i] = i;
            shuffle(ord.begin()+1, ord.end()-1, rng);

            while (optimize(ord));
            // for (int i=0; i<=n+1; i++) cout << ord[i] << " ";
            int tmp = 0;
            for (int i=1; i<=n; i++) tmp += dis[ord[i]][ord[i+1]];
            ans = min(ans, tmp);
        }
        return ans;
    }
};

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);

    int n, q;
    cin >> n >> q;

    vector<vector<int>> dis(n+2, vector<int>(n+2, 0));
    vector<int> fow(n+1), srk(n+1);

    for (int i=1; i<=n; i++) {
        for (int j=1; j<=n; j++) dis[i][j] = 1;
    }

    while (q--) {
        int x;
        char c;
        cin >> x >> c;

        if (c == '+') {
            fow[x]++;
            for (int i=1; i<=n; i++) {
                dis[x][i] = max(dis[x][i], fow[x] - srk[i] + 1);
            }
            dis[x][n+1]++;
        }else{
            srk[x]++;
        }
    }

    // for (int i=1; i<=n+1; i++) {
    //     for (int j=1; j<=n+1; j++) {
    //         cout << dis[i][j] << " ";
    //     }
    //     cout << "\n";
    // }

    Graph G(n, dis);
    cout << G.shortest_hamilton()+1 << "\n";
}

**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
5 10
1 +
2 +
3 +
4 +
5 +
1 -
2 -
3 -
4 -
5 -
```


**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 itn long long
#define int long long
#define ff first
#define ss second
#define pb push_back
const int N=3e5+5;
int x[N];
char c[N];
int dp[(1<<21)+2][21];
int mx[21][21];
signed main(){
    int n,q;
    cin>>n>>q;
    for(int i=1;i<=q;i++){
        cin>>x[i]>>c[i];
        x[i]--;
    }
    for(int i=0;i<n;i++){
        for(itn j=0;j<n;j++){
            int L=0,R=0,l=0,r=0;
            for(int k=1;k<=q;k++){
                if(x[k]==i){
                    if(c[k]=='+'){
                        R++;
                    }
                    else{
                        L++;
                    }
                }
                else if(x[k]==j){
                    if(c[k]=='+'){
                        r++;
                    }
                    else{
                        l++;
                    }
                }
                mx[i][j]=max(mx[i][j],R-l+1);
            }
        }
    }

    for(int i=0;i<(1<<n);i++){
        for(int j=0;j<n;j++)dp[i][j]=1e9;
    }

    for(int i=0;i<n;i++)dp[(1<<i)][i]=mx[i][i];

    for(int i=1;i<(1<<n);i++){
        int k=0;
        for(int j=0;j<n;j++){
            if(i & (1<<j))k++;
        }
        if(k==1)continue;

        for(int j=0;j<n;j++){
            if(i & (1<<j)){ // suuliih
                for(int k=0;k<n;k++){
                    if(k==j or (1<<k)&i==0)continue;
                    dp[i][k]=min(dp[i][k],dp[i-(1<<k)][j]+mx[k][j]);
                }
            }
        }
    }
    int ans=1e9;
    for(int i=0;i<n;i++){
        ans=min(ans,dp[(1<<n)-1][i]);
    }
    cout<<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:
To generate a challenging test input that exposes the bug in the given code, we need to focus on edge cases that can cause issues in the algorithm. Here are some potential edge cases to consider:

1. **Snakes that do not move**: If a snake does not move at all, the algorithm should handle this correctly.
2. **Snakes that only shrink**: If a snake only shrinks, it should not run into any obstacles.
3. **Snakes that only enlarge**: If a snake only enlarges, it should not run into any obstacles.
4. **Snakes that alternate between enlarging and shrinking**: This can cause complex interactions and is a good test case for the algorithm.
5. **Snakes that are close to each other**: This can cause issues if the snakes run into each other.

Given the constraints and the need for a small-scale test input, let's create a test case with 3 snakes and 10 events that includes a mix of the above scenarios.

```plaintext
3 10
1 +
2 +
3 +
1 -
2 -
3 -
1 +
2 +
3 +
1 -
```

### Explanation:
- **Snakes 1, 2, and 3** each start at a different cell.
- **First 3 events**: All snakes enlarge.
- **Next 3 events**: All snakes shrink.
- **Next 3 events**: All snakes enlarge again.
- **Last event**: Snake 1 shrinks.

This test case includes a mix of enlarging and shrinking events, and it ensures that the snakes do not run into each other or the end of the strip. It also tests the algorithm's ability to handle multiple snakes and a sequence of events.


**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:
def solve():
    import sys

    data = sys.stdin.read().strip().split()
    n = int(data[0])
    q = int(data[1])

    # 对每条蛇 i, 收集扩张时间列表expTimes[i], 收缩时间列表shrTimes[i]
    # 全都开1-based, i取[1..n]
    expTimes = [[] for _ in range(n+1)]
    shrTimes = [[] for _ in range(n+1)]

    # 读入事件, t从1到q, 每秒只发生一次
    idx = 2
    for t in range(1, q+1):
        s_i = int(data[idx])
        sign = data[idx+1]
        idx += 2
        if sign == '+':
            expTimes[s_i].append(t)
        else:
            shrTimes[s_i].append(t)

    # 定义函数: 计算 M_{i,j} = max_{0<=t<=q}( expand_i(t)-shrink_j(t) )
    # 通过"合并 expTimes[i] 与 shrTimes[j]"完成
    def calc_M(i, j):
        # 若 i=j, 按题意可视作无效(不会拿来排在一起), 直接返回极小即可
        if i == j:
            return -10**10

        eList = expTimes[i]
        sList = shrTimes[j]
        p, q_ = 0, 0
        cntE, cntS = 0, 0
        maxDiff = cntE - cntS  # time=0时刻的值(无扩张无收缩)

        lenE = len(eList)
        lenS = len(sList)
        eNext = eList[0] if p < lenE else 10**15
        sNext = sList[0] if q_ < lenS else 10**15

        # 逐步合并
        # 因为题意"每秒只发生一次事件", 所以不会出现 eNext == sNext(在 i!=j 情况下)
        # 若真的出现, 也可同时更新 cntE,cntS
        while p < lenE or q_ < lenS:
            if eNext < sNext:
                # 下个事件是 i 的扩张
                # 先更新一次差值(事件发生前)
                if cntE - cntS > maxDiff:
                    maxDiff = cntE - cntS
                cntE += 1
                p += 1
                eNext = eList[p] if p < lenE else 10**15
            elif sNext < eNext:
                # 下个事件是 j 的收缩
                if cntE - cntS > maxDiff:
                    maxDiff = cntE - cntS
                cntS += 1
                q_ += 1
                sNext = sList[q_] if q_ < lenS else 10**15
            else:
                # 若 eNext == sNext(极罕见, 一般仅i=j或同秒双事件才会发生),
                # 这里写个并行更新逻辑, 防止死循环:
                if cntE - cntS > maxDiff:
                    maxDiff = cntE - cntS
                cntE += 1
                cntS += 1
                p += 1
                q_ += 1
                eNext = eList[p] if p < lenE else 10**15
                sNext = sList[q_] if q_ < lenS else 10**15

        # 最后可能还要再看看结尾差值
        if cntE - cntS > maxDiff:
            maxDiff = cntE - cntS

        return maxDiff

    # 预先计算并存 M[i][j]
    M = [[0]*(n+1) for _ in range(n+1)]
    for i_ in range(1, n+1):
        for j_ in range(1, n+1):
            M[i_][j_] = calc_M(i_, j_)

    # 再计算每条蛇的最大扩张次数 E[i] = max expand_i(t)
    # 直接就是 len(expTimes[i]) (总扩张次数), 因为 expand_i(t) = 在时刻t之前+当t发生的扩张数量
    # 其最大值就是蛇i总扩张次数
    E = [0]*(n+1)
    for i_ in range(1, n+1):
        E[i_] = len(expTimes[i_])

    # 接下来做位压缩DP (TSP):
    # dp[mask][last] = 已摆放集合=mask(蛇编号集合),最后(最右)那条蛇=last时,
    #                  "last的左端L_last(0)"的最小可行值
    # mx[mask][last] = 同一状态下, 已摆放蛇的区间里 "L_i(0)+ E[i]" 的最大值(即最右端)
    # 初始: 当mask只含一个蛇i时, dp[mask][i]=1 (放最左格=1)
    # 转移: dp[newMask][j] = dp[mask][last] + 1 + M[last][j]
    # 并更新 mx[newMask][j] = max(mx[mask][last], dp[newMask][j]+ E[j])

    INF = 10**15
    size = 1 << n

    dp = [[INF]*(n+1) for _ in range(size)]
    mx = [[INF]*(n+1) for _ in range(size)]

    # 初始化, 每条蛇单独摆放
    for i_ in range(1,n+1):
        msk = 1<<(i_-1)
        dp[msk][i_] = 1
        mx[msk][i_] = 1 + E[i_]

    for mask in range(size):
        dp_mask = dp[mask]
        mx_mask = mx[mask]
        # 枚举"最后"那条蛇=last
        for last in range(1,n+1):
            Lj0 = dp_mask[last]
            if Lj0 == INF:
                continue
            baseMaxR = mx_mask[last]
            Mlast = M[last]  # 减少索引
            # 给没在mask的蛇 j 做转移
            for j_ in range(1,n+1):
                bitj = 1<<(j_-1)
                if (mask & bitj):
                    continue
                newMask = mask | bitj
                # 新蛇 j 的 L_j(0) = Lj0 + 1 + M[last][j]
                Lj_new = Lj0 + 1 + Mlast[j_]
                if Lj_new < dp[newMask][j_]:
                    dp[newMask][j_] = Lj_new
                    # 更新右端
                    tmp = Lj_new + E[j_]
                    if tmp < baseMaxR:
                        tmp = baseMaxR
                    mx[newMask][j_] = tmp
                else:
                    # 若 Lj_new==dp[newMask][j_], 可能还需比较mx
                    if Lj_new == dp[newMask][j_]:
                        tmp = Lj_new + E[j_]
                        if tmp < baseMaxR:
                            tmp = baseMaxR
                        if tmp < mx[newMask][j_]:
                            mx[newMask][j_] = tmp

    # 找到mask=全体时, 所有last的最小mx即可
    fullMask = (1<<n) - 1
    ans = INF
    dp_full = dp[fullMask]
    mx_full = mx[fullMask]
    for i_ in range(1,n+1):
        if dp_full[i_] < INF:
            val = mx_full[i_]
            if val < ans:
                ans = val

    print(ans)

if __name__ == '__main__':
    solve()

**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 4
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: 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
from os import path
if(path.exists('D:\Codeing\Code Forces\input.txt')):
    sys.stdin = open('D:\Codeing\Code Forces\input.txt','r')
    sys.stdout = open('D:\Codeing\Code Forces\output.txt','w')

# For fast I/O
input = sys.stdin.readline
print = sys.stdout.write
inf = 10**10
def solve():
    n, q = map(int, input().split())
    idx = []
    char = []
    for _ in range(q):
        a,b = input().split()
        idx.append(int(a) - 1)
        char.append(1 if b == '+' else -1)

    # to compute the minimum distance from s to t
    def getDist(s, t):
        curr_pref = 0
        mn_pref = 0
        for i in range(q):
            if idx[i] == t:
                curr_pref += char[i] < 0
            if idx[i] == s:
                curr_pref -= char[i] > 0
            mn_pref = min(mn_pref, curr_pref)
        return -mn_pref + 1

    minDist = [[inf] * n for _ in range(n)]
    for i in range(n):
        for j in range(n):
            minDist[i][j] = getDist(i, j)

    deg = [0] * n
    for e in range(q):
        deg[idx[e]] += char[e] > 0

    # dp[mask][i] --> min cost to visit the set of nodes in `mask`ending at node i.
    # Transition --> cost of moving from i to j
    #     dp[nmask][j] = min(dp[nmask][j], dp[mask][i] + minDist[i][j])
    # Base case:
    # If we start at node `i`, the cost is 1.
    # Final Subproblem --> min of final cost of visiting the last node

    dp = [[inf] * n for _ in range(1 << n)]
    for i in range(n):
        dp[1 << i][i] = 1

    for mask in range(1, 1 << n):
        for i in range(n):
            if dp[mask][i] == inf:  # unreachable states
                continue
            for j in range(n):
                if (mask >> j) & 1:  # visited nodes
                    continue
                nmask = mask | (1 << j)  #  making j visited
                dp[nmask][j] = min(dp[nmask][j], dp[mask][i] + minDist[i][j])

    ans = inf
    last = dp[(1 << n) -1]
    for i in range(n):
        ans = min(ans,last[i]+deg[i])
    print(str(ans))

for __ in range(1):
    solve()

**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 3
1 +
2 +
1 -
```


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

public class Main {

    private static int n , q;
    private static int [][] moves , mnDist;
    private static int mx = (1 << 20) + 5;
    private static int [][] dp = new int[22][mx];
    private static int [] streak ;

    private static int minimumDistance(int x , int y){

        int mn = Integer.MAX_VALUE;
        int sum = 0;
        for (int i = 0; i < q; i++) {
            if (moves[i][0] == x && moves[i][1] == 1) sum--;
            if (moves[i][0] == y && moves[i][1] == -1) sum++;
            mn = Math.min(mn , sum);
        }

        return mnDist[x][y] = -mn+1;
    }

    private static boolean mnwr(int bit , int mask){
        return ((1 << bit) & mask) > 0;
    }
    private static int fullState;
    private static int solve(int last , int mask){
        if (mask == fullState) return streak[last];
       // System.out.println(Integer.toBinaryString(mask) + " " + Integer.toBinaryString(fullState));
        if (dp[last][mask] != -1) return dp[last][mask];
        int ret = Integer.MAX_VALUE;
        for (int i = 0; i < n; ++i){
            if (!mnwr(i , mask)){
                int ch = mnDist[last][i] + solve(i , mask | (1 << i));
                ret = Math.min(ret , ch);
            }
        }
        return dp[last][mask] = ret;
    }

    private static void solve() throws IOException {
        n = sc.nextInt();
        q = sc.nextInt();
        moves = new int[q][2];
        mnDist = new int[n][n];
        streak = new int[20];

        fullState = (1 << n)-1;
        for (int i = 0; i <= n; ++i) for (int j = 0; j <= fullState+2; ++j) {
            dp[i][j] = -1;
        }

        for (int i = 0; i < q; ++i){
            moves[i][0] = sc.nextInt() - 1;
            moves[i][1] = (sc.next().charAt(0) == '+' ? 1 : -1);
            streak[moves[i][0]]+= moves[i][1] > 0 ? 1 : 0;
        }
        for (int i = 0; i < n; ++i){
            for (int j = 0; j < n; ++j){
                if (j != i) minimumDistance(i , j);
            }
        }

        int ans = Integer.MAX_VALUE;
        for (int i = 0; i < n; ++i){
            ans = Math.min(ans ,1+ solve(i , (1 << i)));
        }
        pw.println(ans);
        pw.flush();

    }

    public static void main(String[] args) throws IOException {

        int tc = 1;
        //   tc = sc.nextInt();
        while (tc-- > 0) {
            solve();
        }

    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public Scanner(String s) throws FileNotFoundException {
            br = new BufferedReader(new FileReader(s));
        }

        public String next() throws IOException {
            while (st == null|| !st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        public int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        public long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

        public boolean ready() throws IOException {
            return br.ready();
        }
    }
    static Scanner sc = new Scanner(System.in);
    static PrintWriter pw = new PrintWriter(System.out);

    static class segTree{
        private int n;
        private int[] tree;
        segTree(int n){
            this.n = n + 5;
            while (Integer.bitCount(this.n) != 1) {
                this.n++;
            }
            tree = new int[2 * this.n];

        }

        private int merge(int u , int  v){
            return u + v;
        }

        private void upd (int node , int nl , int nr , int idx , int v ){
            if (nl + 1 == nr){
                tree[node] += v;
                return;
            }
            int md = (nl + nr) >> 1;
            if (idx < md) upd(2 * node + 1 , nl , md , idx, v);
            else upd(2* node + 2 , md, nr, idx, v);
            tree[node] = merge (tree[2* node + 1] , tree[2* node + 2]);
        }

        private int query (int node , int nl , int nr , int ql , int qr){
            if (nl >= ql && qr >= nr) return tree[node];
            int md = (nl + nr ) >> 1;
            if (qr <= md) return query(2* node + 1 , nl , md , ql ,qr);
            if (ql >= md) return query (2* node + 2, md, nr , ql ,qr);
            return merge(query (2* node + 1 , nl , md , ql ,qr), query (2 * node + 2, md, nr , ql , qr));
        }

        public void upd (int idx , int v ){
            upd(0 , 0 , n , idx , v);
        }
        public int  query (int L , int R){
            return query (0 ,0 , n , L , R+ 1);
        }
    }

    public static int existsMultiple(int n, int d) {
        int mod = 9 * n;  // Effective modulus
        int remainder = 0;  // Represents (10^k - 1) % (9n)
        int powerOfTen = 1;  // Represents 10^k % (9n)
        HashSet<Integer> seen = new HashSet<>();  // To detect cycles
        int k = 0;
        while (true) {
            ++k;
            remainder = (remainder + powerOfTen) % mod;
            if (d * remainder  % n == 0) {
                return k; // Found a valid k
            }
            if (seen.contains(remainder)) {
                break;  // Cycle detected, no valid k
            }
            seen.add(remainder);
            powerOfTen = (powerOfTen * 10) % mod;
        }
        return -1;
    }

    private static class Multiset <T extends Number , M extends Number & Comparable<M>>{
        private TreeMap<T, M> map;
        private M inc;
        public Multiset(M inc){
            map = new TreeMap<>();
            this.inc = inc;

        }

        void add(T i){
            map.put(i, map.containsKey(i) ? addNumbers( inc, map.get(i)) : inc);
        }

        void remove(T i){
            M freq = map.get(i);
            if (freq.equals(1)) map.remove(i);
            else map.put(i , subNumbers(freq , inc));
        }

        T getMax(){
            return map.lastKey();
        }

        boolean isEmpty(){
            return map.isEmpty();
        }

        Optional<T> lowerKey(T v){
            return Optional.ofNullable(map.lowerKey(v));
        }

        private M addNumbers(M a , M b){
            return castNumber(a.intValue() + b.intValue());
        }
        private M subNumbers(M a , M b){
            return castNumber(a.intValue() - b.intValue());
        }
        private M castNumber(int n){
            return (M) Integer.valueOf(n);
        }

    }

    public static long mul(long a, long b, long m) {
        return ((a % m) * (b % m)) % m;
    }

    public static long add(long a, long b, long m) {
        return ((a + b) % m + m) % m;
    }

    public static long fp(long b, long p, long m) {
        if (p == 0) return 1;
        long v = fp(b, p >> 1, m);
        v = mul(v, v, m);
        if ((p & 1) == 1) v = mul(v, b, m);
        return v;
    }

    public static long modInv(long n, long m) {
        return fp(n, m - 2, m);
    }

}

**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 given buggy code, we need to focus on edge cases and scenarios that might cause the algorithm to fail. Here are the steps to think through this:

1. **Small Number of Snakes and Events**: Start with a small number of snakes and events to ensure the test input is manageable and complete.

2. **Edge Cases**:
   - **Single Snake**: Test with a single snake to see if the algorithm handles it correctly.
   - **Multiple Snakes**: Test with multiple snakes to see if the algorithm can handle interactions between snakes.
   - **Shrinking and Enlarging**: Ensure the test includes both types of events to check if the algorithm correctly handles these transitions.
   - **Snakes at the Boundaries**: Test with snakes starting at the edges of the strip to see if the algorithm handles boundary conditions correctly.

3. **Interactions**:
   - **Overlapping Snakes**: Test with snakes that might overlap or come close to each other to see if the algorithm can handle potential collisions.

4. **Valid Sequence**: Ensure the sequence of events is valid, meaning a snake of length 1 never shrinks.

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

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

This test input:
- Has 3 snakes and 6 events.
- Includes both enlarging and shrinking events.
- Ensures that the snakes do not overlap or collide, but the sequence is designed to check the algorithm's ability to handle these events correctly.

This input should be sufficient to expose any issues in the buggy code.


**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 java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
public class main {
    static PrintWriter pw = new PrintWriter(new BufferedOutputStream(System.out));
    static FastReader in = new FastReader();
    public static int n;
    public static int[][] dis;
    public static int[] max_dis;
    public static Integer[][] dp;
    public static void main(String[] args) {
        int t = 1;
        while (t-- > 0) {
            n = in.nextInt();
            int q = in.nextInt();
            dis = new int[n][n];
            max_dis = new int[n];
            for (int i = 0; i < dis.length; i++) {
                Arrays.fill(dis[i], 1);
            }
            ArrayList<int[]> queries = new ArrayList<>();
            for (int i = 0; i < q; i++) {
                int l = in.nextInt() - 1;
                int r = in.next().charAt(0) == '+' ? 1 : 0;
                queries.add(new int[]{l, r});
            }
            for (int i = 0; i < q; i++) {
                int l = queries.get(i)[0];
                int r = queries.get(i)[1];
                if (r == 1) {
                    max_dis[l]++;
                }
            }
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    if (i != j) {
                        int count = 1;
                        for (int j2 = 0; j2 < q; j2++) {
                            int l = queries.get(j2)[0];
                            int r = queries.get(j2)[1];
                            if(l != i && l != j) continue;
                            if(l == i && r == 1) count--;
                            if(l == j && r == 0) count++;
                            if(count == 0) {
                                count++;
                                dis[i][j]++;
                            }
                        }
                    }
                }
            }
            int max = (1 << n) - 1;
            dp = new Integer[max + 1][n + 1];
            int ans = (int) 1e9;
            for (int i = 0; i < n; i++) {
                ans = Math.min(ans, 1 + solve((1 << i), max, i));
            }
            pw.println(ans);
        }
        pw.close();
    }
    public static int solve(int mask, int max, int last) {
        if (mask == max) {
            return max_dis[last];
        }
        if (dp[mask][last] != null) {
            return dp[mask][last];
        }
        int ans = (int) 1e9;
        for (int i = 0; i < n; i++) {
            int l = (1 << i);
            if((mask & l) == 0) {
                ans = Math.min(ans, dis[last][i] + solve(mask + l, max, i));
            }
        }
        return dp[mask][last] = ans;
    }
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}

**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 are likely to cause issues. The key points to consider are:

1. **Initial Placement of Snakes**: Ensure that the initial placement of snakes can lead to conflicts or overlaps when the events are processed.
2. **Event Sequence**: Design a sequence of events that can cause the snakes to interact in complex ways, potentially leading to incorrect calculations of distances and maximum positions.
3. **Edge Cases**: Consider edge cases such as snakes at the boundaries of the strip or snakes that shrink and enlarge multiple times.

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

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

### Explanation:
- **Initial Setup**: We have 3 snakes and 6 events.
- **Events**:
  - Snake 1 enlarges, then shrinks.
  - Snake 2 enlarges, then shrinks.
  - Snake 3 enlarges, then shrinks.

### Why This Test Input is Challenging:
1. **Initial Placement**: The snakes are placed in such a way that they can potentially overlap if not managed correctly.
2. **Event Sequence**: The sequence of enlarging and shrinking for each snake can cause the distances and maximum positions to be calculated incorrectly if the code does not handle these transitions properly.
3. **Edge Cases**: The events are symmetric and simple, but they can expose issues in the distance calculation and the dynamic programming approach used in the buggy code.

This test input should help in identifying any issues in the buggy code, particularly in how it handles the sequence of events and the distances between snakes.