**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Showering
time_limit_ms: 2000
memory_limit_mb: 256
Description: As a computer science student, Alex faces a hard challenge — showering. He tries to shower daily, but despite his best efforts there are always challenges. He takes $s$ minutes to shower and a day only has $m$ minutes!

He already has $n$ tasks planned for the day. Task $i$ is represented as an interval $(l_i$, $r_i)$, which means that Alex is busy and can not take a shower in that time interval (at any point in time strictly between $l_i$ and $r_i$). No two tasks overlap.

Given all $n$ time intervals, will Alex be able to shower that day? In other words, will Alex have a free time interval of length at least $s$?

![](CDN_BASE_URL/5d5195053b99e5c6936ccefadc239679)

In the first test case, Alex can shower for the first $3$ minutes of the day and not miss any of the tasks.

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

The first line of each test case contains three integers $n$, $s$, and $m$ ($1 \leq n \leq 2 \cdot 10^5$; $1 \leq s, m \leq 10^9$) — the number of time intervals Alex already has planned, the amount of time Alex takes to take a shower, and the amount of minutes a day has.

Then $n$ lines follow, the $i$-th of which contains two integers $l_i$ and $r_i$ ($0 \leq l_i < r_i \leq m$) — the time interval of the $i$-th task. No two tasks overlap.

Additional constraint on the input: $l_i > r_{i-1}$ for every $i > 1$.

The sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case output "YES" (without quotes) if Alex can take a shower for that given test case, and "NO" (also without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yEs", "yes", and "Yes" will be recognized as a positive response).

Examples:
input:
4
3 3 10
3 5
6 8
9 10
3 3 10
1 2
3 5
6 7
3 3 10
1 2
3 5
6 8
3 4 10
1 2
6 7
8 9
output:
YES
YES
NO
YES

Note:

Buggy Code:
t = int(input())
for _ in range(t):
    n, s, m = list(map(int, input().split()))
    intervals = []
    for i in range(n):
        intervals.append(list(map(int,input().split())))
    for i in range(n):
        if i == 0 and intervals[0][0] - 0 == s:
            print("YES")
            break
        if i == n - 1 and m - intervals[i][1]  == s:
            print("YES")
            break
        if i == n - 1 and m - intervals[i][1]  != s:
            print("NO")
            break
        if intervals[i + 1][0] - intervals[i][1] == s:
            print("YES")
            break

**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 given buggy code, we need to focus on edge cases where the code might fail to correctly determine if Alex can take a shower. The key issues in the code are:

1. It checks for exact equality with `s` for the free intervals, which might miss cases where the free interval is greater than `s`.
2. It does not correctly handle the case where the free interval at the beginning or end of the day is greater than `s`.

Let's construct a test case that highlights these issues:

- **Case 1:** Free interval at the beginning of the day is greater than `s`.
- **Case 2:** Free interval at the end of the day is greater than `s`.
- **Case 3:** Free interval between two tasks is greater than `s`.

Here is a small-scale test input that should expose these issues:

```plaintext
3
3 3 10
0 2
4 6
7 9
3 3 10
1 2
4 6
7 9
3 4 10
1 2
4 6
7 9
```

### Explanation:
1. **First Test Case:**
   - Free interval at the beginning: 0 to 2 (2 minutes)
   - Free interval between tasks: 2 to 4 (2 minutes)
   - Free interval between tasks: 6 to 7 (1 minute)
   - Free interval at the end: 9 to 10 (1 minute)
   - Alex can take a 3-minute shower in the first free interval (0 to 3).

2. **Second Test Case:**
   - Free interval at the beginning: 0 to 1 (1 minute)
   - Free interval between tasks: 2 to 4 (2 minutes)
   - Free interval between tasks: 6 to 7 (1 minute)
   - Free interval at the end: 9 to 10 (1 minute)
   - Alex can take a 3-minute shower in the second free interval (2 to 5).

3. **Third Test Case:**
   - Free interval at the beginning: 0 to 1 (1 minute)
   - Free interval between tasks: 2 to 4 (2 minutes)
   - Free interval between tasks: 6 to 7 (1 minute)
   - Free interval at the end: 9 to 10 (1 minute)
   - Alex can take a 4-minute shower in the second free interval (2 to 6).

This test input should expose the bug in the code by providing free intervals that are greater than `s`, which the code might fail to recognize correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Showering
time_limit_ms: 2000
memory_limit_mb: 256
Description: As a computer science student, Alex faces a hard challenge — showering. He tries to shower daily, but despite his best efforts there are always challenges. He takes $s$ minutes to shower and a day only has $m$ minutes!

He already has $n$ tasks planned for the day. Task $i$ is represented as an interval $(l_i$, $r_i)$, which means that Alex is busy and can not take a shower in that time interval (at any point in time strictly between $l_i$ and $r_i$). No two tasks overlap.

Given all $n$ time intervals, will Alex be able to shower that day? In other words, will Alex have a free time interval of length at least $s$?

![](CDN_BASE_URL/5d5195053b99e5c6936ccefadc239679)

In the first test case, Alex can shower for the first $3$ minutes of the day and not miss any of the tasks.

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

The first line of each test case contains three integers $n$, $s$, and $m$ ($1 \leq n \leq 2 \cdot 10^5$; $1 \leq s, m \leq 10^9$) — the number of time intervals Alex already has planned, the amount of time Alex takes to take a shower, and the amount of minutes a day has.

Then $n$ lines follow, the $i$-th of which contains two integers $l_i$ and $r_i$ ($0 \leq l_i < r_i \leq m$) — the time interval of the $i$-th task. No two tasks overlap.

Additional constraint on the input: $l_i > r_{i-1}$ for every $i > 1$.

The sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case output "YES" (without quotes) if Alex can take a shower for that given test case, and "NO" (also without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yEs", "yes", and "Yes" will be recognized as a positive response).

Examples:
input:
4
3 3 10
3 5
6 8
9 10
3 3 10
1 2
3 5
6 7
3 3 10
1 2
3 5
6 8
3 4 10
1 2
6 7
8 9
output:
YES
YES
NO
YES

Note:

Buggy Code:
days = int(input())
for i in range(days):
    k = 0
    first = str(input()).split()
    quant = int(first[0])
    show = int(first[1])
    daytime = int(first[2])
    deals = []
    time = []
    for tm in range(daytime + 1):
        time.append(0)
    for x in range(quant):
        deals.append(str(input()).split())
    for deal in deals:
        time[int(deal[0])] = int(deal[0])
        time[int(deal[1])] = int(deal[1])
    if time[-1] == 0:
        time.pop(-1)
    for s in time:
        if s == 0:
            k += 1
        if k >= show:
            print('YES')
            break
        if s != 0:
            k = 1
    if k < show:
        print('NO')

**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 maximize the failure probability. Here are the steps to think through this:

1. **Understand the Bug**: The buggy code initializes a `time` array with zeros and then marks the start and end points of each task. It then iterates through this array to count consecutive zeros (free time). The main issue is that it only marks the start and end points of the tasks, not the entire duration, and it resets the counter `k` to 1 instead of 0 when it encounters a non-zero value.

2. **Edge Cases**:
   - **Single Task**: Test cases where there is only one task.
   - **Multiple Tasks**: Test cases with multiple tasks that are close to each other.
   - **Task at the Beginning/End**: Tasks that start or end at the very beginning or end of the day.
   - **Task with Gaps**: Tasks that leave small gaps between them.
   - **Task that Fills Most of the Day**: Tasks that almost fill the entire day, leaving a small gap.

3. **Generate Test Input**:
   - **Single Task at the Beginning**: This should make it easy to find a long free interval.
   - **Single Task at the End**: This should also make it easy to find a long free interval.
   - **Multiple Tasks with Small Gaps**: This should make it harder to find a long free interval.
   - **Tasks that Almost Fill the Day**: This should make it very hard to find a long free interval.

Based on these considerations, here is a small-scale test input that should expose the bug:

```plaintext
4
1 3 10
1 4
1 3 10
6 9
3 3 10
1 4
6 9
8 10
3 4 10
1 3
6 8
9 10
```

### Explanation:
1. **First Test Case**:
   - One task from 1 to 4.
   - Free intervals: [0, 1] and [4, 10].
   - Alex can shower for 3 minutes in the interval [4, 10].

2. **Second Test Case**:
   - One task from 6 to 9.
   - Free intervals: [0, 6] and [9, 10].
   - Alex can shower for 3 minutes in the interval [0, 6].

3. **Third Test Case**:
   - Three tasks: [1, 4], [6, 9], and [8, 10].
   - Free intervals: [0, 1] and [4, 6].
   - Alex cannot shower for 3 minutes in any of these intervals.

4. **Fourth Test Case**:
   - Three tasks: [1, 3], [6, 8], and [9, 10].
   - Free intervals: [0, 1], [3, 6], and [8, 9].
   - Alex can shower for 4 minutes in the interval [3, 6].

This test input should help identify the issues with the buggy code, particularly the incorrect handling of task durations and the reset of the counter.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Showering
time_limit_ms: 2000
memory_limit_mb: 256
Description: As a computer science student, Alex faces a hard challenge — showering. He tries to shower daily, but despite his best efforts there are always challenges. He takes $s$ minutes to shower and a day only has $m$ minutes!

He already has $n$ tasks planned for the day. Task $i$ is represented as an interval $(l_i$, $r_i)$, which means that Alex is busy and can not take a shower in that time interval (at any point in time strictly between $l_i$ and $r_i$). No two tasks overlap.

Given all $n$ time intervals, will Alex be able to shower that day? In other words, will Alex have a free time interval of length at least $s$?

![](CDN_BASE_URL/5d5195053b99e5c6936ccefadc239679)

In the first test case, Alex can shower for the first $3$ minutes of the day and not miss any of the tasks.

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

The first line of each test case contains three integers $n$, $s$, and $m$ ($1 \leq n \leq 2 \cdot 10^5$; $1 \leq s, m \leq 10^9$) — the number of time intervals Alex already has planned, the amount of time Alex takes to take a shower, and the amount of minutes a day has.

Then $n$ lines follow, the $i$-th of which contains two integers $l_i$ and $r_i$ ($0 \leq l_i < r_i \leq m$) — the time interval of the $i$-th task. No two tasks overlap.

Additional constraint on the input: $l_i > r_{i-1}$ for every $i > 1$.

The sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case output "YES" (without quotes) if Alex can take a shower for that given test case, and "NO" (also without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yEs", "yes", and "Yes" will be recognized as a positive response).

Examples:
input:
4
3 3 10
3 5
6 8
9 10
3 3 10
1 2
3 5
6 7
3 3 10
1 2
3 5
6 8
3 4 10
1 2
6 7
8 9
output:
YES
YES
NO
YES

Note:

Buggy Code:
#include<bits/stdc++.h>
#define ll long long
#define lli long long int

#define mxt(a) *max_element(a.begin(),a.end())
#define mnt(a) *min_element(a.begin(),a.end())

#define mxti(a,i,j) *max_element(a.begin()+i+1,a.begin()+j+1)
#define mnti(a,i,j) *min_element(a.begin()+i+1,a.begin()+j+1)

#define acm(v,i,j) accumulate(v.begin()+i+1,v.begin()+j+1,0ll)

//extract sub vector from main vector
//vector<int> sub_vector(main_vector.begin()+start_index,main_vector.begin()+end_index)

#define srt(a) sort(a.begin(),a.end())
#define rev(a) reverse(a.begin(),a.end())

#define sp " "
#define nl '\n'
#define op(a,n) for(ll i=0;i<n;i++)cout<<a[i]<<" "
#define in(a,n) for(ll i=0;i<n;i++)cin>>a[i]

#define mii map<int,int>
#define mis map<int,string>
#define msi map<string,int>
#define mci map<char,int>
#define mic map<int,char>

// pairs
#define pii  pair<int, int>
#define pll  pair<ll, ll>
#define pd  pair<db, db>
#define mkp make_pair
#define f first
#define s second

#define vi vector<int>
#define vll vector<ll>
#define vpi vector<pi>
#define vs vector<string>
#define vc vector<char>
#define vvi vector<vector<int>>
#define vvll vector<vector<ll>>
#define vvs vector<vector<string>>
#define vvc vector<vector<char>>
#define pb push_back

#define fre(x, a) for (auto &x : a)
#define fr(a,c,b) for(ll a=c;a<b;a++)

const int MOD = 998244353;  // 1e9+7;
const int MX = (int)2e5 + 5;
const ll BIG = 1e18;  // not too close to LLONG_MAX

#define PYES cout<<"YES\n"
#define PNO cout<<"NO\n"
#define PYes cout<<"Yes\n"
#define PNo cout<<"No\n"

#define fastIO() ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);

using namespace std;

//--------------------------------------------functions-------------------------------------------------//

ll power(ll a,ll b){
    ll result=1;
    while(b>0){
        if(b%2 == 1){
            result *= a;
        }
        a *= a;
        b /= 2;
    }
    return result;
}
ll gcd(ll x,ll y){
    ll r;
    while(y!=0&&(r=x%y)!=0){
        x=y;y=r;
    }
    return y==0?x:y;
}
ll countSetBits(ll x){ll Count=0;while(x>0){if(x&1) Count++;x=x>>1;}return Count;}
ll mod(ll x,ll M){return ((x%M + M)%M);}
ll add(ll a, ll b,ll M){return mod(mod(a,M)+mod(b,M),M);}
ll mul(ll a, ll b,ll M){return mod(mod(a,M)*mod(b,M),M);}
ll powerM(ll a,ll b,ll M){
    ll res=1ll;
    while(b){
        if(b%2ll==1ll){
            res=(a*res)%M;
        }
        a=(a*a)%M;b/=2ll;
    }
    return res;
}
ll mod_inv(ll a, ll m) {
    ll g = m, r = a, x = 0, y = 1;
    while (r != 0) {
        ll q = g / r;
        g %= r; swap(g, r);
        x -= q * y; swap(x, y);
    }
    return (x%m + m)%m;
}

////------------------------------------sieve of Eratosthenes-----------------------------------------------//
//ll MAXN=3e5;
//vector<ll> fact(MAXN+1);
//void sieve_of_erantothenes()
//{
//    fact[1] = 1;
//    for (ll i = 2; i <= MAXN; i++)
//        fact[i] = i;
//    for (ll i = 4; i <= MAXN; i += 2)
//        fact[i] = 2;
//    for (ll i = 3; i * i <= MAXN; i++){
//        if (fact[i] == i){
//            for (ll j = i * i; j <= MAXN; j += i)
//                if (fact[j] == j)
//                    fact[j] = i;
//        }
//    }
//}
//----------------------------------------nCr mod------------------------------------------------------//
ll nCr(ll n, ll k){
    if(n<k) return 0;
    if(k==0) return 1;
    ll res = 1;
    if (k > n - k) k = n - k;
    for(ll i = 0; i < k; ++i){
        res *= (n - i);
        res /= (i + 1);
    }
    return res;
}

vector<ll> FA,INV;
void fcalc(ll n,ll M){
    FA.resize(n+1);
    FA[0]=1;
    for(ll i=1;i<=n;i++){
        FA[i]=(FA[i-1]*i)%M;
    }
}

void icalc(ll n,ll M){
    INV.resize(n+1);
    INV[n]=mod_inv(FA[n],M);
    for(ll i=n-1;i>=0;i--){
        INV[i]=(INV[i+1]*(i+1))%M;
    }
}
ll nCrM(ll n,ll r,ll M){
    if(r<0) return 0;
    if(n<r) return 0;
    if(r==0) return 1;
    ll a=FA[n];
    a*=INV[r];
    a%=M;
    a*=INV[n-r];
    a%=M;
    return a;
}
// ll rand_int(ll l, ll r){
//     return uniform_int_distribution<ll>(l, r)(rng);
// }

ifstream fin;
ofstream fout;

template <typename T>
ostream& operator <<(ostream& output, const vector<T>& data)
{
    for (const T& x : data)
        output << x <<" ";
    return output;
}

template<typename T>
istream& operator>>(istream& input,vector<T>& data){
    for (auto& item : data) {
        input >> item;
    }
    return input;
}

ll LCM(ll a, ll b){
    return ((a * b) / gcd(a, b));
}

int ceil(int a, int b){// a / b
    return a / b + ((a ^ b) > 0 && a % b);
}

//--------------------------------------------solve----------------------------------------------//
ll fact(ll n)
{
    if (n == 0 || n == 1)
        return 1;
    return n * fact(n - 1);
}
int LSBi(int x){
    int last=0;
    for(int i=0;i<32;i++){
        if(((x>>i)&1)==1)last=i;
    }
    return last; //returns last set bit index

}
ll maxSubArraySum(vll &a, ll size)
{
    ll max_so_far = INT_MIN, max_ending_here = 0;

    for (ll i = 0; i < size; i++) {
        max_ending_here = max_ending_here + a[i];
        if (max_so_far < max_ending_here)
            max_so_far = max_ending_here;

        if (max_ending_here < 0)
            max_ending_here = 0;
    }
    return max_so_far;
}

//  int dx[4] = {-1, 0, 1, 0};
//  int dy[4] = {0, -1, 0, 1}; // for every grid problem!!

// bool inp(int i, int j, int n, int m) { // FN TO CHECK IF GIVEN index i,j inside grid
//     return (0 <= i && i < n && 0 <= j && j < m);
// }

void solve() {
    ll t;
    cin >> t;
    while (t--) {
        ll n, s, m;
        cin >> n >> s >> m;
        vll a(2 * n);
        in(a, 2 * n);
        vll b(m, 1); // initialize b with 1s

        for (ll j = 0; j < 2 * n; j += 2) {
            ll d = a[j];
            ll e = a[j + 1];
            for (ll k = d; k < e; k++) {
                if (k < m) b[k] = 0; // ensure k is within bounds of b
            }
        }

        ll ans = 0;
        ll mn = 0;
        ll sum = 0;
        fr(i, 0, m) {
            if (b[i] == 0) {
                mn = 0;
                sum = 0;
            } else {
                sum += b[i];
                ans = max(ans, sum - mn);
                mn = min(mn, sum);
            }
        }

        if (ans >= s) cout << "YES" << nl;
        else cout << "NO" << nl;
    }
}

int main() {
    fastIO();
    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
1
1 10 10
0 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Showering
time_limit_ms: 2000
memory_limit_mb: 256
Description: As a computer science student, Alex faces a hard challenge — showering. He tries to shower daily, but despite his best efforts there are always challenges. He takes $s$ minutes to shower and a day only has $m$ minutes!

He already has $n$ tasks planned for the day. Task $i$ is represented as an interval $(l_i$, $r_i)$, which means that Alex is busy and can not take a shower in that time interval (at any point in time strictly between $l_i$ and $r_i$). No two tasks overlap.

Given all $n$ time intervals, will Alex be able to shower that day? In other words, will Alex have a free time interval of length at least $s$?

![](CDN_BASE_URL/5d5195053b99e5c6936ccefadc239679)

In the first test case, Alex can shower for the first $3$ minutes of the day and not miss any of the tasks.

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

The first line of each test case contains three integers $n$, $s$, and $m$ ($1 \leq n \leq 2 \cdot 10^5$; $1 \leq s, m \leq 10^9$) — the number of time intervals Alex already has planned, the amount of time Alex takes to take a shower, and the amount of minutes a day has.

Then $n$ lines follow, the $i$-th of which contains two integers $l_i$ and $r_i$ ($0 \leq l_i < r_i \leq m$) — the time interval of the $i$-th task. No two tasks overlap.

Additional constraint on the input: $l_i > r_{i-1}$ for every $i > 1$.

The sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case output "YES" (without quotes) if Alex can take a shower for that given test case, and "NO" (also without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yEs", "yes", and "Yes" will be recognized as a positive response).

Examples:
input:
4
3 3 10
3 5
6 8
9 10
3 3 10
1 2
3 5
6 7
3 3 10
1 2
3 5
6 8
3 4 10
1 2
6 7
8 9
output:
YES
YES
NO
YES

Note:

Buggy Code:
#include <bits/stdc++.h>

#define ll long long
#define el endl
#define UwU ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(0);

using namespace std;

ll sum;
vector<ll> v;

//
//int cnt;
//map<ll,bool> vis;
//map<ll,vector<ll>> adj;
//ll farnode = 1;
//ll maxdepth = 0;
//vector<ll> path;

//void dfs(ll node, ll dep) {
//    vis[node] = 1;
//    if (dep > maxdepth) {
//        maxdepth = dep;
//        farnode = node;
//    }
//    path.push_back(node);
//
//    for (auto i: adj[node]) {
//        if (vis[i] == 0) {
//            dfs(i, dep + 1);
//        }
//    }
//
//}
//void dfs(ll node){
//    vis[node] = 1;
//    path.push_back(node);
//
//    for(auto i:adj[node]){
//        if(vis[i]==0){
//            dfs(i);
//        }
//    }
//}

void x(int num){
    int y = num;
    if(num == 1){
        return ;
    }

    x(num-1);
    cout<<num<<" ";
}

  struct TreeNode {
      int val;
      TreeNode *left;
      TreeNode *right;
      TreeNode() : val(0), left(nullptr), right(nullptr) {}
      TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
      TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
  };
int counter=0;
void averageOfSubtreehelp(TreeNode* root,int sum,int nodesnumber) {
    if (!root){
        return;
    }

    if (root->left) {
        averageOfSubtreehelp(root->left, sum, nodesnumber);
    }
    if (root->right) {
        averageOfSubtreehelp(root->right, sum, nodesnumber);
    }
    sum += root->val;
    nodesnumber += 1;
    if(sum/nodesnumber==root->val){
        counter++;
    }
}

int averageOfSubtree(TreeNode* root) {
    averageOfSubtreehelp(root,0,0);
    return counter;
}

#include <iostream>
#include <vector>
#include <fstream>

using namespace std;

bool vis[10005];
vector<int> adj[10005];
vector<vector<int>> allPaths; // To store all paths

void dfs(int node, vector<int>& path, int n) {
    vis[node] = true;
    path.push_back(node); // Add current node to the path

    if (node == n - 1) { // If the current node is the destination node, add the path to allPaths
        allPaths.push_back(path);
        path.pop_back();
        for (int i = 0; i < 10005; i++) {
            vis[i] = 0;
        }
        return;
    }

    for (auto i : adj[node]) {
        if (!vis[i]) {
            dfs(i, path, n);
        }
    }

    // Backtrack: remove the current node from the path to explore other paths
    path.pop_back();
}
void swapElements(std::vector<int>& vec, size_t l, size_t r) {
    if (l < vec.size() && r < vec.size()) {
        std::swap(vec[l], vec[r]);
    } else {
        std::cerr << "Error: Index out of bounds." << std::endl;
    }
}
bool hasConsecutiveValues(const map<int, int>& mp, int value, int count) {
    for(auto it = mp.begin(); it != mp.end(); ++it) {
        auto currentIt = it;
        bool allEqual = true;

        for(int i = 0; i < count; ++i) {
            if (currentIt == mp.end() || currentIt->second != value) {
                allEqual = false;
                break;
            }
            ++currentIt;
        }

        if (allEqual) {
            return true;
        }
    }

    return false;
}
int main() {

#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif

//    int n, m;
//    cin >> n >> m; // Assuming n is the number of nodes
//
//    for (int i = 0; i < m; i++) {
//        int x, y;
//        cin >> x >> y;
//        adj[x].push_back(y);
//    }
//
//    vector<int> path;
//    dfs(0, path, n); // Start DFS from node 0 to find all paths to node n - 1
//
//    // Output all paths
//    for (const auto& path : allPaths) {
//        for (int node : path) {
//            cout << node << " ";
//        }
//        cout << endl;
//    }
int t;
cin>>t;
    while (t--){
        int n,s,m;
        cin>>n>>s>>m;
        int l,r;
        map<int,int>mp;
        for (int i = 0; i < m; ++i) {
            mp[i] = 0;
        }

        while (n--) {
            cin >> l >> r;
            m = m - (r - l);
            for (int i = l; i < r; i++) {
                mp[i]++;
            }
        }

        if (hasConsecutiveValues(mp, 0, s)) {
            cout << "YES" << endl;
        } else {
            cout << "NO" << endl;
        }
    }

//    int x1,x2,y1,y2;
//    cin>>x1>>x2>>y1>>y2;
//    int cnt=0;
//
//    int suneet_wins = 0, slavic_wins = 0;
//    if (x1 > y1) suneet_wins++; else if (x1 < y1) slavic_wins++;
//    if (x2 > y2) suneet_wins++; else if (x2 < y2) slavic_wins++;
//    if (suneet_wins > slavic_wins) cnt++;
//
//    suneet_wins = 0; slavic_wins = 0;
//    if (x1 > y2) suneet_wins++; else if (x1 < y2) slavic_wins++;
//    if (x2 > y1) suneet_wins++; else if (x2 < y1) slavic_wins++;
//    if (suneet_wins > slavic_wins) cnt++;
//
//    suneet_wins = 0; slavic_wins = 0;
//    if (x2 > y1) suneet_wins++; else if (x2 < y1) slavic_wins++;
//    if (x1 > y2) suneet_wins++; else if (x1 < y2) slavic_wins++;
//    if (suneet_wins > slavic_wins) cnt++;
//
//    suneet_wins = 0; slavic_wins = 0;
//    if (x2 > y2) suneet_wins++; else if (x2 < y2) slavic_wins++;
//    if (x1 > y1) suneet_wins++; else if (x1 < y1) slavic_wins++;
//    if (suneet_wins > slavic_wins) cnt++;
//
//    cout << cnt << endl;

//    TreeNode
////    ll n, m;
////    cin >> m;
////    ll sta;
////    for (ll i = 0; i < m; i++) {
////
////        ll x, y;
////        cin >> x >> y;
////        if(i == 0){
////        sta = x;}
////        adj[x].push_back(y);
////        adj[y].push_back(x);
////
////    }
////    dfs(sta,0);
////
////
////    vis.clear();
////    path.clear();
////    dfs(farnode, 0);
//////    cout << maxdepth;
////    for (int i = 0; i < path.size(); i++) {
////        cout << path[i];
////        if (i != path.size() - 1) {
////            cout << " ";
////        }
////    }
//
//    int t;
//    cin >> t;
//
//    while(t--) {
//        string s;
//        cin >> s;
//
//        vector<int> ve(s.size());
//        for(int i = 0; i < s.size(); i++) {
//            ve[i] = s[i] - 'a';
//        }
//
//        sort(ve.begin(), ve.end());
//
//        bool boo = true;
//        for(int i = 1; i < ve.size(); i++) {
//            if(ve[i] != ve[i-1] + 1) {
//                boo = false;
//                break;
//            }
//        }
//
//        if(boo) {
//            cout << "YES";
//        } else {
//            cout << "NO";
//        }
//        cout << 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 buggy code, we need to focus on edge cases and scenarios that might cause the code to fail. The buggy code uses a map to track the availability of time slots and checks for consecutive free slots of length `s`. Here are some key points to consider:

1. **Edge Cases**:
   - When `s` is equal to the total available time `m`.
   - When `s` is very small compared to `m`.
   - When there are no tasks at all.
   - When tasks are very close to each other, leaving very small gaps.

2. **Scenarios to Test**:
   - A single task that covers almost the entire day, leaving a small gap.
   - Multiple tasks that cover the entire day with no gaps.
   - Multiple tasks with small gaps between them.
   - No tasks at all.

Given these considerations, let's create a small-scale test input that is complete and valid:

```plaintext
4
1 9 10
1 1
3 1 10
1 2
2 3
3 4
2 2 10
1 5
6 10
1 10 10
1 10
```

### Explanation:
1. **Test Case 1**:
   - `n = 1`, `s = 9`, `m = 10`
   - Task: (1, 1)
   - There is a single task that takes 1 minute, leaving 9 minutes available for the shower. This should return "YES".

2. **Test Case 2**:
   - `n = 3`, `s = 1`, `m = 10`
   - Tasks: (1, 2), (2, 3), (3, 4)
   - There are three tasks, each taking 1 minute, with no gaps between them. This should return "NO" because there is no single free minute.

3. **Test Case 3**:
   - `n = 2`, `s = 2`, `m = 10`
   - Tasks: (1, 5), (6, 10)
   - There are two tasks, each taking 4 minutes, with a 1-minute gap between them. This should return "NO" because there is no 2-minute free interval.

4. **Test Case 4**:
   - `n = 1`, `s = 10`, `m = 10`
   - Task: (1, 10)
   - There is a single task that covers the entire day. This should return "NO" because there is no free time.

This test input covers a range of edge cases and should help expose the bug in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Showering
time_limit_ms: 2000
memory_limit_mb: 256
Description: As a computer science student, Alex faces a hard challenge — showering. He tries to shower daily, but despite his best efforts there are always challenges. He takes $s$ minutes to shower and a day only has $m$ minutes!

He already has $n$ tasks planned for the day. Task $i$ is represented as an interval $(l_i$, $r_i)$, which means that Alex is busy and can not take a shower in that time interval (at any point in time strictly between $l_i$ and $r_i$). No two tasks overlap.

Given all $n$ time intervals, will Alex be able to shower that day? In other words, will Alex have a free time interval of length at least $s$?

![](CDN_BASE_URL/5d5195053b99e5c6936ccefadc239679)

In the first test case, Alex can shower for the first $3$ minutes of the day and not miss any of the tasks.

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

The first line of each test case contains three integers $n$, $s$, and $m$ ($1 \leq n \leq 2 \cdot 10^5$; $1 \leq s, m \leq 10^9$) — the number of time intervals Alex already has planned, the amount of time Alex takes to take a shower, and the amount of minutes a day has.

Then $n$ lines follow, the $i$-th of which contains two integers $l_i$ and $r_i$ ($0 \leq l_i < r_i \leq m$) — the time interval of the $i$-th task. No two tasks overlap.

Additional constraint on the input: $l_i > r_{i-1}$ for every $i > 1$.

The sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case output "YES" (without quotes) if Alex can take a shower for that given test case, and "NO" (also without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yEs", "yes", and "Yes" will be recognized as a positive response).

Examples:
input:
4
3 3 10
3 5
6 8
9 10
3 3 10
1 2
3 5
6 7
3 3 10
1 2
3 5
6 8
3 4 10
1 2
6 7
8 9
output:
YES
YES
NO
YES

Note:

Buggy Code:
import java.util.Scanner;

public class Shower {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();  // Number of test cases

        while (t-- > 0) {
            int n = sc.nextInt();  // Number of showers
            int s = sc.nextInt();  // Minimum required time for a shower
            int m = sc.nextInt();  // Total available time

            // Difference array to mark busy times
            int[] diff = new int[m + 1];  // Extra space for boundary condition

            // Process each shower interval
            for (int i = 0; i < n; i++) {
                int l = sc.nextInt();  // Start time
                int r = sc.nextInt();  // End time
                if (l < m) {
                    diff[l]++;
                    if (r < m) {
                        diff[r]--;
                    }
                }
            }

            // Compute continuous free period without extra memory
            int currentBusy = 0;
            int freePeriod = 0;
            boolean found = false;

            for (int i = 0; i < m; i++) {
                currentBusy += diff[i];
                if (currentBusy == 0) {
                    freePeriod++;
                    if (freePeriod >= s) {
                        System.out.println("Yes");
                        found = true;
                        break;
                    }
                } else {
                    freePeriod = 0;  // Reset free period counter
                }
            }

            if (!found) {
                System.out.println("No");
            }
        }

        sc.close();  // Close the scanner
    }
}

**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
3 2 10
1 2
4 5
7 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Showering
time_limit_ms: 2000
memory_limit_mb: 256
Description: As a computer science student, Alex faces a hard challenge — showering. He tries to shower daily, but despite his best efforts there are always challenges. He takes $s$ minutes to shower and a day only has $m$ minutes!

He already has $n$ tasks planned for the day. Task $i$ is represented as an interval $(l_i$, $r_i)$, which means that Alex is busy and can not take a shower in that time interval (at any point in time strictly between $l_i$ and $r_i$). No two tasks overlap.

Given all $n$ time intervals, will Alex be able to shower that day? In other words, will Alex have a free time interval of length at least $s$?

![](CDN_BASE_URL/5d5195053b99e5c6936ccefadc239679)

In the first test case, Alex can shower for the first $3$ minutes of the day and not miss any of the tasks.

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

The first line of each test case contains three integers $n$, $s$, and $m$ ($1 \leq n \leq 2 \cdot 10^5$; $1 \leq s, m \leq 10^9$) — the number of time intervals Alex already has planned, the amount of time Alex takes to take a shower, and the amount of minutes a day has.

Then $n$ lines follow, the $i$-th of which contains two integers $l_i$ and $r_i$ ($0 \leq l_i < r_i \leq m$) — the time interval of the $i$-th task. No two tasks overlap.

Additional constraint on the input: $l_i > r_{i-1}$ for every $i > 1$.

The sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case output "YES" (without quotes) if Alex can take a shower for that given test case, and "NO" (also without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yEs", "yes", and "Yes" will be recognized as a positive response).

Examples:
input:
4
3 3 10
3 5
6 8
9 10
3 3 10
1 2
3 5
6 7
3 3 10
1 2
3 5
6 8
3 4 10
1 2
6 7
8 9
output:
YES
YES
NO
YES

Note:

Buggy Code:
import java.util.Arrays;
import java.util.Scanner;

public class Solution{
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t= sc.nextInt();
        while(t-- >0){
            int n = sc.nextInt();
            int s = sc.nextInt();
            int m = sc.nextInt();
            int[] arr = new int[m+1];
            int temp = n;
            while(temp -- > 0){
                int l1 = sc.nextInt();
                int l2 = sc.nextInt();
                for(int i = l1+1;i <= l2;i++){
                    arr[i] = 1;
                }
            }
            int contiguos = 0;
            boolean flag = true;
            for(int i = 1;i <= m;){
                while(i <= m && arr[i] == 0){
                    ++contiguos;
                    i++;
                }
                if(contiguos >= s) {
                    System.out.println("YES");
                    flag = false;
                    break;
                }
                contiguos = 0;
                i++;
            }
            if(flag) System.out.println("NO");
        }
    }
}

**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
2 5 10
0 4
6 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kousuke's Assignment
time_limit_ms: 2000
memory_limit_mb: 256
Description: After a trip with Sakurako, Kousuke was very scared because he forgot about his programming assignment. In this assignment, the teacher gave him an array $a$ of $n$ integers and asked him to calculate the number of non- overlapping segments of the array $a$, such that each segment is considered beautiful.

A segment $[l,r]$ is considered beautiful if $a_l + a_{l+1} + \dots + a_{r-1} + a_r=0$.

For a fixed array $a$, your task is to compute the maximum number of non- overlapping beautiful segments.

Input: The first line of input contains the number $t$ ($1 \le t \le 10^4$) — the number of test cases. Each test case consists of $2$ lines.

  * The first line contains one integer $n$ ($1 \le n \le 10^5$) — the length of the array.   * The second line contains $n$ integers $a_i$ ($-10^5 \le a_i \le 10^5$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ across all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, output a single integer: the maximum number of non- overlapping beautiful segments.

Examples:
input:
3
5
2 1 -3 2 1
7
12 -4 4 43 -3 -5 8
6
0 -4 0 3 0 1
output:
1
2
3

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
using ll=long long;
void solve()
{
  int n;
  cin>>n;
  vector<int> a(n);
  for(int i=0;i<n;i++) {
    cin>>a[i];
    if(i>0) a[i]+=a[i-1];
  }
  // for(int i=0;i<n;i++) cout<<a[i]<<" ";
  // cout<<endl;
  int ans=0;
  map<int,int> mp;
  for(int i=0;i<n;i++) {
    mp[a[i]]++;
    if(mp[a[i]]>1) {
      ans++;
      if(!mp.empty()) mp.clear();
    }
    if(a[i]==0 && i!=0) {
      ans++;
      if(!mp.empty()) mp.clear();
    }
    //cout<<a[i]<<" ";
  }
  cout<<ans<<endl;
}
int main()
{
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin>>t;
  while(t--) 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:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider the following points:

1. **Edge Cases:**
   - Arrays with all elements being zero.
   - Arrays with a single zero.
   - Arrays with multiple zeros.
   - Arrays with negative and positive numbers that sum to zero in multiple segments.

2. **Scenarios to Maximize Failure:**
   - Arrays where the sum of elements in multiple segments equals zero.
   - Arrays where the sum of elements in a segment equals zero, but the segment is not the maximum non-overlapping segment.
   - Arrays where the sum of elements in a segment equals zero, but the segment is not the first or last segment.

Given these considerations, let's create a small-scale test input that is complete and valid:

```plaintext
3
5
0 0 0 0 0
7
-1 1 -1 1 -1 1 -1
6
1 -1 2 -2 3 -3
```

### Explanation:

1. **Test Case 1:**
   - Array: `[0, 0, 0, 0, 0]`
   - This array has five zeros. Each zero can be considered a beautiful segment, and the maximum number of non-overlapping beautiful segments is 5.

2. **Test Case 2:**
   - Array: `[-1, 1, -1, 1, -1, 1, -1]`
   - This array has pairs of `-1` and `1` that sum to zero. The maximum number of non-overlapping beautiful segments is 3 (pairs: `[-1, 1]`, `[-1, 1]`, `[-1, 1]`).

3. **Test Case 3:**
   - Array: `[1, -1, 2, -2, 3, -3]`
   - This array has pairs of `1` and `-1`, `2` and `-2`, and `3` and `-3` that sum to zero. The maximum number of non-overlapping beautiful segments is 3 (pairs: `[1, -1]`, `[2, -2]`, `[3, -3]`).

These test cases are designed to expose the bug in the given code by testing various edge cases and scenarios where the sum of elements in multiple segments equals zero.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kousuke's Assignment
time_limit_ms: 2000
memory_limit_mb: 256
Description: After a trip with Sakurako, Kousuke was very scared because he forgot about his programming assignment. In this assignment, the teacher gave him an array $a$ of $n$ integers and asked him to calculate the number of non- overlapping segments of the array $a$, such that each segment is considered beautiful.

A segment $[l,r]$ is considered beautiful if $a_l + a_{l+1} + \dots + a_{r-1} + a_r=0$.

For a fixed array $a$, your task is to compute the maximum number of non- overlapping beautiful segments.

Input: The first line of input contains the number $t$ ($1 \le t \le 10^4$) — the number of test cases. Each test case consists of $2$ lines.

  * The first line contains one integer $n$ ($1 \le n \le 10^5$) — the length of the array.   * The second line contains $n$ integers $a_i$ ($-10^5 \le a_i \le 10^5$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ across all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, output a single integer: the maximum number of non- overlapping beautiful segments.

Examples:
input:
3
5
2 1 -3 2 1
7
12 -4 4 43 -3 -5 8
6
0 -4 0 3 0 1
output:
1
2
3

Note:

Buggy Code:
#include <bits/stdc++.h>
#define int long long int
#define mod 1000000007
using namespace std;
struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }
    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};
long long power(long long x, long long y, long long p)
{
    long long res = 1;
    x = x % p;

    while (y > 0) {

        if (y & 1)
            res = (res * x) % p;

        y = y >> 1;
        x = (x * x) % p;
    }
    return res;
}
vector<int> printDivisors(int n)
{
    vector<int> temp;
    for (int i=1; i<=sqrt(n); i++)
    {
        if (n%i == 0)
        {
            if (n/i == i)
                temp.push_back(i);
            else
            {
                temp.push_back(i);
                temp.push_back(n/i);
            }
        }
    }
    return temp;
}

void f(vector<int>& v, int k, int& cnt){
    int ORS = 0;
    for(int x = 0;x < v.size();x++){
        int OR = 0;
        for(int y = x;y < v.size();y++){
            OR |= v[y];
            ORS += OR;
            cout << OR << " ";
        }
    }
    cout << "\n";
    cout << "SUM : " << ORS << "\n";
    cnt += (ORS <= k);
}

bool static cmp(pair<int,int> m1, pair<int,int>  m2) {
    if (m1.second < m2.second) return true;
    else if (m1.second > m2.second) return false;
    else if (m1.second < m2.second) return true;
    return false;
}
int findLongestChain(vector<pair<int,int>>& pairs) {
    sort(begin(pairs),end(pairs),cmp);

    int prevMeeting = 0 , maxi = 1 ;
    for(int i = 1 ; i < pairs.size() ; ++i){
        if(pairs[i].first > pairs[prevMeeting].second) prevMeeting = i , ++maxi ;
    }
    return maxi ;
}
void solve(){
    int n;
    cin >> n;
    vector<int> v(n), pref(n,0);
    for(int i = 0;i < n;i++){
        cin >> v[i];
        if(i > 0)
            pref[i] = pref[i-1] + v[i];
        else pref[0] = v[0];
    }
    map<int,int> p;
    p[0] = -1;
    int lc = -1;
    int cnt = 0;
    vector<pair<int,int>> pr;
    for(int i = 0;i < n;i++){
        if(p.find(pref[i]) != p.end()){
            pr.push_back({p[pref[i]] + 1, i});
            cnt++;
        }
        p[pref[i]] = i;
    }
    cout << findLongestChain(pr) << "\n";
}

signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int q=1;
    cin>>q;
    while(q--)
    {
        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:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider the following points:

1. **Edge Cases**: The code should handle edge cases such as arrays with all elements being zero, arrays with a single element, and arrays where no segment sums to zero.
2. **Overlapping Segments**: The code should correctly handle non-overlapping segments and not count overlapping segments multiple times.
3. **Prefix Sum Handling**: The code uses prefix sums to identify segments that sum to zero. We need to ensure that the prefix sum logic is robust.
4. **Large Input**: While the test input should be small-scale, it should still be complex enough to expose the bug.

Let's consider a test case where the array has multiple segments that sum to zero, but the segments are not non-overlapping. This will help us test the logic for identifying non-overlapping segments.

### Test Input
```plaintext
3
5
1 -1 1 -1 1
6
0 0 0 0 0 0
7
1 2 -3 1 2 -3 0
```

### Explanation
1. **Test Case 1**: `1 -1 1 -1 1`
   - Segments that sum to zero: `[1, -1]`, `[1, -1]`, `[1, -1]`
   - Correct output: 1 (since they are overlapping)
   - Expected output: 1

2. **Test Case 2**: `0 0 0 0 0 0`
   - Segments that sum to zero: `[0]`, `[0, 0]`, `[0, 0, 0]`, etc.
   - Correct output: 6 (since all single elements and all combinations sum to zero)
   - Expected output: 6

3. **Test Case 3**: `1 2 -3 1 2 -3 0`
   - Segments that sum to zero: `[1, 2, -3]`, `[1, 2, -3, 0]`
   - Correct output: 2 (since they are non-overlapping)
   - Expected output: 2

This test input will help identify issues with the code's logic for handling non-overlapping segments and prefix sums.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kousuke's Assignment
time_limit_ms: 2000
memory_limit_mb: 256
Description: After a trip with Sakurako, Kousuke was very scared because he forgot about his programming assignment. In this assignment, the teacher gave him an array $a$ of $n$ integers and asked him to calculate the number of non- overlapping segments of the array $a$, such that each segment is considered beautiful.

A segment $[l,r]$ is considered beautiful if $a_l + a_{l+1} + \dots + a_{r-1} + a_r=0$.

For a fixed array $a$, your task is to compute the maximum number of non- overlapping beautiful segments.

Input: The first line of input contains the number $t$ ($1 \le t \le 10^4$) — the number of test cases. Each test case consists of $2$ lines.

  * The first line contains one integer $n$ ($1 \le n \le 10^5$) — the length of the array.   * The second line contains $n$ integers $a_i$ ($-10^5 \le a_i \le 10^5$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ across all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, output a single integer: the maximum number of non- overlapping beautiful segments.

Examples:
input:
3
5
2 1 -3 2 1
7
12 -4 4 43 -3 -5 8
6
0 -4 0 3 0 1
output:
1
2
3

Note:

Buggy Code:
#include <bits/stdc++.h>
#define int long long
#define pb push_back
#define endl "\n"
#define vi vector<int>
#define vpi vector<pair<int,int>>
#define mod 1000000007
#define yes cout << "YES\n"
#define no cout << "NO\n"
using namespace std;
int32_t main(){
ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
int T;
cin >> T;
while(T--){
    int n,ans=0;
    cin >> n;
    vi v(n),diff;
    for(auto&x:v) cin >> x;
    vi pre(n);
    map<int,vi> m;
    pre[0]=v[0];
    for(int i=1;i<n;i++) pre[i]=pre[i-1]+v[i];
    for(int i=0;i<n;i++){
        if(!(i==0 and pre[i]==0)){
            m[pre[i]].pb(i);
        }
        else ans++;
    }
    for(auto&x:m){
        if(x.first==0) diff.pb(x.second[0]+1);
        for(int i=0;i<x.second.size()-1;i++){
            diff.pb(x.second[i+1]-x.second[i]);
        }
    }
    int su=0;
    sort(diff.begin(),diff.end());
    for(int i=0;i<diff.size();i++){
        su+=diff[i];
        if(su>n) break;
        else ans++;
    }
    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 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 maximize the failure probability. The key issues to consider are:

1. **Edge Case: Empty Array or Single Element Array**: These should be handled correctly.
2. **Edge Case: All Elements are Zero**: This should result in the maximum number of non-overlapping segments.
3. **Edge Case: Segments with Zero Sum**: We need to ensure that the code correctly identifies and counts all possible non-overlapping segments with a zero sum.
4. **Edge Case: Overlapping Segments**: Ensure that the code does not count overlapping segments.

Let's create a test input that includes these edge cases:

```plaintext
3
1
0
5
0 0 0 0 0
9
1 -1 2 -2 3 -3 4 -4 0
```

### Explanation:
1. **Test Case 1**: `1 0`
   - This tests the case where the array has a single element, which is zero. The expected output is `1`.

2. **Test Case 2**: `5 0 0 0 0 0`
   - This tests the case where all elements are zero. The expected output is `5` because each element can form a beautiful segment.

3. **Test Case 3**: `9 1 -1 2 -2 3 -3 4 -4 0`
   - This tests a more complex scenario where there are multiple segments with a zero sum. The expected output is `4`:
     - Segments: `[1, -1]`, `[2, -2]`, `[3, -3]`, `[4, -4]`

This test input should help expose any issues in the buggy code, particularly in handling edge cases and correctly identifying non-overlapping segments.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kousuke's Assignment
time_limit_ms: 2000
memory_limit_mb: 256
Description: After a trip with Sakurako, Kousuke was very scared because he forgot about his programming assignment. In this assignment, the teacher gave him an array $a$ of $n$ integers and asked him to calculate the number of non- overlapping segments of the array $a$, such that each segment is considered beautiful.

A segment $[l,r]$ is considered beautiful if $a_l + a_{l+1} + \dots + a_{r-1} + a_r=0$.

For a fixed array $a$, your task is to compute the maximum number of non- overlapping beautiful segments.

Input: The first line of input contains the number $t$ ($1 \le t \le 10^4$) — the number of test cases. Each test case consists of $2$ lines.

  * The first line contains one integer $n$ ($1 \le n \le 10^5$) — the length of the array.   * The second line contains $n$ integers $a_i$ ($-10^5 \le a_i \le 10^5$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ across all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, output a single integer: the maximum number of non- overlapping beautiful segments.

Examples:
input:
3
5
2 1 -3 2 1
7
12 -4 4 43 -3 -5 8
6
0 -4 0 3 0 1
output:
1
2
3

Note:

Buggy Code:
t = int(input())
for _ in range(t):
    n = int(input())
    arr = list(map(int, input().split()))
    sumSet = set()
    sum = 0
    res = 0
    for i in arr:
        sum += i
        if sum in sumSet or i ==0:
            res += 1
            sum = 0
            sumSet.clear()
            continue
        sumSet.add(sum)
    print(res)

**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 and scenarios where the code might fail. The buggy code has a flaw in how it handles the condition for a beautiful segment. Specifically, it resets the sum and the set when it encounters a zero element, which can lead to incorrect results.

Here are the steps to think through the test input:

1. **Zero Elements**: The code incorrectly handles zero elements by treating them as individual beautiful segments. We need to create a test case where this behavior leads to an incorrect result.

2. **Multiple Zero Elements**: Test cases with multiple zero elements in different positions to see how the code handles them.

3. **Sum Reset**: Create a test case where the sum of elements in a segment equals zero, but the code incorrectly resets the sum and set due to the presence of zero elements.

4. **Non-overlapping Segments**: Ensure the test case includes multiple non-overlapping beautiful segments to test the code's ability to count them correctly.

Based on these considerations, here is a challenging test input:

```plaintext
3
5
0 1 -1 0 0
6
0 1 -1 0 1 -1
7
0 1 -1 0 1 -1 0
```

### Explanation:
1. **First Test Case**:
   - Array: `[0, 1, -1, 0, 0]`
   - Beautiful segments: `[0]`, `[1, -1]`, `[0]`, `[0]`
   - Expected output: 4
   - Buggy code output: 5 (incorrectly treats each zero as a separate segment)

2. **Second Test Case**:
   - Array: `[0, 1, -1, 0, 1, -1]`
   - Beautiful segments: `[0]`, `[1, -1]`, `[0]`, `[1, -1]`
   - Expected output: 4
   - Buggy code output: 6 (incorrectly treats each zero as a separate segment)

3. **Third Test Case**:
   - Array: `[0, 1, -1, 0, 1, -1, 0]`
   - Beautiful segments: `[0]`, `[1, -1]`, `[0]`, `[1, -1]`, `[0]`
   - Expected output: 5
   - Buggy code output: 7 (incorrectly treats each zero as a separate segment)

This test input should effectively expose the bug in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kousuke's Assignment
time_limit_ms: 2000
memory_limit_mb: 256
Description: After a trip with Sakurako, Kousuke was very scared because he forgot about his programming assignment. In this assignment, the teacher gave him an array $a$ of $n$ integers and asked him to calculate the number of non- overlapping segments of the array $a$, such that each segment is considered beautiful.

A segment $[l,r]$ is considered beautiful if $a_l + a_{l+1} + \dots + a_{r-1} + a_r=0$.

For a fixed array $a$, your task is to compute the maximum number of non- overlapping beautiful segments.

Input: The first line of input contains the number $t$ ($1 \le t \le 10^4$) — the number of test cases. Each test case consists of $2$ lines.

  * The first line contains one integer $n$ ($1 \le n \le 10^5$) — the length of the array.   * The second line contains $n$ integers $a_i$ ($-10^5 \le a_i \le 10^5$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ across all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, output a single integer: the maximum number of non- overlapping beautiful segments.

Examples:
input:
3
5
2 1 -3 2 1
7
12 -4 4 43 -3 -5 8
6
0 -4 0 3 0 1
output:
1
2
3

Note:

Buggy Code:
def solution():
    n = int(input())
    a = list(map(int, input().split()))
    lookup = {0}
    result = prefix = 0
    for x in a:
        prefix += x
        if prefix in lookup:
            result += 1
            lookup.clear()
        else:
            lookup.add(prefix)
    return result

for _ in range(int(input())):
    print(solution())

**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
10
1 -1 1 -1 1 -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: Kousuke's Assignment
time_limit_ms: 2000
memory_limit_mb: 256
Description: After a trip with Sakurako, Kousuke was very scared because he forgot about his programming assignment. In this assignment, the teacher gave him an array $a$ of $n$ integers and asked him to calculate the number of non- overlapping segments of the array $a$, such that each segment is considered beautiful.

A segment $[l,r]$ is considered beautiful if $a_l + a_{l+1} + \dots + a_{r-1} + a_r=0$.

For a fixed array $a$, your task is to compute the maximum number of non- overlapping beautiful segments.

Input: The first line of input contains the number $t$ ($1 \le t \le 10^4$) — the number of test cases. Each test case consists of $2$ lines.

  * The first line contains one integer $n$ ($1 \le n \le 10^5$) — the length of the array.   * The second line contains $n$ integers $a_i$ ($-10^5 \le a_i \le 10^5$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ across all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, output a single integer: the maximum number of non- overlapping beautiful segments.

Examples:
input:
3
5
2 1 -3 2 1
7
12 -4 4 43 -3 -5 8
6
0 -4 0 3 0 1
output:
1
2
3

Note:

Buggy Code:
def subArray(arr, n):
    b = []
    skip = -1
    for i in range(0,n):

        for j in range(i,n):
            c = []
            for k in range(i,j+1):
                if k <= skip:
                    continue
                else:
                    c.append(arr[k])
            if sum(c)==0 and len(c)>0:
                b.append(c)
                skip = j
    return b

t = int(input())
for i in range(t):
    n = int(input())
    arr = list(map(int,input().split()))
    b = subArray(arr, n)
    print(max(len(b),arr.count(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
1
10
-5 3 2 -2 0 0 0 5 -3 -2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kousuke's Assignment
time_limit_ms: 2000
memory_limit_mb: 256
Description: After a trip with Sakurako, Kousuke was very scared because he forgot about his programming assignment. In this assignment, the teacher gave him an array $a$ of $n$ integers and asked him to calculate the number of non- overlapping segments of the array $a$, such that each segment is considered beautiful.

A segment $[l,r]$ is considered beautiful if $a_l + a_{l+1} + \dots + a_{r-1} + a_r=0$.

For a fixed array $a$, your task is to compute the maximum number of non- overlapping beautiful segments.

Input: The first line of input contains the number $t$ ($1 \le t \le 10^4$) — the number of test cases. Each test case consists of $2$ lines.

  * The first line contains one integer $n$ ($1 \le n \le 10^5$) — the length of the array.   * The second line contains $n$ integers $a_i$ ($-10^5 \le a_i \le 10^5$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ across all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, output a single integer: the maximum number of non- overlapping beautiful segments.

Examples:
input:
3
5
2 1 -3 2 1
7
12 -4 4 43 -3 -5 8
6
0 -4 0 3 0 1
output:
1
2
3

Note:

Buggy Code:
import java.util.*;

public class C5 {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            int a[] = new int[n];

            for (int i = 0; i < n; i++) {
                a[i] = sc.nextInt();
            }
            Map<Integer, Integer> map = new HashMap<>();

            map.put(0, -1);

            int sum = 0, ans = 0, last = -2;

            for (int i = 0; i < n; i++) {
                sum += a[i];

                if (map.containsKey(sum) && map.get(sum) > last) {
                    ans++;
                    sum = 0;

                    last = i;

                    map.put(0, i);
                } else {
                    map.put(sum, i);
                }
            }

            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
10
1 -1 1 -1 1 -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: Kousuke's Assignment
time_limit_ms: 2000
memory_limit_mb: 256
Description: After a trip with Sakurako, Kousuke was very scared because he forgot about his programming assignment. In this assignment, the teacher gave him an array $a$ of $n$ integers and asked him to calculate the number of non- overlapping segments of the array $a$, such that each segment is considered beautiful.

A segment $[l,r]$ is considered beautiful if $a_l + a_{l+1} + \dots + a_{r-1} + a_r=0$.

For a fixed array $a$, your task is to compute the maximum number of non- overlapping beautiful segments.

Input: The first line of input contains the number $t$ ($1 \le t \le 10^4$) — the number of test cases. Each test case consists of $2$ lines.

  * The first line contains one integer $n$ ($1 \le n \le 10^5$) — the length of the array.   * The second line contains $n$ integers $a_i$ ($-10^5 \le a_i \le 10^5$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ across all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, output a single integer: the maximum number of non- overlapping beautiful segments.

Examples:
input:
3
5
2 1 -3 2 1
7
12 -4 4 43 -3 -5 8
6
0 -4 0 3 0 1
output:
1
2
3

Note:

Buggy Code:
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int T=sc.nextInt();
        while (T-- > 0) {
            int n=sc.nextInt();
            int[] arr=new int[n];
            for (int i=0;i<n;i++) {
                arr[i]=sc.nextInt();
            }
            Long sum=0L;
            Map<Long,Integer> map=new HashMap<>();
            int ans=0;
            map.put(0L,0);
            int last=0;
            for (int i = 0; i < arr.length; i++) {
                sum+=arr[i];
                if (map.containsKey(sum)) {
                    if (map.get(sum) >= last) {
                        ans++;
                        last = i;
                    }
                } else {
                    map.put(sum, i);
                }
            }
            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
10
1 -1 1 -1 1 -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: Kousuke's Assignment
time_limit_ms: 2000
memory_limit_mb: 256
Description: After a trip with Sakurako, Kousuke was very scared because he forgot about his programming assignment. In this assignment, the teacher gave him an array $a$ of $n$ integers and asked him to calculate the number of non- overlapping segments of the array $a$, such that each segment is considered beautiful.

A segment $[l,r]$ is considered beautiful if $a_l + a_{l+1} + \dots + a_{r-1} + a_r=0$.

For a fixed array $a$, your task is to compute the maximum number of non- overlapping beautiful segments.

Input: The first line of input contains the number $t$ ($1 \le t \le 10^4$) — the number of test cases. Each test case consists of $2$ lines.

  * The first line contains one integer $n$ ($1 \le n \le 10^5$) — the length of the array.   * The second line contains $n$ integers $a_i$ ($-10^5 \le a_i \le 10^5$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ across all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, output a single integer: the maximum number of non- overlapping beautiful segments.

Examples:
input:
3
5
2 1 -3 2 1
7
12 -4 4 43 -3 -5 8
6
0 -4 0 3 0 1
output:
1
2
3

Note:

Buggy Code:
def solve():
    n = int(input())
    nums = list(map(int,input().split(' ')))

    if n == 1:
        if nums[0] == 0:
            print(1)
        else:
            print(0)

    psum = [0]
    for i in range(n):
        psum.append(psum[-1] + nums[i])

    elements = set([0])

    res =  0
    ind = 1
    while ind < n + 1:
        if psum[ind] in elements:
            res += 1
            elements = set([psum[ind]])
            ind += 1
        else:
            elements.add(psum[ind])
            ind += 1

    print(res)

times = int(input())
for _ in range(times):
    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:
To expose the bug in the provided code, we need to focus on scenarios where the prefix sums might not be handled correctly. Specifically, we should look for cases where the code might prematurely reset the set of elements or miss counting beautiful segments.

Here are the key points to consider:
1. **Multiple Beautiful Segments**: Ensure the code can handle multiple beautiful segments that are non-overlapping.
2. **Single Beautiful Segment**: Test cases where there is only one beautiful segment.
3. **No Beautiful Segments**: Test cases where there are no beautiful segments.
4. **Edge Cases**: Test cases with small arrays, arrays with all zeros, and arrays with negative and positive values that sum to zero.

Given these considerations, a challenging test input that can expose the bug is:

```plaintext
3
5
1 -1 1 -1 1
6
1 -1 1 -1 1 -1
7
1 -1 1 -1 1 -1 1
```

### Explanation:
1. **Test Case 1**:
   - Array: `[1, -1, 1, -1, 1]`
   - Prefix Sums: `[0, 1, 0, 1, 0, 1]`
   - Expected Output: `2`
   - This test case has multiple beautiful segments: `[1, -1]` and `[1, -1]`.

2. **Test Case 2**:
   - Array: `[1, -1, 1, -1, 1, -1]`
   - Prefix Sums: `[0, 1, 0, 1, 0, 1, 0]`
   - Expected Output: `3`
   - This test case has multiple beautiful segments: `[1, -1]`, `[1, -1]`, and `[1, -1]`.

3. **Test Case 3**:
   - Array: `[1, -1, 1, -1, 1, -1, 1]`
   - Prefix Sums: `[0, 1, 0, 1, 0, 1, 0, 1]`
   - Expected Output: `3`
   - This test case has multiple beautiful segments: `[1, -1]`, `[1, -1]`, and `[1, -1]`.

These test cases are designed to ensure that the code correctly identifies and counts multiple non-overlapping beautiful segments, which is a common scenario where the provided code might fail.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kousuke's Assignment
time_limit_ms: 2000
memory_limit_mb: 256
Description: After a trip with Sakurako, Kousuke was very scared because he forgot about his programming assignment. In this assignment, the teacher gave him an array $a$ of $n$ integers and asked him to calculate the number of non- overlapping segments of the array $a$, such that each segment is considered beautiful.

A segment $[l,r]$ is considered beautiful if $a_l + a_{l+1} + \dots + a_{r-1} + a_r=0$.

For a fixed array $a$, your task is to compute the maximum number of non- overlapping beautiful segments.

Input: The first line of input contains the number $t$ ($1 \le t \le 10^4$) — the number of test cases. Each test case consists of $2$ lines.

  * The first line contains one integer $n$ ($1 \le n \le 10^5$) — the length of the array.   * The second line contains $n$ integers $a_i$ ($-10^5 \le a_i \le 10^5$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ across all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, output a single integer: the maximum number of non- overlapping beautiful segments.

Examples:
input:
3
5
2 1 -3 2 1
7
12 -4 4 43 -3 -5 8
6
0 -4 0 3 0 1
output:
1
2
3

Note:

Buggy Code:
#ProblemA

t = int(input())
for q in range(t):
    M=[]

    n=int(input())
    ch=input()
    L=[int(i)for i in ch.split()]
    I=[L[0]]*n
    d=dict()
    d[L[0]]=0

    nb=0
    if L[0]==0:
        nb+=1
    ind=0

    for i in range(1,n):
        I[i]=L[i]+I[i-1]
        if I[i] in d:

            if d[I[i]]>=ind:
                nb+=1
                ind = i
        d[I[i]]=i

    print(nb)

**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
1
0
2
0 0
4
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: Kousuke's Assignment
time_limit_ms: 2000
memory_limit_mb: 256
Description: After a trip with Sakurako, Kousuke was very scared because he forgot about his programming assignment. In this assignment, the teacher gave him an array $a$ of $n$ integers and asked him to calculate the number of non- overlapping segments of the array $a$, such that each segment is considered beautiful.

A segment $[l,r]$ is considered beautiful if $a_l + a_{l+1} + \dots + a_{r-1} + a_r=0$.

For a fixed array $a$, your task is to compute the maximum number of non- overlapping beautiful segments.

Input: The first line of input contains the number $t$ ($1 \le t \le 10^4$) — the number of test cases. Each test case consists of $2$ lines.

  * The first line contains one integer $n$ ($1 \le n \le 10^5$) — the length of the array.   * The second line contains $n$ integers $a_i$ ($-10^5 \le a_i \le 10^5$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ across all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, output a single integer: the maximum number of non- overlapping beautiful segments.

Examples:
input:
3
5
2 1 -3 2 1
7
12 -4 4 43 -3 -5 8
6
0 -4 0 3 0 1
output:
1
2
3

Note:

Buggy Code:
#ifdef LOCAL
#define _GLIBCXX_DEBUG
#endif

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;

//using namespace __gnu_pbds;
//template<class c, class cmp = less<c> > using ordered_set = tree<c, null_type, cmp, rb_tree_tag, tree_order_statistics_node_update>;

//#pragma GCC optimize("O3, unroll-loops")
//#pragma GCC target("avx2,bmi,bmi2,popcnt,lzcnt")
#define int long long
#define str string
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define ff first
#define ss second
#define ll long long
#define db long double
#define ld long double
#define ull unsigned long long
#define sz(a) (int) a.size()
#define debug(x) cerr << (#x) << ": " << x << endl

template<typename T>
istream &operator>>(istream &in, vector<T> &arr) {
    for (auto &i: arr) in >> i;
    return in;
}

template<typename T>
ostream &operator<<(ostream &out, vector<T> &arr) {
    for (auto i: arr) out << i << ' ';
    out << '\n';
    return out;
}

template<typename T1, typename T2>
ostream &operator<<(ostream &out, pair<T1, T2> p) {
    out << p.first << ' ' << p.second;
    return out;
}

template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
    in >> p.first >> p.second;
    return in;
}

void fastios() {
    ios::sync_with_stdio(0);
    cin.tie(nullptr);
}

const ll INF = 1e9;
const db EPS = 1e-6;
const int MOD = 1e9 + 7;
const int MOD2 = 998244353;
const int MAXN = 1e6 + 1;
//const int p = 123;
//const int q = 1789
mt19937 rnd(time(0));

void solve();

signed main() {
    fastios();
#ifdef LOCAL
    freopen("inp.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    double start_time = clock();
#endif
    int ttt = 1;
    cin >> ttt;
    while (ttt--) {
        solve();
#ifdef LOCAL
        cout << "----------------------\n";
#endif
    }
#ifdef LOCAL
    cerr << "Time: " << (clock() - start_time) / CLOCKS_PER_SEC;
#endif
}

void solve() {
    int n;
    cin >> n;
    vector<int> a(n), pref(n + 1);
    cin >> a;
    for (int i = 0; i < n; ++i)
        pref[i + 1] = pref[i] + a[i];
    vector<int> dp(n + 1);
    map<int, pair<int, int>> tmp;
    tmp[0] = {0, 0};
    for (int i = 1; i <= n; ++i) {
        if (tmp.find(pref[i]) == tmp.end()) {
            dp[i] = dp[i - 1];
            tmp[pref[i]] = {dp[i], i};
        } else {
            dp[i] = dp[tmp[pref[i]].ss] + 1;
            tmp[pref[i]] = {dp[i], i};
        }
    }
    cout << *max_element(all(dp)) << '\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:
To expose the bug in the provided code, we need to focus on edge cases that can highlight potential issues. The code aims to find the maximum number of non-overlapping beautiful segments, where a beautiful segment has a sum of zero. The code uses a prefix sum and a map to keep track of the prefix sums and their corresponding positions.

### Edge Cases to Consider:
1. **Single Element Array**: An array with a single element.
2. **All Elements Zero**: An array where all elements are zero.
3. **Repeating Prefix Sums**: An array where the prefix sums repeat multiple times.
4. **Negative and Positive Elements**: An array with a mix of negative and positive elements that sum to zero in multiple segments.
5. **No Beautiful Segments**: An array where no segment sums to zero.

### Test Input:
Let's create a test input that includes a mix of these edge cases to maximize the failure probability:

```plaintext
4
1
0
3
0 0 0
5
-1 1 -1 1 -1
6
1 -1 1 -1 1 -1
4
1 2 3 4
```

### Explanation:
1. **Single Element Array (0)**:
   - This tests the case where the array has a single element, which is zero.
   - Expected output: 1

2. **All Elements Zero**:
   - This tests the case where all elements are zero.
   - Expected output: 3

3. **Repeating Prefix Sums**:
   - This tests the case where the prefix sums repeat multiple times.
   - Expected output: 3

4. **Negative and Positive Elements**:
   - This tests the case where the array has a mix of negative and positive elements that sum to zero in multiple segments.
   - Expected output: 3

5. **No Beautiful Segments**:
   - This tests the case where no segment sums to zero.
   - Expected output: 0

This test input should help expose any issues in the code's handling of edge cases and its logic for finding non-overlapping beautiful segments.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kousuke's Assignment
time_limit_ms: 2000
memory_limit_mb: 256
Description: After a trip with Sakurako, Kousuke was very scared because he forgot about his programming assignment. In this assignment, the teacher gave him an array $a$ of $n$ integers and asked him to calculate the number of non- overlapping segments of the array $a$, such that each segment is considered beautiful.

A segment $[l,r]$ is considered beautiful if $a_l + a_{l+1} + \dots + a_{r-1} + a_r=0$.

For a fixed array $a$, your task is to compute the maximum number of non- overlapping beautiful segments.

Input: The first line of input contains the number $t$ ($1 \le t \le 10^4$) — the number of test cases. Each test case consists of $2$ lines.

  * The first line contains one integer $n$ ($1 \le n \le 10^5$) — the length of the array.   * The second line contains $n$ integers $a_i$ ($-10^5 \le a_i \le 10^5$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ across all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, output a single integer: the maximum number of non- overlapping beautiful segments.

Examples:
input:
3
5
2 1 -3 2 1
7
12 -4 4 43 -3 -5 8
6
0 -4 0 3 0 1
output:
1
2
3

Note:

Buggy Code:
import java.util.*;

public class D_Kousuke_s_Assignment {
    static Scanner sc = new Scanner(System.in);

    public static void solve() {
        int n = sc.nextInt();
        HashMap<Long, Integer> last = new HashMap<>();
        last.put(0L, -1);

        int ans = 0;
        long sum = 0;

        TreeMap<Integer, Integer> end = new TreeMap<>();
        end.put(-1, 0);

        for(int i = 0; i < n; i++) {
            sum += sc.nextInt();
            if(last.containsKey(sum)) {
                int prevIndex = last.get(sum);
                Map.Entry<Integer, Integer> prev = end.floorEntry(prevIndex);
                if(prev != null && prev.getKey() <= prevIndex) {
                    int newAns = prev.getValue() + 1;
                    end.put(i, newAns);
                    ans = Math.max(ans, newAns);
                }
            }

            last.put(sum, i);
        }

        System.out.println(ans);
    }

    public static void main(String[] args) {
        int t = sc.nextInt();
        while(t-- > 0) 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:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider edge cases and scenarios that are likely to cause issues. Here are some potential areas to focus on:

1. **Duplicate Prefix Sums**: The code uses a `HashMap` to track the last occurrence of each prefix sum. If there are multiple segments with the same prefix sum, the code might not handle them correctly.
2. **Overlapping Segments**: The code should ensure that segments do not overlap. We need to test if the code correctly handles non-overlapping segments.
3. **Negative and Positive Numbers**: The code should handle arrays with both negative and positive numbers correctly.
4. **Zero Sum Segments at the Beginning or End**: The code should handle segments that start at the beginning or end of the array.

Let's construct a test case that includes these elements:

1. **Duplicate Prefix Sums**: We can create an array where multiple segments have the same prefix sum.
2. **Overlapping Segments**: We can create an array where there are multiple valid segments, but the code might incorrectly count overlapping segments.
3. **Negative and Positive Numbers**: We can include both negative and positive numbers in the array.
4. **Zero Sum Segments at the Beginning or End**: We can include segments that start at the beginning or end of the array.

Here is a small-scale test input that covers these scenarios:

```plaintext
3
5
1 -1 1 -1 1
7
1 2 -3 4 -4 3 -2
6
0 1 -1 2 -2 0
```

Explanation:
1. **First Test Case**: `1 -1 1 -1 1`
   - Segments: `[1, -1]`, `[1, -1]`, `[1, -1]`
   - Expected output: `3` (three non-overlapping segments)

2. **Second Test Case**: `1 2 -3 4 -4 3 -2`
   - Segments: `[1, 2, -3]`, `[4, -4]`, `[3, -2]`
   - Expected output: `3` (three non-overlapping segments)

3. **Third Test Case**: `0 1 -1 2 -2 0`
   - Segments: `[0]`, `[1, -1]`, `[2, -2]`, `[0]`
   - Expected output: `4` (four non-overlapping segments)

This test input should expose any issues in the buggy code related to handling duplicate prefix sums, overlapping segments, and zero sum segments at the beginning or end of the array.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kousuke's Assignment
time_limit_ms: 2000
memory_limit_mb: 256
Description: After a trip with Sakurako, Kousuke was very scared because he forgot about his programming assignment. In this assignment, the teacher gave him an array $a$ of $n$ integers and asked him to calculate the number of non- overlapping segments of the array $a$, such that each segment is considered beautiful.

A segment $[l,r]$ is considered beautiful if $a_l + a_{l+1} + \dots + a_{r-1} + a_r=0$.

For a fixed array $a$, your task is to compute the maximum number of non- overlapping beautiful segments.

Input: The first line of input contains the number $t$ ($1 \le t \le 10^4$) — the number of test cases. Each test case consists of $2$ lines.

  * The first line contains one integer $n$ ($1 \le n \le 10^5$) — the length of the array.   * The second line contains $n$ integers $a_i$ ($-10^5 \le a_i \le 10^5$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ across all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, output a single integer: the maximum number of non- overlapping beautiful segments.

Examples:
input:
3
5
2 1 -3 2 1
7
12 -4 4 43 -3 -5 8
6
0 -4 0 3 0 1
output:
1
2
3

Note:

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

ll func(int i, vector<int>&a, ll sum, ll count){
    if(i<0) return count;
    ll tsum = sum + a[i];
    if(tsum==0) return func(i-1, a, tsum, count+1);
    else return max({func(i-1, a, tsum, count), func(i-1, a, 0, count)});
}

int main() {
	// your code goes here
	int t;
	cin>>t;

	while(t--){
	    int n;
        cin>>n;
        vector<int> a(n);
        for(int &i: a) cin>>i;
        cout<<func(n-1, a, 0, 0)<<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 expose the bug in the provided code, we need to consider the following aspects:

1. **Non-Overlapping Segments**: The code should correctly identify non-overlapping segments where the sum is zero.
2. **Recursive Overhead**: The code uses recursion, which can lead to stack overflow or incorrect results if not handled properly.
3. **Edge Cases**: The code might not handle edge cases such as all elements being zero, or segments that overlap in a way that the recursive function fails to recognize.

Let's construct a small-scale test input that can expose these issues:

### Test Case 1: All Elements are Zero
This case should return the number of elements as the answer since each element itself forms a beautiful segment.

### Test Case 2: Multiple Non-Overlapping Segments
This case should return the correct number of non-overlapping segments.

### Test Case 3: Overlapping Segments
This case should test the code's ability to handle overlapping segments correctly.

### Test Case 4: Negative and Positive Values
This case should test the code's ability to handle a mix of negative and positive values.

### Test Input
```plaintext
4
5
0 0 0 0 0
7
1 -1 2 -2 3 -3 4
6
1 2 -3 3 -3 3
10
-1 1 -2 2 -3 3 -4 4 -5 5
```

### Explanation:
1. **Test Case 1**: All elements are zero. The expected output is `5`.
2. **Test Case 2**: Multiple non-overlapping segments: `[1, -1]`, `[2, -2]`, `[3, -3]`, `[4]`. The expected output is `4`.
3. **Test Case 3**: Overlapping segments: `[1, 2, -3]`, `[3, -3, 3]`. The expected output is `2`.
4. **Test Case 4**: Mixed values: `[-1, 1]`, `[-2, 2]`, `[-3, 3]`, `[-4, 4]`, `[-5, 5]`. The expected output is `5`.

This test input should help identify the issues in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kousuke's Assignment
time_limit_ms: 2000
memory_limit_mb: 256
Description: After a trip with Sakurako, Kousuke was very scared because he forgot about his programming assignment. In this assignment, the teacher gave him an array $a$ of $n$ integers and asked him to calculate the number of non- overlapping segments of the array $a$, such that each segment is considered beautiful.

A segment $[l,r]$ is considered beautiful if $a_l + a_{l+1} + \dots + a_{r-1} + a_r=0$.

For a fixed array $a$, your task is to compute the maximum number of non- overlapping beautiful segments.

Input: The first line of input contains the number $t$ ($1 \le t \le 10^4$) — the number of test cases. Each test case consists of $2$ lines.

  * The first line contains one integer $n$ ($1 \le n \le 10^5$) — the length of the array.   * The second line contains $n$ integers $a_i$ ($-10^5 \le a_i \le 10^5$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ across all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, output a single integer: the maximum number of non- overlapping beautiful segments.

Examples:
input:
3
5
2 1 -3 2 1
7
12 -4 4 43 -3 -5 8
6
0 -4 0 3 0 1
output:
1
2
3

Note:

Buggy Code:
import java.util.*;
public class assignment {
    static long st[];
    public static void main(String args[]){
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        while(n>0){
            int b=sc.nextInt();
            st=new long[b+1];
            st[b]=0;
            long a[]=new long[b];
            for(int i=0;i<b;i++)
                a[i]=sc.nextInt();
            System.out.println(find(a));
            n--;
        }
    }
    public static long find(long a[]){
        int b=a.length;
        for(int i=b-1;i>=0;i--){
            long op1=0,sum=0;
            for(int j=i;j<b;j++){
                sum+=a[j];
                if(sum==0)
                    op1=Math.max(op1,1+st[j+1]);
                else
                    op1=Math.max(op1,st[j+1]);
                //System.out.println(a[i]+" "+st[i]);
            }
            st[i]=op1;
        }
        return st[0];
    }
    public static long find(long a[],int j){
        if(j>=a.length)
            return 0;
        if(st[j]!=-1)
            return st[j];
        int b=a.length;
        long op1=0,sum=0;
        for(int i=j;i<b;i++){
            sum+=a[i];
            if(sum==0)
                op1=Math.max(op1,1+find(a,i+1));
            else
                op1=Math.max(op1,find(a,i+1));
        }
        st[j]=op1;
        return op1;
    }
}

**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 and scenarios that are likely to cause issues. Here are the key points to consider:

1. **Edge Cases:**
   - Arrays with a single element.
   - Arrays with all elements being zero.
   - Arrays with a mix of positive and negative elements that sum to zero in multiple segments.

2. **Performance Issues:**
   - Large input sizes that might cause performance issues due to the nested loops in the `find` method.

3. **Logical Errors:**
   - The code uses a dynamic programming approach but might have issues with the way it updates the `st` array.

Given these points, let's create a small-scale test input that includes a mix of these scenarios:

```plaintext
3
1
0
3
0 0 0
6
1 -1 1 -1 1 -1
```

### Explanation:
1. **Test Case 1:**
   - Input: `1 0`
   - Expected Output: `1`
   - This tests the case where the array has a single element that is zero.

2. **Test Case 2:**
   - Input: `3 0 0 0`
   - Expected Output: `3`
   - This tests the case where the array has multiple zeros, which should result in multiple beautiful segments.

3. **Test Case 3:**
   - Input: `6 1 -1 1 -1 1 -1`
   - Expected Output: `3`
   - This tests the case where the array has alternating positive and negative elements that sum to zero in multiple segments.

These test cases should help expose any issues in the buggy code, particularly in how it handles multiple beautiful segments and the edge case of a single element.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kousuke's Assignment
time_limit_ms: 2000
memory_limit_mb: 256
Description: After a trip with Sakurako, Kousuke was very scared because he forgot about his programming assignment. In this assignment, the teacher gave him an array $a$ of $n$ integers and asked him to calculate the number of non- overlapping segments of the array $a$, such that each segment is considered beautiful.

A segment $[l,r]$ is considered beautiful if $a_l + a_{l+1} + \dots + a_{r-1} + a_r=0$.

For a fixed array $a$, your task is to compute the maximum number of non- overlapping beautiful segments.

Input: The first line of input contains the number $t$ ($1 \le t \le 10^4$) — the number of test cases. Each test case consists of $2$ lines.

  * The first line contains one integer $n$ ($1 \le n \le 10^5$) — the length of the array.   * The second line contains $n$ integers $a_i$ ($-10^5 \le a_i \le 10^5$) — the elements of the array $a$.

It is guaranteed that the sum of $n$ across all test cases does not exceed $3\cdot 10^5$.

Output: For each test case, output a single integer: the maximum number of non- overlapping beautiful segments.

Examples:
input:
3
5
2 1 -3 2 1
7
12 -4 4 43 -3 -5 8
6
0 -4 0 3 0 1
output:
1
2
3

Note:

Buggy Code:
#include <iostream>
#include <deque>
#include <stack>
#include <algorithm>
#include <map>
#include <cmath>
#include <climits>
#include <vector>

using namespace std;

int main()
{
	int tt; cin >> tt;
	int n;
	int temp;
	while (tt--)
	{
		int cnt = 0;
		int sum = 0;
		cin >> n;
		map<int, int> mp;
		for (int i = 0; i < n; i++)
		{
			cin >> temp;
			if (temp == 0)
			{
				cnt++;
				mp.clear();
				sum = 0;
				continue;
			}
			sum += temp;
			if (mp[sum] == 1)
			{
				cnt++;
				mp.clear();
				sum = 0;
				continue;
			}

			if (sum == 0)
			{
				cnt++;
				mp.clear();
				sum = 0;
				continue;
			}

			else mp[sum] = 1;

		}
		cout << cnt << "\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:
To create a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can highlight the issues in the logic. The primary issues in the buggy code are:

1. **Incorrect Handling of Zero Sums**: The code incorrectly resets the map and sum when a zero sum is found, which can lead to missing valid segments.
2. **Incorrect Map Usage**: The map is used to track the sums, but it is cleared prematurely, which can cause valid segments to be missed.

Given these issues, we should create a test case that includes:

- Multiple zero values.
- Segments that sum to zero but are not detected due to the map being cleared prematurely.
- Overlapping segments that should not be counted as valid.

A small-scale test input that can expose these issues is:

```plaintext
3
5
0 0 0 0 0
8
1 -1 1 -1 1 -1 1 -1
6
1 -1 1 0 -1 1
```

Explanation:
1. **Test Case 1 (5 0 0 0 0 0)**:
   - The array consists entirely of zeros. The correct output should be 5, but the buggy code will likely reset the map and sum prematurely, leading to an incorrect result.

2. **Test Case 2 (8 1 -1 1 -1 1 -1 1 -1)**:
   - The array has multiple segments that sum to zero (e.g., [1, -1], [1, -1], etc.). The correct output should be 4, but the buggy code may miss some segments due to the map being cleared prematurely.

3. **Test Case 3 (6 1 -1 1 0 -1 1)**:
   - The array includes a zero and multiple segments that sum to zero. The correct output should be 3, but the buggy code may miss some segments due to the map being cleared prematurely.

This test input should effectively expose the bugs in the provided code.