**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Deterministic Heap (Easy Version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the easy version of the problem. The difference between the two versions is the definition of deterministic max-heap, time limit, and constraints on $n$ and $t$. You can make hacks only if both versions of the problem are solved.

Consider a perfect binary tree with size $2^n - 1$, with nodes numbered from $1$ to $2^n-1$ and rooted at $1$. For each vertex $v$ ($1 \le v \le 2^{n - 1} - 1$), vertex $2v$ is its left child and vertex $2v + 1$ is its right child. Each node $v$ also has a value $a_v$ assigned to it.

Define the operation $\mathrm{pop}$ as follows:

  1. initialize variable $v$ as $1$;    2. repeat the following process until vertex $v$ is a leaf (i.e. until $2^{n - 1} \le v \le 2^n - 1$);      1. among the children of $v$, choose the one with the larger value on it and denote such vertex as $x$; if the values on them are equal (i.e. $a_{2v} = a_{2v + 1}$), you can choose any of them;      2. assign $a_x$ to $a_v$ (i.e. $a_v := a_x$);      3. assign $x$ to $v$ (i.e. $v := x$);    3. assign $-1$ to $a_v$ (i.e. $a_v := -1$).

Then we say the $\mathrm{pop}$ operation is deterministic if there is a unique way to do such operation. In other words, $a_{2v} \neq a_{2v + 1}$ would hold whenever choosing between them.

A binary tree is called a max-heap if for every vertex $v$ ($1 \le v \le 2^{n - 1} - 1$), both $a_v \ge a_{2v}$ and $a_v \ge a_{2v + 1}$ hold.

A max-heap is deterministic if the $\mathrm{pop}$ operation is deterministic to the heap when we do it for the first time.

Initially, $a_v := 0$ for every vertex $v$ ($1 \le v \le 2^n - 1$), and your goal is to count the number of different deterministic max- heaps produced by applying the following operation $\mathrm{add}$ exactly $k$ times:

  * Choose an integer $v$ ($1 \le v \le 2^n - 1$) and, for every vertex $x$ on the path between $1$ and $v$, add $1$ to $a_x$.

Two heaps are considered different if there is a node which has different values in the heaps.

Since the answer might be large, print it modulo $p$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains three integers $n, k, p$ ($1 \le n, k \le 500$, $10^8 \le p \le 10^9$, $p$ is a prime).

It is guaranteed that the sum of $n$ and the sum of $k$ over all test cases does not exceed $500$.

Output: For each test case, output a single line containing an integer: the number of different deterministic max-heaps produced by applying the aforementioned operation $\mathrm{add}$ exactly $k$ times, modulo $p$.

Examples:
input:
7
1 13 998244353
2 1 998244353
3 2 998244853
3 3 998244353
3 4 100000037
4 2 100000039
4 3 100000037
output:
1
2
12
52
124
32
304

input:
1
500 500 100000007
output:
76297230

input:
6
87 63 100000037
77 77 100000039
100 200 998244353
200 100 998244353
32 59 998244853
1 1 998244353
output:
26831232
94573603
37147649
847564946
727060898
1

Note: For the first testcase, there is only one way to generate $a$, and such sequence is a deterministic max-heap, so the answer is $1$.

For the second testcase, if we choose $v = 1$ and do the operation, we would have $a = [1, 0, 0]$, and since $a_2 = a_3$, we can choose either of them when doing the first $\mathrm{pop}$ operation, so such heap is not a deterministic max-heap.

And if we choose $v = 2$, we would have $a = [1, 1, 0]$, during the first $\mathrm{pop}$, the following would happen:

  * initialize $v$ as $1$    * since $a_{2v} > a_{2v + 1}$, choose $2v$ as $x$, then $x = 2$    * assign $a_x$ to $a_v$, then $a = [1, 1, 0]$    * assign $x$ to $v$, then $v = 2$    * since $v$ is a leaf, assign $-1$ to $a_v$, then $a = [1, -1, 0]$

Since the first $\mathrm{pop}$ operation is deterministic, this is a deterministic max-heap. Also, if we choose $v = 3$, $a$ would be a deterministic max-heap, so the answer is $2$.

Buggy Code:
#include<bits/stdc++.h>
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,popcnt,lzcnt")
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;
                            // comparator
template<typename T>
using pbds = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;// find_by_order, order_of_key
using ll = long long;
using ull = unsigned long long;
using lld = long double;
#define all(a) (a).begin(), (a).end()
#define sz(a) static_cast<int> ((a).size())
#define pqg priority_queue<ll,vector<ll>>
#define pql priority_queue<ll, vector<ll>, greater<ll>>
#define ff first
#define ss second
#ifndef ONLINE_JUDGE
#define debug(x) cerr << #x <<" "; _print(x); cerr << endl;
#else
#define debug(x)
#endif
void _print(ll t) {cerr << t;}
void _print(int t) {cerr << t;}
void _print(string t) {cerr << t;}
void _print(char t) {cerr << t;}
void _print(lld t) {cerr << t;}
void _print(double t) {cerr << t;}
void _print(ull t) {cerr << t;}
template <class T, class V> void _print(pair <T, V> p);
template <class T> void _print(vector <T> v);
template <class T> void _print(set <T> v);
template <class T, class V> void _print(map <T, V> v);
template <class T> void _print(multiset <T> v);
template <class T, class V> void _print(pair <T, V> p) {cerr << "{"; _print(p.ff); cerr << ","; _print(p.ss); cerr << "}";}
template <class T> void _print(vector <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T,unsigned size> void _print(array <T,size> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(set <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(multiset <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T, class V> void _print(map <T, V> v) {cerr << "[ "; for (auto i : v) {_print(i); cerr << " ";} cerr << "]";}
const ll mod1 = 1e9+7, mod2 = 998244353;
const ll INF = 1e9, INF2 = 1e18;
const int mx = 2e5+10;
int dp[505][505][2];
int pf[505][2];
void solve()
{
    int n,k,p;
    cin >> n >> k >> p;
    // we have to produce deterministic max heaps
    // op -> always a max heap
    // 0 -> deterministic
    for (int i = 0; i < n+1; i++)
    {
        for (int j = 0; j < k+1; j++)
        {
            dp[i][j][0]=0;
            dp[i][j][1]=0;
        }

    }

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

    for (int i = 1; i < n; i++)
    {
        // build pf on previous level
        for (int j = 0; j < k+1; j++)
        {
            pf[j][0]=0;pf[j][1]=0;
        }

        pf[0][0] = dp[i-1][0][0];
        pf[0][1] = dp[i-1][0][1];
        for (int j = 1; j <=k; j++)
        {
            pf[j][0] = (pf[j-1][0] + dp[i-1][j][0]) % p;
            pf[j][1] = (pf[j-1][1] + dp[i-1][j][1]) % p;
        }
        for (int j = 0; j <=k; j++)
        {
            for (int x = 0; x <=j; x++)
            {
                dp[i][j][1] += 1ll*(dp[i-1][x][1] * pf[j-x][1])%p;
                dp[i][j][1] %= p;
                if(x == 0) continue;
                // deterministic with j operations
                // for left child
                dp[i][j][0] += 1ll*(dp[i-1][x][0] * pf[min(x-1<0?0:x-1,j-x)][1])%p;
                dp[i][j][0] %= p;
                // for right child
                dp[i][j][0] += 1ll*(dp[i-1][x][0] * pf[min(x-1<0?0:x-1,j-x)][1])%p;
                dp[i][j][0] %= p;

            }
        }

    }
    cout << dp[n-1][k][0] << endl;
}
int32_t main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(NULL);
    #ifndef ONLINE_JUDGE
        freopen("input.txt","r",stdin);
        freopen("output.txt","w",stdout);
        freopen("Error.txt", "w", stderr);
    #endif
    int t=1;
    cin >> t;
    while(t--) solve();

    // for(int i=1;i<=t;i++)
    // {
    //     cout << "Case #" << i << ": " << (solve()) << endl;
    // }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
3 3 100000037
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Deterministic Heap (Easy Version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the easy version of the problem. The difference between the two versions is the definition of deterministic max-heap, time limit, and constraints on $n$ and $t$. You can make hacks only if both versions of the problem are solved.

Consider a perfect binary tree with size $2^n - 1$, with nodes numbered from $1$ to $2^n-1$ and rooted at $1$. For each vertex $v$ ($1 \le v \le 2^{n - 1} - 1$), vertex $2v$ is its left child and vertex $2v + 1$ is its right child. Each node $v$ also has a value $a_v$ assigned to it.

Define the operation $\mathrm{pop}$ as follows:

  1. initialize variable $v$ as $1$;    2. repeat the following process until vertex $v$ is a leaf (i.e. until $2^{n - 1} \le v \le 2^n - 1$);      1. among the children of $v$, choose the one with the larger value on it and denote such vertex as $x$; if the values on them are equal (i.e. $a_{2v} = a_{2v + 1}$), you can choose any of them;      2. assign $a_x$ to $a_v$ (i.e. $a_v := a_x$);      3. assign $x$ to $v$ (i.e. $v := x$);    3. assign $-1$ to $a_v$ (i.e. $a_v := -1$).

Then we say the $\mathrm{pop}$ operation is deterministic if there is a unique way to do such operation. In other words, $a_{2v} \neq a_{2v + 1}$ would hold whenever choosing between them.

A binary tree is called a max-heap if for every vertex $v$ ($1 \le v \le 2^{n - 1} - 1$), both $a_v \ge a_{2v}$ and $a_v \ge a_{2v + 1}$ hold.

A max-heap is deterministic if the $\mathrm{pop}$ operation is deterministic to the heap when we do it for the first time.

Initially, $a_v := 0$ for every vertex $v$ ($1 \le v \le 2^n - 1$), and your goal is to count the number of different deterministic max- heaps produced by applying the following operation $\mathrm{add}$ exactly $k$ times:

  * Choose an integer $v$ ($1 \le v \le 2^n - 1$) and, for every vertex $x$ on the path between $1$ and $v$, add $1$ to $a_x$.

Two heaps are considered different if there is a node which has different values in the heaps.

Since the answer might be large, print it modulo $p$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains three integers $n, k, p$ ($1 \le n, k \le 500$, $10^8 \le p \le 10^9$, $p$ is a prime).

It is guaranteed that the sum of $n$ and the sum of $k$ over all test cases does not exceed $500$.

Output: For each test case, output a single line containing an integer: the number of different deterministic max-heaps produced by applying the aforementioned operation $\mathrm{add}$ exactly $k$ times, modulo $p$.

Examples:
input:
7
1 13 998244353
2 1 998244353
3 2 998244853
3 3 998244353
3 4 100000037
4 2 100000039
4 3 100000037
output:
1
2
12
52
124
32
304

input:
1
500 500 100000007
output:
76297230

input:
6
87 63 100000037
77 77 100000039
100 200 998244353
200 100 998244353
32 59 998244853
1 1 998244353
output:
26831232
94573603
37147649
847564946
727060898
1

Note: For the first testcase, there is only one way to generate $a$, and such sequence is a deterministic max-heap, so the answer is $1$.

For the second testcase, if we choose $v = 1$ and do the operation, we would have $a = [1, 0, 0]$, and since $a_2 = a_3$, we can choose either of them when doing the first $\mathrm{pop}$ operation, so such heap is not a deterministic max-heap.

And if we choose $v = 2$, we would have $a = [1, 1, 0]$, during the first $\mathrm{pop}$, the following would happen:

  * initialize $v$ as $1$    * since $a_{2v} > a_{2v + 1}$, choose $2v$ as $x$, then $x = 2$    * assign $a_x$ to $a_v$, then $a = [1, 1, 0]$    * assign $x$ to $v$, then $v = 2$    * since $v$ is a leaf, assign $-1$ to $a_v$, then $a = [1, -1, 0]$

Since the first $\mathrm{pop}$ operation is deterministic, this is a deterministic max-heap. Also, if we choose $v = 3$, $a$ would be a deterministic max-heap, so the answer is $2$.

Buggy Code:
// #pragma GCC optimize(2, "Ofast", "inline")
#include <bits/stdc++.h>
using namespace std;
// #define int long long
#define ll long long
typedef pair<int, int> pii;
#define endl "\n"
int mod = 998244353; //(ll)1e18,double浮点数只有15位，1e15后面+的小数会被省略
const int N = 3e5 + 10;

map<array<int, 3>, int> mp;
int cal(int n, int k, int fl)
{
  if (mp.count({n, k, fl})) return mp[{n, k, fl}];
  if (n<= 1)
    {
      return mp[{n, k, fl}] = 1;
    }
  int ans = 0;
  for (int j = 0; j <= k; ++j)
    for (int i = 0; i <= k - j; ++i)
      {
        int a = i, b = k - i - j;
        if (a < b) swap(a, b);
        if (fl)
          {
            if (a == b) continue;
            ans = (ans + cal(n - 1, a, 1) * 1ll *
                             cal(n - 1, b, 0) % mod) %
                  mod;
          }
        else
          {
            ans = (ans + cal(n - 1, a, 0) * 1ll *
                             cal(n - 1, b, 0) % mod) %
                  mod;
          }
      }
  return mp[{n, k, fl}] = ans;
}

int n, k;
void mysolve()
{
  cin >> n >> k >> mod;
  if (n == 1) return cout << 1 << endl, void();
  mp.clear();
  int ans = cal(n, k, 1);
  cout << ans << endl;
}

signed main()
{
#ifdef DEBUG
  freopen("in.in", "r", stdin);
  freopen("out.out", "w", stdout);
#endif
  std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0); // 使用read请把解绑注释了
  int t = 1;
  cin >> t;
  while (t--)
    {
      mysolve();
    }
  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
2 2 998244353
```

### Explanation:
This test case is designed to expose potential issues in the buggy code, especially in the handling of small-scale inputs where the structure of the heap is simple and the number of operations is limited.

1. **Test Case Description:**
   - **n = 2**: The tree has 3 nodes (1, 2, 3).
   - **k = 2**: We perform the `add` operation 2 times.
   - **p = 998244353**: The modulo value.

2. **Expected Behavior:**
   - The tree structure is as follows:
     - Node 1 (root) with children 2 and 3.
   - Possible `add` operations:
     - `add(1)`: This increments the values of nodes 1, 2, and 3.
     - `add(2)`: This increments the values of nodes 1 and 2.
     - `add(3)`: This increments the values of nodes 1 and 3.

3. **Possible Heaps:**
   - After 2 `add` operations, the possible values for the nodes are:
     - `[2, 1, 1]` (if `add(1)` and `add(2)` or `add(1)` and `add(3)` are performed)
     - `[1, 2, 1]` (if `add(2)` and `add(3)` are performed)
   - Both of these heaps are deterministic max-heaps because there are no equal values at the same level that would cause a non-deterministic choice during the `pop` operation.

4. **Expected Output:**
   - The number of different deterministic max-heaps is 2.

This test case is small and manageable, making it easy to verify the correctness of the solution. It also ensures that the code correctly handles the deterministic condition and the counting of different heaps.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Deterministic Heap (Easy Version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the easy version of the problem. The difference between the two versions is the definition of deterministic max-heap, time limit, and constraints on $n$ and $t$. You can make hacks only if both versions of the problem are solved.

Consider a perfect binary tree with size $2^n - 1$, with nodes numbered from $1$ to $2^n-1$ and rooted at $1$. For each vertex $v$ ($1 \le v \le 2^{n - 1} - 1$), vertex $2v$ is its left child and vertex $2v + 1$ is its right child. Each node $v$ also has a value $a_v$ assigned to it.

Define the operation $\mathrm{pop}$ as follows:

  1. initialize variable $v$ as $1$;    2. repeat the following process until vertex $v$ is a leaf (i.e. until $2^{n - 1} \le v \le 2^n - 1$);      1. among the children of $v$, choose the one with the larger value on it and denote such vertex as $x$; if the values on them are equal (i.e. $a_{2v} = a_{2v + 1}$), you can choose any of them;      2. assign $a_x$ to $a_v$ (i.e. $a_v := a_x$);      3. assign $x$ to $v$ (i.e. $v := x$);    3. assign $-1$ to $a_v$ (i.e. $a_v := -1$).

Then we say the $\mathrm{pop}$ operation is deterministic if there is a unique way to do such operation. In other words, $a_{2v} \neq a_{2v + 1}$ would hold whenever choosing between them.

A binary tree is called a max-heap if for every vertex $v$ ($1 \le v \le 2^{n - 1} - 1$), both $a_v \ge a_{2v}$ and $a_v \ge a_{2v + 1}$ hold.

A max-heap is deterministic if the $\mathrm{pop}$ operation is deterministic to the heap when we do it for the first time.

Initially, $a_v := 0$ for every vertex $v$ ($1 \le v \le 2^n - 1$), and your goal is to count the number of different deterministic max- heaps produced by applying the following operation $\mathrm{add}$ exactly $k$ times:

  * Choose an integer $v$ ($1 \le v \le 2^n - 1$) and, for every vertex $x$ on the path between $1$ and $v$, add $1$ to $a_x$.

Two heaps are considered different if there is a node which has different values in the heaps.

Since the answer might be large, print it modulo $p$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains three integers $n, k, p$ ($1 \le n, k \le 500$, $10^8 \le p \le 10^9$, $p$ is a prime).

It is guaranteed that the sum of $n$ and the sum of $k$ over all test cases does not exceed $500$.

Output: For each test case, output a single line containing an integer: the number of different deterministic max-heaps produced by applying the aforementioned operation $\mathrm{add}$ exactly $k$ times, modulo $p$.

Examples:
input:
7
1 13 998244353
2 1 998244353
3 2 998244853
3 3 998244353
3 4 100000037
4 2 100000039
4 3 100000037
output:
1
2
12
52
124
32
304

input:
1
500 500 100000007
output:
76297230

input:
6
87 63 100000037
77 77 100000039
100 200 998244353
200 100 998244353
32 59 998244853
1 1 998244353
output:
26831232
94573603
37147649
847564946
727060898
1

Note: For the first testcase, there is only one way to generate $a$, and such sequence is a deterministic max-heap, so the answer is $1$.

For the second testcase, if we choose $v = 1$ and do the operation, we would have $a = [1, 0, 0]$, and since $a_2 = a_3$, we can choose either of them when doing the first $\mathrm{pop}$ operation, so such heap is not a deterministic max-heap.

And if we choose $v = 2$, we would have $a = [1, 1, 0]$, during the first $\mathrm{pop}$, the following would happen:

  * initialize $v$ as $1$    * since $a_{2v} > a_{2v + 1}$, choose $2v$ as $x$, then $x = 2$    * assign $a_x$ to $a_v$, then $a = [1, 1, 0]$    * assign $x$ to $v$, then $v = 2$    * since $v$ is a leaf, assign $-1$ to $a_v$, then $a = [1, -1, 0]$

Since the first $\mathrm{pop}$ operation is deterministic, this is a deterministic max-heap. Also, if we choose $v = 3$, $a$ would be a deterministic max-heap, so the answer is $2$.

Buggy Code:
def solve():
    n, k, p = map(int, input().split())
    dp1 = [1]*(k+1)
    dp2 = [1]*(k+1)
    dp3 = [0]
    for a in dp2:
        dp3.append(dp3[-1]+a)
    for i in range(n-1):
        ndp1 = [0]*(k+1)
        ndp2 = [0]*(k+1)
        ndp3 = [0]
        ndp1[0] = ndp2[0] = 1
        for j in range(1, k+1):
            for a in range(j+1):
                ndp1[j] = (ndp1[j]+2*dp1[a]*dp3[min(a, j-a+1)])%p
                ndp2[j] = (ndp2[j]+dp2[a]*dp3[j-a+1])%p
        for a in ndp2:
            ndp3.append((ndp3[-1]+a)%p)
        dp1, dp2, dp3 = ndp1, ndp2, ndp3
    return dp1[-1]%p

import sys
input = lambda: sys.stdin.readline().rstrip()
t = int(input())
for i in range(t):
    print(solve())

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

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

Think step by step.

Response:
```plaintext
1
3 3 998244353
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Deterministic Heap (Easy Version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the easy version of the problem. The difference between the two versions is the definition of deterministic max-heap, time limit, and constraints on $n$ and $t$. You can make hacks only if both versions of the problem are solved.

Consider a perfect binary tree with size $2^n - 1$, with nodes numbered from $1$ to $2^n-1$ and rooted at $1$. For each vertex $v$ ($1 \le v \le 2^{n - 1} - 1$), vertex $2v$ is its left child and vertex $2v + 1$ is its right child. Each node $v$ also has a value $a_v$ assigned to it.

Define the operation $\mathrm{pop}$ as follows:

  1. initialize variable $v$ as $1$;    2. repeat the following process until vertex $v$ is a leaf (i.e. until $2^{n - 1} \le v \le 2^n - 1$);      1. among the children of $v$, choose the one with the larger value on it and denote such vertex as $x$; if the values on them are equal (i.e. $a_{2v} = a_{2v + 1}$), you can choose any of them;      2. assign $a_x$ to $a_v$ (i.e. $a_v := a_x$);      3. assign $x$ to $v$ (i.e. $v := x$);    3. assign $-1$ to $a_v$ (i.e. $a_v := -1$).

Then we say the $\mathrm{pop}$ operation is deterministic if there is a unique way to do such operation. In other words, $a_{2v} \neq a_{2v + 1}$ would hold whenever choosing between them.

A binary tree is called a max-heap if for every vertex $v$ ($1 \le v \le 2^{n - 1} - 1$), both $a_v \ge a_{2v}$ and $a_v \ge a_{2v + 1}$ hold.

A max-heap is deterministic if the $\mathrm{pop}$ operation is deterministic to the heap when we do it for the first time.

Initially, $a_v := 0$ for every vertex $v$ ($1 \le v \le 2^n - 1$), and your goal is to count the number of different deterministic max- heaps produced by applying the following operation $\mathrm{add}$ exactly $k$ times:

  * Choose an integer $v$ ($1 \le v \le 2^n - 1$) and, for every vertex $x$ on the path between $1$ and $v$, add $1$ to $a_x$.

Two heaps are considered different if there is a node which has different values in the heaps.

Since the answer might be large, print it modulo $p$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains three integers $n, k, p$ ($1 \le n, k \le 500$, $10^8 \le p \le 10^9$, $p$ is a prime).

It is guaranteed that the sum of $n$ and the sum of $k$ over all test cases does not exceed $500$.

Output: For each test case, output a single line containing an integer: the number of different deterministic max-heaps produced by applying the aforementioned operation $\mathrm{add}$ exactly $k$ times, modulo $p$.

Examples:
input:
7
1 13 998244353
2 1 998244353
3 2 998244853
3 3 998244353
3 4 100000037
4 2 100000039
4 3 100000037
output:
1
2
12
52
124
32
304

input:
1
500 500 100000007
output:
76297230

input:
6
87 63 100000037
77 77 100000039
100 200 998244353
200 100 998244353
32 59 998244853
1 1 998244353
output:
26831232
94573603
37147649
847564946
727060898
1

Note: For the first testcase, there is only one way to generate $a$, and such sequence is a deterministic max-heap, so the answer is $1$.

For the second testcase, if we choose $v = 1$ and do the operation, we would have $a = [1, 0, 0]$, and since $a_2 = a_3$, we can choose either of them when doing the first $\mathrm{pop}$ operation, so such heap is not a deterministic max-heap.

And if we choose $v = 2$, we would have $a = [1, 1, 0]$, during the first $\mathrm{pop}$, the following would happen:

  * initialize $v$ as $1$    * since $a_{2v} > a_{2v + 1}$, choose $2v$ as $x$, then $x = 2$    * assign $a_x$ to $a_v$, then $a = [1, 1, 0]$    * assign $x$ to $v$, then $v = 2$    * since $v$ is a leaf, assign $-1$ to $a_v$, then $a = [1, -1, 0]$

Since the first $\mathrm{pop}$ operation is deterministic, this is a deterministic max-heap. Also, if we choose $v = 3$, $a$ would be a deterministic max-heap, so the answer is $2$.

Buggy Code:
import sys
import threading

def main():
    import sys
    import math

    sys.setrecursionlimit(1 << 25)
    t = int(sys.stdin.readline())
    for _ in range(t):
        n, k, p = map(int, sys.stdin.readline().split())

        # Precompute factorial up to k
        fact = [1] * (k +1)
        for i in range(1, k +1):
            fact[i] = fact[i -1] * i %p
        # Compute inverse factorial
        inv_fact = [1] * (k +1)
        inv_fact[k] = pow(fact[k], p -2, p)
        for i in range(k -1, -1, -1):
            inv_fact[i] = inv_fact[i +1] * (i +1) %p

        # Precompute power2
        power2 = [1] * (n +2)
        for h in range(1, n +1):
            power2[h] = power2[h -1] *2 %p

        # Function to compute C(m,k} mod p
        def comb_mod_p(m, a, p):
            if a ==0:
                return 1
            res =1
            for i in range(a):
                term = (m -i) %p
                if term ==0:
                    return 0
                res = res * term %p
            res = res * inv_fact[a] %p
            return res

        # Initialize f_prev for h=1
        f_prev = [1] * (k +1)

        for h in range(2, n +1):
            f_new = [0] * (k +1)
            for current_k in range(0, k +1):
                sum_total =0
                for c_root in range(0, current_k +1):
                    rem = current_k -c_root
                    if rem <1:
                        continue
                    a_right_max = (rem -1) //2
                    for a_right in range(0, a_right_max +1):
                        a_left = rem -a_right
                        if a_left <=a_right:
                            continue
                        m = (power2[h -1] + a_right -2) %p
                        if power2[h -1] + a_right -2 <0:
                            m +=p
                        c = comb_mod_p(power2[h -1] +a_right -2, a_right, p)
                        if c ==0:
                            continue
                        sum_total = (sum_total + f_prev[a_left] * c) %p
                f_new[current_k] = (2 * sum_total) %p
            f_prev = f_new
        if n ==1:
            print(1 if k >=0 else 0)
        else:
            print(f_prev[k] if n >=1 else 0)

threading.Thread(target=main,).start()

**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 1 998244353
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Deterministic Heap (Easy Version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the easy version of the problem. The difference between the two versions is the definition of deterministic max-heap, time limit, and constraints on $n$ and $t$. You can make hacks only if both versions of the problem are solved.

Consider a perfect binary tree with size $2^n - 1$, with nodes numbered from $1$ to $2^n-1$ and rooted at $1$. For each vertex $v$ ($1 \le v \le 2^{n - 1} - 1$), vertex $2v$ is its left child and vertex $2v + 1$ is its right child. Each node $v$ also has a value $a_v$ assigned to it.

Define the operation $\mathrm{pop}$ as follows:

  1. initialize variable $v$ as $1$;    2. repeat the following process until vertex $v$ is a leaf (i.e. until $2^{n - 1} \le v \le 2^n - 1$);      1. among the children of $v$, choose the one with the larger value on it and denote such vertex as $x$; if the values on them are equal (i.e. $a_{2v} = a_{2v + 1}$), you can choose any of them;      2. assign $a_x$ to $a_v$ (i.e. $a_v := a_x$);      3. assign $x$ to $v$ (i.e. $v := x$);    3. assign $-1$ to $a_v$ (i.e. $a_v := -1$).

Then we say the $\mathrm{pop}$ operation is deterministic if there is a unique way to do such operation. In other words, $a_{2v} \neq a_{2v + 1}$ would hold whenever choosing between them.

A binary tree is called a max-heap if for every vertex $v$ ($1 \le v \le 2^{n - 1} - 1$), both $a_v \ge a_{2v}$ and $a_v \ge a_{2v + 1}$ hold.

A max-heap is deterministic if the $\mathrm{pop}$ operation is deterministic to the heap when we do it for the first time.

Initially, $a_v := 0$ for every vertex $v$ ($1 \le v \le 2^n - 1$), and your goal is to count the number of different deterministic max- heaps produced by applying the following operation $\mathrm{add}$ exactly $k$ times:

  * Choose an integer $v$ ($1 \le v \le 2^n - 1$) and, for every vertex $x$ on the path between $1$ and $v$, add $1$ to $a_x$.

Two heaps are considered different if there is a node which has different values in the heaps.

Since the answer might be large, print it modulo $p$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains three integers $n, k, p$ ($1 \le n, k \le 500$, $10^8 \le p \le 10^9$, $p$ is a prime).

It is guaranteed that the sum of $n$ and the sum of $k$ over all test cases does not exceed $500$.

Output: For each test case, output a single line containing an integer: the number of different deterministic max-heaps produced by applying the aforementioned operation $\mathrm{add}$ exactly $k$ times, modulo $p$.

Examples:
input:
7
1 13 998244353
2 1 998244353
3 2 998244853
3 3 998244353
3 4 100000037
4 2 100000039
4 3 100000037
output:
1
2
12
52
124
32
304

input:
1
500 500 100000007
output:
76297230

input:
6
87 63 100000037
77 77 100000039
100 200 998244353
200 100 998244353
32 59 998244853
1 1 998244353
output:
26831232
94573603
37147649
847564946
727060898
1

Note: For the first testcase, there is only one way to generate $a$, and such sequence is a deterministic max-heap, so the answer is $1$.

For the second testcase, if we choose $v = 1$ and do the operation, we would have $a = [1, 0, 0]$, and since $a_2 = a_3$, we can choose either of them when doing the first $\mathrm{pop}$ operation, so such heap is not a deterministic max-heap.

And if we choose $v = 2$, we would have $a = [1, 1, 0]$, during the first $\mathrm{pop}$, the following would happen:

  * initialize $v$ as $1$    * since $a_{2v} > a_{2v + 1}$, choose $2v$ as $x$, then $x = 2$    * assign $a_x$ to $a_v$, then $a = [1, 1, 0]$    * assign $x$ to $v$, then $v = 2$    * since $v$ is a leaf, assign $-1$ to $a_v$, then $a = [1, -1, 0]$

Since the first $\mathrm{pop}$ operation is deterministic, this is a deterministic max-heap. Also, if we choose $v = 3$, $a$ would be a deterministic max-heap, so the answer is $2$.

Buggy Code:
import java.io.*;
import java.util.*;

public class Solution {
    static final int DEFAULT_MOD = 1000000007; // Assuming a default MOD value

    public static void solveProblem(BufferedReader reader) throws IOException {
        String[] inputs = reader.readLine().split(" ");
        int levelCount = Integer.parseInt(inputs[0]);
        int maxTotal = Integer.parseInt(inputs[1]);
        int moduloValue = Integer.parseInt(inputs[2]);

        Vector<Integer> validSequences = new Vector<>(maxTotal + 1);
        Vector<Integer> invalidSequences = new Vector<>(maxTotal + 1);
        for (int i = 0; i <= maxTotal; i++) {
            validSequences.add(1);
            invalidSequences.add(1);
        }

        for (int currentLevel = 2; currentLevel <= levelCount; currentLevel++) {
            Vector<Integer> newValidSequences = new Vector<>(Collections.nCopies(maxTotal + 1, 0));
            Vector<Integer> newInvalidSequences = new Vector<>(Collections.nCopies(maxTotal + 1, 0));

            for (int firstSum = 0; firstSum <= maxTotal; firstSum++) {
                for (int secondSum = 0; secondSum <= maxTotal; secondSum++) {
                    if (firstSum + secondSum > maxTotal)
                        break;
                    newInvalidSequences.set(firstSum + secondSum,
                        (newInvalidSequences.get(firstSum + secondSum) +
                        (int)((long)invalidSequences.get(firstSum) * invalidSequences.get(secondSum) % moduloValue)) % moduloValue);

                    if (firstSum != secondSum) {
                        newValidSequences.set(firstSum + secondSum,
                            (newValidSequences.get(firstSum + secondSum) +
                            (int)((long)validSequences.get(Math.max(firstSum, secondSum)) *
                            invalidSequences.get(Math.min(firstSum, secondSum)) % moduloValue)) % moduloValue);
                    }
                }
            }

            for (int i = 1; i <= maxTotal; i++) {
                newValidSequences.set(i, (newValidSequences.get(i) + newValidSequences.get(i - 1)) % moduloValue);
                newInvalidSequences.set(i, (newInvalidSequences.get(i) + newInvalidSequences.get(i - 1)) % moduloValue);
            }

            validSequences = newValidSequences;
            invalidSequences = newInvalidSequences;
        }

        System.out.println(validSequences.get(maxTotal));
    }

    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        int testCount = Integer.parseInt(reader.readLine());

        while (testCount-- > 0) {
            solveProblem(reader);
        }
    }
}

**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 3 998244353
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Deterministic Heap (Easy Version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the easy version of the problem. The difference between the two versions is the definition of deterministic max-heap, time limit, and constraints on $n$ and $t$. You can make hacks only if both versions of the problem are solved.

Consider a perfect binary tree with size $2^n - 1$, with nodes numbered from $1$ to $2^n-1$ and rooted at $1$. For each vertex $v$ ($1 \le v \le 2^{n - 1} - 1$), vertex $2v$ is its left child and vertex $2v + 1$ is its right child. Each node $v$ also has a value $a_v$ assigned to it.

Define the operation $\mathrm{pop}$ as follows:

  1. initialize variable $v$ as $1$;    2. repeat the following process until vertex $v$ is a leaf (i.e. until $2^{n - 1} \le v \le 2^n - 1$);      1. among the children of $v$, choose the one with the larger value on it and denote such vertex as $x$; if the values on them are equal (i.e. $a_{2v} = a_{2v + 1}$), you can choose any of them;      2. assign $a_x$ to $a_v$ (i.e. $a_v := a_x$);      3. assign $x$ to $v$ (i.e. $v := x$);    3. assign $-1$ to $a_v$ (i.e. $a_v := -1$).

Then we say the $\mathrm{pop}$ operation is deterministic if there is a unique way to do such operation. In other words, $a_{2v} \neq a_{2v + 1}$ would hold whenever choosing between them.

A binary tree is called a max-heap if for every vertex $v$ ($1 \le v \le 2^{n - 1} - 1$), both $a_v \ge a_{2v}$ and $a_v \ge a_{2v + 1}$ hold.

A max-heap is deterministic if the $\mathrm{pop}$ operation is deterministic to the heap when we do it for the first time.

Initially, $a_v := 0$ for every vertex $v$ ($1 \le v \le 2^n - 1$), and your goal is to count the number of different deterministic max- heaps produced by applying the following operation $\mathrm{add}$ exactly $k$ times:

  * Choose an integer $v$ ($1 \le v \le 2^n - 1$) and, for every vertex $x$ on the path between $1$ and $v$, add $1$ to $a_x$.

Two heaps are considered different if there is a node which has different values in the heaps.

Since the answer might be large, print it modulo $p$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 500$). The description of the test cases follows.

The first line of each test case contains three integers $n, k, p$ ($1 \le n, k \le 500$, $10^8 \le p \le 10^9$, $p$ is a prime).

It is guaranteed that the sum of $n$ and the sum of $k$ over all test cases does not exceed $500$.

Output: For each test case, output a single line containing an integer: the number of different deterministic max-heaps produced by applying the aforementioned operation $\mathrm{add}$ exactly $k$ times, modulo $p$.

Examples:
input:
7
1 13 998244353
2 1 998244353
3 2 998244853
3 3 998244353
3 4 100000037
4 2 100000039
4 3 100000037
output:
1
2
12
52
124
32
304

input:
1
500 500 100000007
output:
76297230

input:
6
87 63 100000037
77 77 100000039
100 200 998244353
200 100 998244353
32 59 998244853
1 1 998244353
output:
26831232
94573603
37147649
847564946
727060898
1

Note: For the first testcase, there is only one way to generate $a$, and such sequence is a deterministic max-heap, so the answer is $1$.

For the second testcase, if we choose $v = 1$ and do the operation, we would have $a = [1, 0, 0]$, and since $a_2 = a_3$, we can choose either of them when doing the first $\mathrm{pop}$ operation, so such heap is not a deterministic max-heap.

And if we choose $v = 2$, we would have $a = [1, 1, 0]$, during the first $\mathrm{pop}$, the following would happen:

  * initialize $v$ as $1$    * since $a_{2v} > a_{2v + 1}$, choose $2v$ as $x$, then $x = 2$    * assign $a_x$ to $a_v$, then $a = [1, 1, 0]$    * assign $x$ to $v$, then $v = 2$    * since $v$ is a leaf, assign $-1$ to $a_v$, then $a = [1, -1, 0]$

Since the first $\mathrm{pop}$ operation is deterministic, this is a deterministic max-heap. Also, if we choose $v = 3$, $a$ would be a deterministic max-heap, so the answer is $2$.

Buggy Code:
#include<bits/stdc++.h>
#define MP make_pair
#define F first
#define S second

using namespace std;

typedef long long LL;
const int N = 509;
int n, k, p;

int main()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d %d %d", &n, &k, &p);
        vector <vector<int> > dp(n + 1, vector<int>(k + 1)), f(n + 1, vector<int>(k + 1));
        for (int i = 0; i <= k; i++) dp[1][i] = f[1][i] = 1;
        for (int i = 2; i <= n; i++)
        {
            for (int j = 0; j <= k; j++)
            {
                vector <int> sum(k + 1);
                sum[0] = 1;
                for (int l = 1; l <= j; l++)
                    sum[l] = (sum[l - 1] + f[i - 1][l]) % p;
                for (int l = 1; l <= j; l++)
                    dp[i][j] = (dp[i][j] + 1ll * sum[min(j - l, l - 1)] * dp[i - 1][l] % p) % p;
                dp[i][j] = dp[i][j] * 2 % p;
            }
            f[i][0] = 1;
            for (int j = 1; j <= k; j++)
            {
                vector <int> sum(k + 1);
                sum[0] = 1;
                for (int l = 1; l <= j; l++)
                    sum[l] = (sum[l - 1] + f[i - 1][l]) % p;
                for (int l = 1; l <= j; l++)
                    f[i][j] = (f[i][j] + 1ll * sum[min(j - l, l - 1)] * f[i - 1][l] % p) % p;
                f[i][j] = f[i][j] * 2 % p;
                for (int l = 0; l * 2 <= j; l++)
                    f[i][j] = (f[i][j] + 1ll * f[i - 1][l] * f[i - 1][l] % p) % p;
            }
        }
        printf("%d\n", dp[n][k]);
    }
    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
3 3 100000007
```


**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 <iostream>
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  int t, n, s, m, l = 0, r = 0;
  cin >> t;

  while (t--) {
    cin >> n >> s >> m;
    bool ans = 0;
    for (int i = 1; i <= n; ++i) {
      int lt, rt;
      cin >> lt >> rt;
      if (ans) continue;
      if (lt - r >= s) ans = 1;
      l = lt;
      r = rt;
    }
    if (m - r >= s) ans = 1;
    cout << (ans ? "YES\n" : "NO\n");
  }
  return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
5 3 15
0 2
3 5
6 8
9 11
12 14
```


**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 int
#define fst ios_base:: sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define PI 2*acos(0)
#define yes cout<<"YES"<<endl;
#define no cout<<"NO"<<endl;
#define ln "\n";
#define all(a) a.begin(), a.end()
#define allr(a) a.rbegin(), a.rend()
#define srt(v) sort(v.begin(),v.end());
#define rsrt(v) sort(v.rbegin(),v.rend());
#define rvs(v) reverse(v.begin(),v.end());
#define pb push_back
#define tst ll t;cin>>t;while(t--)
#define UNIQUE(X) (X).erase(unique(all(X)), (X).end())
#define distinct(v) v.resize(unique(v.begin(),v.end()) - v.begin());
#define ff first
#define ss second
#define mod 1000000007
#define gcd(a,b) __gcd(a,b)
#define lcm(a,b) (a*b)/gcd(a,b)
#define inf INT_MAX
/* ascii value
A=65,Z=90,a=97,z=122
Digit Count=log10(x)+1
s.rfind('B')->last occurance<->s.find('B')->first occurance
*/
using namespace std;
void solve(){

    ll n,s,m,k;
    cin>>n>>s>>m;
    ll ans=-1;
    while(n--)
    {
        ll l,r;
        cin>>l>>r;
        if(ans==-1)
        ans=l;
        ans=max(r-l,ans);
        ans=max(l-k,ans);
        k=r;

    }
    ans=max(ans,m-k);
    if(ans>=s)yes
    else no

}
signed main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
int tt = 1;
cin>>tt;
while (tt--)
solve();
}

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

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

Think step by step.

Response:
```plaintext
1
3 3 10
0 1
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:
#include<bits/stdc++.h>
using namespace std;
int main()
{
    int t;
    cin >> t;
    while(t--){
        int n, s, m;
        cin >> n >> s >> m;
        vector<int> l(n+1);
        vector<int> r(n);
        bool findAns = false;
        for(int i = 1; i <= n; i++){
            cin >> l[i] >> r[i];
        }
        l[n+1] = m;
        r[0] = 0;
        for(int i = 1; i <= n+1; i++){
            if(l[i] - r[i-1] >= s){
                cout << "YES" << endl;
                findAns = true;
                break;
            }
        }
        if(!findAns) cout << "NO" << endl;
    }
}

**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 5
5 10
```


**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:
N = int(input())
for _ in range(N):
    n, s, m = map(int,input().split())
    arr = [1 for __ in range(m)]
    for i in range(n):
        l, r = map(int,input().split())
        for j in range(l-1,r):
            arr[j] = 0
    flag = False
    cou = 0
    for i in arr:
        cou += 1
        if cou >= s:
            flag = True
            break
        if i == 0:
            cou = 0
    if flag:
        print("YES")
    else:
        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:
```plaintext
1
5 3 10
1 2
3 4
5 6
7 8
9 10
```


**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:
for _ in range(int(input())):
    n,s,m = map(int, input().split())
    flag=0
    init_x=0
    for i in range(n):
        x,y=map(int, input().split())
        if i==0:
            init_x=x
            check=y
        if x-check>=s:
            flag=1
            break
        check=y
    if init_x>=s or m-check>=s:
        flag=1
    if flag==0:
        print("NO")
    else :
        print("YES")

**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 code might incorrectly determine whether Alex can take a shower. The key points to consider are:

1. **Initial Free Time:** The code checks if the initial free time before the first task is long enough for Alex to take a shower.
2. **Free Time Between Tasks:** The code checks if there is a free interval of at least `s` minutes between consecutive tasks.
3. **Final Free Time:** The code checks if the free time after the last task is long enough for Alex to take a shower.

The bug in the code is that it only checks the free time between tasks and at the ends, but it does not correctly handle the case where the free time is exactly `s` minutes. Specifically, the condition `if x-check>=s:` should be `if x-check>s:` to ensure that the free time is strictly greater than `s`.

To create a test case that exposes this bug, we can use a scenario where the free time is exactly `s` minutes between tasks. Here is a small-scale test input that should fail the buggy code:

```plaintext
2
3 3 10
1 2
5 6
8 9
3 4 10
1 2
5 6
8 9
```

### Explanation:
1. **Test Case 1:**
   - `n = 3`, `s = 3`, `m = 10`
   - Tasks: (1, 2), (5, 6), (8, 9)
   - Free intervals: [0, 1), [2, 5), [6, 8), [9, 10)
   - The free interval [2, 5) is exactly 3 minutes long, but the buggy code will incorrectly output "NO" because it uses `>=` instead of `>`.
   - Correct output: "YES"

2. **Test Case 2:**
   - `n = 3`, `s = 4`, `m = 10`
   - Tasks: (1, 2), (5, 6), (8, 9)
   - Free intervals: [0, 1), [2, 5), [6, 8), [9, 10)
   - The free interval [2, 5) is exactly 3 minutes long, which is not enough for a 4-minute shower.
   - Correct output: "NO"

This test input should effectively expose the bug in the provided 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:
# tasks = [[3,5], [6,8], [9,10]]
def showering_possible(s, m, tasks):
    # print(tasks)
    # print(s)
    n = len(tasks)
    # task_list = [0 for _ in range(m)]
    for i in range(n):
        # for i in range(eachTask[0], eachTask[1]):
        #     task_list[i] = 1
        if i == 0 and tasks[i][0] - 0 >= s:
            return "YES"
        elif i < n-1 and tasks[i][0] - tasks[i-1][1] >= s:
            return "YES"
        elif i == n-1 and m - tasks[i][1] >= s:
            return "YES"
    return "NO"
    # print(task_list)
    # consec_0 = 0
    # for i in range(m):
    #     if task_list[i] == 0:
    #         consec_0 += 1
    #     if task_list[i] == 1:
    #         consec_0 = 0
    #     if consec_0 >= s:
    #         return "YES"
    # return "NO"

# print(sum_of_digits(77))
t = int(input())
while t:
    t -= 1
    n, s, m = [int(x) for x in input().split()]
    tasks = []
    while n:
        n -= 1
        tasks.append([int(x) for x in input().split()])
    print(showering_possible(s, m, tasks))
# print(showering_possible(3, 10, [[3,5], [6,8], [9,10]]))

**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 3 10
0 1
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.*;
import java.io.*;
public class C1999{
	static FastScanner fs = null;
	public static void main(String[] args) {
		fs = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		int t = fs.nextInt();
		while(t-- >0){
			int n = fs.nextInt();
			int s = fs.nextInt();
			int m = fs.nextInt();
			ArrayList<Pair> list = new ArrayList<>();
			for(int i = 0;i<n;i++){
				int x = fs.nextInt();
				int y = fs.nextInt();
				list.add(new Pair(x,y));
			}
			// Collections.sort(list,(o1,o2)->{
			// 	return o1.x-o2.x;
			// });
			// for(Pair p:list){
			// 	System.out.println(p.x+" -- "+p.y);
			// }
			boolean res = false;
			for(int i = 0;i<n;i++){
				Pair p = list.get(i);
				int x = p.x;
				int y = p.y;
				int dif = 0;
				if(i==0){
					dif = x;
				}
				if(i==n-1){
					dif = Math.max(dif,m-y);
				}
				if(i>0 &&i<(n-1)){
					dif = x-list.get(i-1).y;
				}
				if(dif>=s){
					res = true;
					break;
				}
			}
			if(res){
				out.println("YES");
			}
			else{
				out.println("NO");
			}
		}
		out.close();
	}
	static class Pair{
		int x;
		int y;
		Pair(int x,int y){
			this.x = x;
			this.y = y;
		}
	}
	static void sort(int[] a) {
		ArrayList<Integer> l=new ArrayList<>();
		for (int i:a) l.add(i);
		Collections.sort(l);
		for (int i=0; i<a.length; i++) a[i]=l.get(i);
	}
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
	}
}

**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 code, we need to focus on edge cases where the code might fail. The code seems to check for available time slots before and after each task, but it might miss some edge cases or handle them incorrectly.

### Step-by-Step Analysis

1. **Edge Case: No Tasks**:
   - If there are no tasks, the code should always return "YES" if `s` is less than or equal to `m`.

2. **Edge Case: Tasks Cover the Entire Day**:
   - If the tasks cover the entire day (from 0 to `m`), the code should return "NO".

3. **Edge Case: Single Task**:
   - If there is a single task, the code should check the time before and after the task.

4. **Edge Case: Multiple Tasks with Small Gaps**:
   - If there are multiple tasks with small gaps, the code should correctly identify if any gap is large enough for a shower.

5. **Edge Case: Large `m` and Small `s`**:
   - If `m` is large and `s` is small, the code should handle this efficiently.

### Challenging Test Input

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

```plaintext
4
0 3 10
3 3 10
1 2
3 5
6 7
2 2 10
1 2
8 9
3 5 10
1 2
3 5
6 8
```

### Explanation

1. **Test Case 1: No Tasks**:
   - `n = 0`, `s = 3`, `m = 10`
   - There are no tasks, so Alex can shower for 3 minutes.
   - Expected output: "YES"

2. **Test Case 2: Multiple Tasks with Small Gaps**:
   - `n = 3`, `s = 3`, `m = 10`
   - Tasks: (1, 2), (3, 5), (6, 7)
   - Gaps: 0-1, 2-3, 5-6, 7-10
   - The gap from 7 to 10 is 3 minutes, so Alex can shower.
   - Expected output: "YES"

3. **Test Case 3: Multiple Tasks with Small Gaps**:
   - `n = 2`, `s = 2`, `m = 10`
   - Tasks: (1, 2), (8, 9)
   - Gaps: 0-1, 2-8, 9-10
   - The gap from 2 to 8 is 6 minutes, so Alex can shower.
   - Expected output: "YES"

4. **Test Case 4: Tasks Cover the Entire Day**:
   - `n = 3`, `s = 5`, `m = 10`
   - Tasks: (1, 2), (3, 5), (6, 8)
   - Gaps: 0-1, 2-3, 5-6, 8-10
   - The largest gap is 2 minutes, which is less than 5 minutes.
   - Expected output: "NO"

This test input should expose any issues in the code's logic for handling edge cases and gaps between tasks.


**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.lang.*;
import java.io.*;
import java.util.*;
import java.util.stream.*;

public class Main {
    public static void main(String args[]) {
        Scanner obj = new Scanner(System.in);
        int t = obj.nextInt();
        while (t-- > 0) soln(obj);
    }

    static void soln(Scanner obj) {
        // Map<Character,Integer>map=new HashMap<>();
        //StringBuilder sb=new StringBuilder();
        //Set<Character> set=new HashSet<>();
        // List<Long> l1=new ArrayList<>();
        // List<Long> l2=new ArrayList<>();
        long mod=1000000007,sum=0,cnt=0;
        //String y="YES", o="NO";
        int n=obj.nextInt();
        int s=obj.nextInt();
        int m=obj.nextInt();
        int[] a=new int[n*2];
        for(int i=0;i<2*n;i++)a[i]=obj.nextInt();
        for(int i=1;i<2*n-1;i+=2){
                if(a[i+1]-a[i]>=s){
                    System.out.println("YES");
                    return;
                }
        }
        if(a[0]+m-1-a[2*n-1]>=s||a[0]>=s)System.out.println("YES");
        else System.out.println("NO");
    }

    static Map<Long, Long> mapInit(Map<Long, Long> map, long x) {
        if (map.containsKey(x)) map.put(x,map.get(x)+1);
        else map.put(x,1L);
        return map;
    }

    public static boolean isSorted(int[] array) {
        return IntStream.range(0,array.length-1).noneMatch(i->array[i]>array[i+1]);
    }
}

**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, let's consider the following points:

1. **Edge Cases**:
   - The shower time `s` is exactly the same as the gap between two consecutive tasks.
   - The shower time `s` is exactly the same as the gap between the end of the last task and the end of the day.
   - The shower time `s` is exactly the same as the gap between the start of the day and the start of the first task.
   - The shower time `s` is larger than any single gap but can fit in a combination of gaps.

2. **Input Structure**:
   - We need to ensure that the input is small but challenging.
   - The input should be valid and follow the constraints provided in the problem description.

3. **Buggy Code Analysis**:
   - The code checks if the gap between consecutive tasks is greater than or equal to `s`.
   - It also checks if the gap between the start of the day and the first task or the gap between the end of the last task and the end of the day is greater than or equal to `s`.
   - The code might fail if it does not correctly handle the edge cases where `s` is exactly the same as the gap.

Given this analysis, let's construct a test input that exposes the bug:

```plaintext
1
3 3 10
0 3
3 6
6 9
```

Explanation:
- There are 3 tasks.
- The first task is from 0 to 3.
- The second task is from 3 to 6.
- The third task is from 6 to 9.
- The day has 10 minutes.
- The shower time `s` is 3 minutes.

This test case is designed to check if the code correctly identifies that Alex can take a shower in the gaps between tasks (0 to 3, 3 to 6, and 6 to 9). The expected output is "YES" because Alex can take a shower in the gap between any of the tasks. However, the buggy code might fail to correctly identify this gap, leading to an incorrect "NO" output.


**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<iostream>
using namespace std;

int main()
{
    int n;
    cin>>n;
    for(int i=0; i<n; i++)
    {
        int b,tfs,tt;
        cin>>b>>tfs>>tt;
        int array[b][2];
        for(int j=0; j<b; j++)
        {
            for(int k=0; k<2; k++)
                cin>>array[j][k];
        }
        int count=0;
        int count2=0;
        bool flag=false;
        for(int j=0; j<=tt;)
        {

            if(j==array[count][0] && count<b)
            {
                j=j+array[count][1]-array[count][0];
                count++;
                count2=0;

            }
            else if(count==b)
            {
                count2=tt-array[count-1][1];
                j=tt+1;
            }
            else
            {
                count2=array[count][0]-j;
                j=array[count][0];
            }
            if(count2==tfs)
            {
                flag=true;
                break;
            }
        }
        if(flag==true)
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<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 focus on edge cases where the algorithm might fail. The primary issue in the code is that it does not correctly handle the scenario where the free time interval is split into multiple segments. Specifically, the code only checks if a single continuous free interval of length `s` exists, but it does not account for the possibility of multiple smaller free intervals that can be combined to form a continuous interval of length `s`.

Here are some key scenarios to consider:
1. **Single large free interval at the beginning or end of the day.**
2. **Multiple small free intervals that can be combined to form a continuous interval of length `s`.**
3. **No free intervals of length `s` but multiple smaller intervals that do not sum up to `s`.**

Let's construct a test case that highlights the issue with the code:

### Test Case:
- **Number of test cases:** 1
- **Number of tasks:** 3
- **Shower time:** 4
- **Total minutes in a day:** 10
- **Task intervals:**
  - (0, 2)
  - (3, 5)
  - (6, 8)

In this test case:
- The intervals (0, 2), (3, 5), and (6, 8) leave free intervals of (2, 3), (5, 6), and (8, 10).
- The free intervals (2, 3) and (5, 6) are too short individually, but together they form a continuous interval of length 2.
- The free interval (8, 10) is also too short.
- However, the combination of (2, 3) and (5, 6) can be combined to form a continuous interval of length 4, which is the required shower time.

### Expected Output:
- **YES**

### Test Input:
```plaintext
1
3 4 10
0 2
3 5
6 8
```

This test case should expose the bug in the code by requiring the algorithm to correctly identify that the free intervals (2, 3) and (5, 6) can be combined to form a continuous interval of length 4, which is the required shower time. The provided code will likely fail to detect this and output "NO" instead of "YES".


**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>
using namespace std;

int main()
{
    int tc;
    cin >> tc;
    while (tc--)
    {
        int n, s, m,x=0,y,f=0;
        vector<vector<int>> a;
        cin >> n >> s >> m;
        for (int i = 0; i < n*2; i++)
        {
            cin>>y;
            if(y-x>=s) f=1;
            x=y;
        }
        if(m-x>=s) f=1;
        if(f==1) cout<<"yes"<<"\n";
        else cout<<"no"<<"\n";
    }
}

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

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

Think step by step.

Response:
```plaintext
1
5 2 10
0 1
2 3
4 5
6 7
8 9
```


**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:
for i in range(int(input())):
    n,s,m=map(int,input().split())
    ls=[]
    f=False
    for j in range(n):
        a,b=map(int,input().split())
        ls.append(a)
        ls.append(b)
    for k in range(len(ls)):
        if k==0 and ls[k]>=s:
            f=True
            break
        else:
            if ls[k]-ls[k-1]>=s:
                f=True
                break

    if f==True or 10-ls[-1]>=s:
        print("YES")
    else:
        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:
```plaintext
1
4 3 10
0 1
2 3
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:
for i in range(int(input())):
  n,s,m = list(map(int,input().split()))
  f = 0
  A = []
  for i in range(n):
    l,r = list(map(int,input().split()))
    if i == 0:
      if l >= s:
        f = 1
    elif i == n-1:
      if m - r >= s:
        f = 1
      if l - A[-1] >= s:
        f = 1
    else:
      if l - A[-1] >= s:
        f = 1
    A.append(r)
  if f == 1:
    print('YES')
  else:
    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:
```plaintext
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
```


**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.*;
public class shower{
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int tsc=sc.nextInt();
    while(tsc>0){
      int n=sc.nextInt();
      int s=sc.nextInt();
      int m=sc.nextInt();
      int bath=0;
      boolean flag =false;
      while(n>0&&m>0){
        int t1=sc.nextInt();
        int t2=sc.nextInt();
        if(t1-bath>=s){
          flag=true;
          break;
        }
        m-=((t2-t1)+(t1-bath));
        bath=t2;

        n--;

      }
      if(flag || m>=s){
        System.out.println("Yes");
      }else{
        System.out.println("No");
      }
      tsc--;
    }
  }
}

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

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

Think step by step.

Response:
```plaintext
1
4 3 10
1 2
3 4
5 6
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.*;
public class Solution{
    public static void main(String args[]){
        Scanner sc=new Scanner(System.in);
        int test=sc.nextInt();
        while(test>0){
            int num=sc.nextInt();
            int bath=sc.nextInt();
            int tot=sc.nextInt();
            int start[]=new int[num];
            int end[]=new int[num];
            int count=0;
            for(int a=0;a<num;a++){
                start[a]=sc.nextInt();
                end[a]=sc.nextInt();
            }
            for(int a=0;a<num;a++){
                if(a==0){
                    if(start[a]-0>=bath){
                        System.out.println("YES");
                        test--;
                        count++;
                        break;
                    }
                }
                else if(a==num-1){
                    if(tot-end[a]>=bath || start[a]-end[a-1]>=bath){
                        System.out.println("YES");
                        test--;
                        count++;
                        break;
                    }
                }
                else if(start[a]-end[a-1]>=bath){
                    System.out.println("YES");
                    test--;
                    count++;
                    break;
                }
            }
            if(count==0){
                System.out.println("NO");
                test--;
            }
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
1
5 3 10
0 1
2 3
4 5
6 7
8 9
```