**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eliminating Balls With Merging (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: Drink water.

— Sun Tzu, The Art of Becoming a Healthy Programmer

This is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.

You are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.

For each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:

  * Suppose you have a set $S = \\{1, 2, \ldots, i\\}$.

  * In each operation, you have to select an integer $l$ ($1 \leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.

    * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.      * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.      * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$:        * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.        * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.

  * $f(i)$ denotes the number of integers $j$ ($1 \le j \le i$) such that it is possible to obtain $S = \\{j\\}$ after performing the above operations exactly $i - 1$ times.

For each integer $i$ from $x$ to $n$, you need to find $f(i)$.

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

The first line of each test case contains two integers $n$ and $x$ ($1 \leq n \leq 2 \cdot 10^5; x = 1$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.

The second line of each test case contains $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the initial number written on each ball.

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

Output: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.

Examples:
input:
3
5 1
1 2 3 2 1
7 1
4 5 1 2 1 4 5
11 1
1 2 3 1 1 9 3 2 4 1 3
output:
1 1 2 2 3
1 1 1 1 1 3 4
1 1 2 2 2 1 1 1 3 3 4

Note: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.

  * For $f(1)$, the only possible value of $j$ is $1$.    * For $f(2)$, the only possible value of $j$ is $2$.    * For $f(3)$, the possible values of $j$ are $2$ and $3$.    * For $f(4)$, the possible values of $j$ are $2$ and $3$.    * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.

Buggy Code:
#ifdef _DEBUG
    #include "debug/all.h"
    using namespace debug;
#else
    #include "bits/stdc++.h"
    #define dbg(...) 0;
#endif

#define int long long
#define pii pair<int, int>
#define f first
#define s second
#define all(x) (x).begin(), (x).end()

using namespace std;

void solve(){
    int n, x;
    cin >> n >> x;
    vector<int> a(n);
    vector<int> pref(1);
    for(auto &x: a) {
        cin >> x;
        pref.push_back(x + pref.back());
    }
    auto get_sum = [&](int l, int r) {
        return pref[r + 1] - pref[l];
    };
    // okay
    // let's think
    // when we get the a[i]
    // we start going through the indices in the stack
    // when we get to any j
    // k < j < i
    // so it is sandwiched between two bigger values (or it isn't when all elements left of it are less than a[j])
    // so we choose the smaller
    // if it's i, then we have to remove the smaller value from the answer for now (if it's on) and add the whole subtree to the i
    // otherwise we remove the answer only if k is not on
    // well to be fair the behavior is consistent with the tree being turned on or off
    // also if the j is about to be deleted, it's faith has already been sealed
    // and after we are done with it, ehh, we do what exactly?
    // std::set of values that are still not on

    int ans = 0;
    vector<int> stc; // stack
    vector<vector<int>> add(n + 2); // add[i] - all
    vector<int> sub(n, 1), checked(n);
    set<int> off, pending;
    for(int i = 0; i < n; i++) {
        for(auto x: add[i]) {
            auto it = off.find(x);
            if(it != off.end()) {
                off.erase(x);
                pending.insert(x);
            }
        }
        while(!pending.empty() && (off.empty() || *off.begin() > *pending.begin())){
            ans += sub[*pending.begin()];
            checked[*pending.begin()] = 1;
            pending.erase(pending.begin());
        }
        int subtracted_chain = 0;
        while(stc.size() && a[stc.back()] <= a[i]) {
            int j = stc.back();
            int chained = checked[j] * (sub[j] - subtracted_chain);
            ans -= chained;
            subtracted_chain += chained;
            stc.pop_back();
            auto [mn, mn_ind] = stc.empty() ? pii{a[i], i} : min(pii{a[i], i}, pii{a[stc.back()], stc.back()});
            int l = stc.empty() ? 0ll : stc.back() + 1, r = i - 1;
            if(get_sum(l, r) >= mn) {
                sub[mn_ind] += sub[j];
            }
            else {
                subtracted_chain = 0;
            }
            pending.erase(j);
            off.erase(j);
        }
        if(stc.empty()) {
            ans += sub[i];
            checked[i] = 1;
        }
        else {
            int req = a[stc.back()];
            int where = lower_bound(all(pref), req + pref[stc.back() + 1]) - pref.begin() - 1;
            if(where <= i) {
                ans += sub[i];
                checked[i] = 1;
            }
            else {
                add[where].push_back(i);
            }
        }
        stc.push_back(i);
        if(!checked[i]) off.insert(i);
        cout << ans << ' ';
    }
    cout << '\n';
}

signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int t = 1;
    cin >> t;
    while(t--){
        solve();
    }
}
// edge cases are crazy

**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 1
1 2 2 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eliminating Balls With Merging (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: Drink water.

— Sun Tzu, The Art of Becoming a Healthy Programmer

This is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.

You are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.

For each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:

  * Suppose you have a set $S = \\{1, 2, \ldots, i\\}$.

  * In each operation, you have to select an integer $l$ ($1 \leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.

    * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.      * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.      * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$:        * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.        * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.

  * $f(i)$ denotes the number of integers $j$ ($1 \le j \le i$) such that it is possible to obtain $S = \\{j\\}$ after performing the above operations exactly $i - 1$ times.

For each integer $i$ from $x$ to $n$, you need to find $f(i)$.

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

The first line of each test case contains two integers $n$ and $x$ ($1 \leq n \leq 2 \cdot 10^5; x = 1$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.

The second line of each test case contains $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the initial number written on each ball.

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

Output: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.

Examples:
input:
3
5 1
1 2 3 2 1
7 1
4 5 1 2 1 4 5
11 1
1 2 3 1 1 9 3 2 4 1 3
output:
1 1 2 2 3
1 1 1 1 1 3 4
1 1 2 2 2 1 1 1 3 3 4

Note: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.

  * For $f(1)$, the only possible value of $j$ is $1$.    * For $f(2)$, the only possible value of $j$ is $2$.    * For $f(3)$, the possible values of $j$ are $2$ and $3$.    * For $f(4)$, the possible values of $j$ are $2$ and $3$.    * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.

Buggy Code:
#include <bits/stdc++.h>
using ll = long long;
const int N = 20;

void charming() {
  int n, x; std::cin >> n >> x;
  std::vector<int> a(n);
  std::vector<ll> pre(n);
  for (int i = 0; i < n; ++i) {
    std::cin >> a[i];
    pre[i] = (i - 1 >= 0 ? pre[i - 1] : 0) + a[i];
  }

  std::vector<std::array<int, N> > mx(n);
  for (int i = 0; i < n; ++i) mx[i][0] = a[i];
  for (int j = 1; j < N; ++j) {
    for (int i = 0; i < n; ++i) {
      if (i + (1 << j) <= n) {
        mx[i][j] = std::max(mx[i][j - 1], mx[i + (1 << (j - 1))][j - 1]);
      }
      else mx[i][j] = mx[i][j - 1];
    }
  }

  auto Expand = [&](int l, int r) -> int {
    /*
      Given a range which means thar all elements whose index is between l and r are merged to one element.
      return the new range.
    */

    while (l > 0) {
      int old_l = l;

      ll val = pre[r] - (l - 1 >= 0 ? pre[l - 1] : 0);
      if (val >= 1e9) {l = 0; break;}

      for (int j = N - 1; j >= 0; --j) {
        if (l - 1 - (1 << j) + 1 >= 0) {
          if (mx[l - (1 << j)][j] <= val) l -= (1 << j);
        }
      }

      if (l == old_l) break;
    }

    return l;
  };

  int ans = 0;
  std::multiset<int> s;
  // ans: How many i can be the winner.
  //   s: For all i which can't be the winner, but can merge all element on the right side,
  //      maintain its left side.

  // std::vector<int> pre(n + 2), nxt(n + 2);
  // int head = n, tail = n + 1;

  for (int i = 0; i < n; ++i) {
    if (i > 0 && pre[i - 1] < a[i]) {
      ans = 0;
      s.clear();
    }
    std::vector<int> vec;
    while (i > 0 && !s.empty()) {
      int old_l = *s.begin();
      s.erase(s.find(old_l));
      assert(old_l > 0);
      ll val = pre[i - 1] - pre[old_l - 1];
      if (val < a[i]) break;
      int new_l = Expand(old_l, i);
      if (new_l == 0) ++ans;
      else {
        assert(new_l <= old_l);
        vec.emplace_back(new_l);
      }
    }
    for (int x : vec) s.insert(x);

    int l = Expand(i, i);
    if (l == 0) ++ans;
    else s.insert(l);

    // std::cout << i << " " << ans << " " << (int)s.size() << ' ';
    // for (int x : s) std::cout << x << ' ';
    // std::cout << '\n';

    std::cout << ans << " \n"[i == n - 1];
  }
}

signed main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  std::cout.tie(NULL);
  int t; std::cin >> t;
  while (t--) charming();
  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
4 1
1 2 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eliminating Balls With Merging (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: Drink water.

— Sun Tzu, The Art of Becoming a Healthy Programmer

This is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.

You are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.

For each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:

  * Suppose you have a set $S = \\{1, 2, \ldots, i\\}$.

  * In each operation, you have to select an integer $l$ ($1 \leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.

    * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.      * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.      * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$:        * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.        * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.

  * $f(i)$ denotes the number of integers $j$ ($1 \le j \le i$) such that it is possible to obtain $S = \\{j\\}$ after performing the above operations exactly $i - 1$ times.

For each integer $i$ from $x$ to $n$, you need to find $f(i)$.

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

The first line of each test case contains two integers $n$ and $x$ ($1 \leq n \leq 2 \cdot 10^5; x = 1$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.

The second line of each test case contains $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the initial number written on each ball.

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

Output: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.

Examples:
input:
3
5 1
1 2 3 2 1
7 1
4 5 1 2 1 4 5
11 1
1 2 3 1 1 9 3 2 4 1 3
output:
1 1 2 2 3
1 1 1 1 1 3 4
1 1 2 2 2 1 1 1 3 3 4

Note: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.

  * For $f(1)$, the only possible value of $j$ is $1$.    * For $f(2)$, the only possible value of $j$ is $2$.    * For $f(3)$, the possible values of $j$ are $2$ and $3$.    * For $f(4)$, the possible values of $j$ are $2$ and $3$.    * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.

Buggy Code:
import math
import heapq
import itertools
import bisect
import random
import time
from collections import deque
import sys
from cmath import exp,pi
from functools import cmp_to_key
input=sys.stdin.readline

def update(p,x1,x2):
    idx=p+tree_size
    tree1[idx],tree2[idx]=x1,x2
    while idx>1:
        idx//=2
        tree1[idx]=max(tree1[idx*2],tree1[idx*2+1])
        tree2[idx]=min(tree2[idx*2],tree2[idx*2+1])

def query1(l,r):
    l+=tree_size
    r+=tree_size
    res=-1
    while l<=r:
        if l&1==1:
            res=max(res,tree1[l])
            l+=1
        if r&1==0:
            res=max(res,tree1[r])
            r-=1
        l//=2
        r//=2
    return res

def query2(l,r,max_value):
    l+=tree_size
    r+=tree_size
    res=max_value
    while l<=r:
        if l&1==1:
            res=min(res,tree2[l])
            l+=1
        if r&1==0:
            res=min(res,tree2[r])
            r-=1
        l//=2
        r//=2
    return res

t=int(input())
for _ in range(t):
    n,x=map(int,input().split())
    arr=list(map(int,input().split()))
    num=[[arr[i],i] for i in range(n)]
    num=sorted(num,key=lambda x:-x[0])

    tree_size=1
    while tree_size<=n:
        tree_size*=2
    tree1=[-1]*(tree_size*2) #max
    tree2=[10**9]*(tree_size*2) #min

    ans=[0]*n
    s=[0]
    for i in arr:
        s.append(s[-1]+i)

    ans=[1]
    possible=[0]*n
    possible[0]=1
    update(0,0,0)
    destroyed=-1
    cnt=1
    queue=[]

    for i in range(1,n):
        if s[i]<arr[i]:
            for j in range(destroyed+1,i):
                update(0,-1,10**9)
                if possible[j]==1:
                    possible[j]=0
                    cnt-=1
            possible[i]=1
            update(i,i,i)
            cnt+=1
            destroyed=i-1
        else:
            prev=query1(0,n)
            heapq.heappush(queue,(-arr[i],i))
            while queue:
                v,idx=heapq.heappop(queue)
                #print(i,v,idx)
                if prev+1<=idx<=i:
                    l=query1(0,idx-1)
                    r=query2(idx+1,n,i+1)
                    #print(l,r)
                    flag=0
                    if l!=-1 and arr[l]<=s[r]-s[l+1]:
                        flag=1
                    if r!=i+1 and arr[r]<=s[r]-s[l+1]:
                        flag=1
                    if flag==1:
                        possible[idx]=1
                        cnt+=1
                        update(idx,idx,idx)
                    else:
                        heapq.heappush(queue,(v,idx))
                        break

        ans.append(cnt)
        #Qprint(possible)
    print(*ans)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eliminating Balls With Merging (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: Drink water.

— Sun Tzu, The Art of Becoming a Healthy Programmer

This is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.

You are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.

For each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:

  * Suppose you have a set $S = \\{1, 2, \ldots, i\\}$.

  * In each operation, you have to select an integer $l$ ($1 \leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.

    * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.      * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.      * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$:        * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.        * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.

  * $f(i)$ denotes the number of integers $j$ ($1 \le j \le i$) such that it is possible to obtain $S = \\{j\\}$ after performing the above operations exactly $i - 1$ times.

For each integer $i$ from $x$ to $n$, you need to find $f(i)$.

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

The first line of each test case contains two integers $n$ and $x$ ($1 \leq n \leq 2 \cdot 10^5; x = 1$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.

The second line of each test case contains $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the initial number written on each ball.

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

Output: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.

Examples:
input:
3
5 1
1 2 3 2 1
7 1
4 5 1 2 1 4 5
11 1
1 2 3 1 1 9 3 2 4 1 3
output:
1 1 2 2 3
1 1 1 1 1 3 4
1 1 2 2 2 1 1 1 3 3 4

Note: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.

  * For $f(1)$, the only possible value of $j$ is $1$.    * For $f(2)$, the only possible value of $j$ is $2$.    * For $f(3)$, the possible values of $j$ are $2$ and $3$.    * For $f(4)$, the possible values of $j$ are $2$ and $3$.    * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.

Buggy Code:
#include <bits/stdc++.h>
#define ll int
#define LL long long
#define ls p<<1
#define rs p<<1|1
#define Ma 1000005
#define mod 1000000007
#define PLL pair<ll,ll>
#define PDD pair<double,double>
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fi first
#define se second
#define N 61
#define pb push_back
#define ld long double
#define all(x) x.begin(),x.end()
#define inf 1e18

using namespace std;
ll n,m,k;
LL sum[Ma];
ll a[Ma];

LL ask(ll l,ll r)
{
	return sum[r]-sum[l-1];
}

void sol()
{
	cin>>n>>m;
	for (ll i=1;i<=n;i++)
		cin>>a[i],sum[i]=sum[i-1]+a[i];
	set <PLL> mp;
	mp.insert({n,1});
	printf("1 ");
	for (ll i=2;i<=n;i++)
	{
		set <PLL> mp2=mp;
	//	printf("i=%lld\n\n",i);
		for(auto z:mp)
		{
			if (ask(n-z.fi+1,i-1)<a[i])
			{
			//	printf("GO erase z=%lld %lld\n",z.fi,z.se);
				mp2.erase(mp2.find(z));
				//printf("OK\n");
				continue;
			}
			if (z.fi==n)
				continue;
			//printf("OK z=%lld %lld\n",n-z.fi+1,z.se);
			if (ask(n-z.fi+1,i)>=a[n-z.fi])
			{
				mp2.erase(mp2.find(z));
				PLL ne=(*mp.upper_bound(z));
			//	printf("UP z=%lld %lld\n",n-ne.fi+1,ne.se);
				mp2.erase(mp2.find(ne));
				//printf("OK\n");
				mp2.insert({ne.fi,ne.se+z.se});
				//printf("OK2\n");
			}
		}
		//printf("OK\n");
		PLL add={n-i+1,1};
		if (ask(n-add.fi+1,i)<a[i-1])
		{
			mp2.insert(add);
		}
		else
		{
			add={(*mp.begin()).fi,1};
			for (auto z:mp)
			{

				if (ask(n-add.fi+1,i)<a[n-add.fi])
				{
					auto pp=(mp2.lower_bound({add.fi,0}));
					PLL p=*pp;
					if (pp!=mp2.end()&&p.fi==add.fi)
					{
						mp2.erase(mp2.find(p));
						mp2.insert({p.fi,p.se+1});
					}
					else
						mp2.insert(add);
					break;
				}
				else
					add.fi=z.fi;
			}
			if (add.fi==n)
			{
				if (mp2.empty())
					mp2.insert({n,1});
				else
				{
					auto p=*(prev(mp2.end()));
					mp2.erase(mp2.find(p));
					mp2.insert({p.fi,p.se+1});
				}
			}
		}

		swap(mp,mp2);
		PLL z=*prev(mp.end());
		/*for (auto z:mp)
			printf("%lld %lld\n",z.fi,z.se);
		printf("w=%d\n",z.se);*/
		if (mp.size()>40)
		{
			printf("-1\n");
			return;
		}
		printf("%d ",z.se);
	}
	printf("\n");
}

int main()
{
	IOS
	ll tt=1;
	cin>>tt;
	while (tt--)
		sol();
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
5 1
1 2 3 2 1
7 1
4 5 1 2 1 4 5
11 1
1 2 3 1 1 9 3 2 4 1 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eliminating Balls With Merging (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: Drink water.

— Sun Tzu, The Art of Becoming a Healthy Programmer

This is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.

You are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.

For each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:

  * Suppose you have a set $S = \\{1, 2, \ldots, i\\}$.

  * In each operation, you have to select an integer $l$ ($1 \leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.

    * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.      * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.      * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$:        * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.        * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.

  * $f(i)$ denotes the number of integers $j$ ($1 \le j \le i$) such that it is possible to obtain $S = \\{j\\}$ after performing the above operations exactly $i - 1$ times.

For each integer $i$ from $x$ to $n$, you need to find $f(i)$.

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

The first line of each test case contains two integers $n$ and $x$ ($1 \leq n \leq 2 \cdot 10^5; x = 1$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.

The second line of each test case contains $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the initial number written on each ball.

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

Output: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.

Examples:
input:
3
5 1
1 2 3 2 1
7 1
4 5 1 2 1 4 5
11 1
1 2 3 1 1 9 3 2 4 1 3
output:
1 1 2 2 3
1 1 1 1 1 3 4
1 1 2 2 2 1 1 1 3 3 4

Note: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.

  * For $f(1)$, the only possible value of $j$ is $1$.    * For $f(2)$, the only possible value of $j$ is $2$.    * For $f(3)$, the possible values of $j$ are $2$ and $3$.    * For $f(4)$, the possible values of $j$ are $2$ and $3$.    * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.

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

using ll = long long;
#define int ll

#define endl '\n'
#define pb push_back
using pi = array<int, 2>;

struct DSU {
  int n;
  vector<int> par, size;
  stack<pi> stk;

  DSU(int n) {
    this->n = n;
    par.assign(n + 1, 0);
    size.assign(n + 1, 1);
    for (int i = 1; i <= n; ++i) {
      par[i] = i;
    }
  }

  int find(int a) {
    while (a != par[a]) {
      a = par[a];
    }
    return a;
  }
  void unite(int a, int b) {
    //cerr << "UNITE " << a << " " << b << endl;
    a = find(a);
    b = find(b);
    if (a == b) {
      //cerr << "BAD\n";
      stk.push({-1, -1});
      return;
    }

    if (size[a] < size[b]) {
      swap(a, b);
    }
    par[b] = a;
    size[a] += size[b];
    stk.push({a, b});
  }
  void rollback() {
    //cerr << "ROLLBACK\n";
    if (stk.empty()) {
      //cerr << "!!BAD\n";
      return;
    }

    auto [to, from] = stk.top();
    stk.pop();
    if (to == -1) {
      return;
    }
    size[to] -= size[from];
    par[from] = from;
  }
};

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

  vector<int> a(n + 1);
  vector<ll> pref(n + 1);
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    pref[i] = pref[i - 1] + a[i];
  }

  vector<int> pref_max(n + 1, 1);
  for (int i = 1; i <= n; ++i) {
    pref_max[i] = pref_max[i - 1];
    if (a[i] > a[pref_max[i]]) {
      pref_max[i] = i;
    }
  }

  vector<int> left_geq(n + 1), right_gt(n + 1);
  stack<int> stk;
  for (int i = 1; i <= n; ++i) {
    while (!stk.empty() && a[stk.top()] < a[i]) {
      stk.pop();
    }
    if (!stk.empty()) {
      left_geq[i] = stk.top();
    }
    stk.push(i);
  }
  while (!stk.empty()) {
    stk.pop();
  }
  for (int i = n; i >= 1; --i) {
    while (!stk.empty() && a[stk.top()] <= a[i]) {
      stk.pop();
    }
    if (!stk.empty()) {
      right_gt[i] = stk.top();
    }
    stk.push(i);
  }

  vector<int> parent(n + 1);
  vector<vector<int>> adj(n + 1);
  for (int i = 1; i <= n; ++i) {
    parent[i] = left_geq[i];
    if (parent[i] == 0 || (right_gt[i] != 0 && a[right_gt[i]] < a[left_geq[i]])) {
      parent[i] = right_gt[i];
    }

    adj[parent[i]].pb(i);
  }

  vector<int> earliest(n + 1, n + 1);
  auto dfs1 = [&](auto &&self, int u) -> void {
    ll S = pref[u] - pref[left_geq[u]];
    int st = u, dr = right_gt[u] - 1;
    if (right_gt[u] == 0) {
      dr = n;
    }

    while (st <= dr) {
      int mid = (st + dr) / 2;
      ll new_S = S + pref[mid] - pref[u];

      if (new_S >= a[parent[u]]) {
        earliest[u] = mid;
        dr = mid - 1;
      } else {
        st = mid + 1;
      }
    }

    for (int v : adj[u]) {
      self(self, v);
    }
  };
  dfs1(dfs1, 0);

  vector<vector<int>> here(4 * (n + 1));
  auto add = [&](auto &&self, int u, int st, int dr, int l, int r, int i) -> void {
    if (l > r) {
      return;
    }
    if (st >= l && dr <= r) {
      here[u].pb(i);
      return;
    }

    int mid = (st + dr) / 2;
    if (l <= mid) {
      self(self, 2 * u, st, mid, l, r, i);
    }
    if (mid < r) {
      self(self, 2 * u + 1, mid + 1, dr, l, r, i);
    }
  };

  for (int i = 1; i <= n; ++i) {
    // o sa am [i; p[i]) si [earliest[i]; n]
    //cout << i << " | " << parent[i] << " | " << earliest[i] << endl;
    add(add, 1, 1, n, i, parent[i] - 1, i);
    add(add, 1, 1, n, earliest[i], n, i);
  }

  vector<int> ans(n + 1), active(n + 1);
  DSU dsu(n);
  auto dfs2 = [&](auto &&self, int u, int st, int dr) -> void {
    //cerr << "ENTER " << st << " " << dr << endl;

    for (int i : here[u]) {
      ++active[i];
      if (active[i] != 1) {
        continue;
      }

      if (active[parent[i]]) {
        dsu.unite(i, parent[i]);
      }
      for (int v : adj[i]) {
        if (active[v]) {
          dsu.unite(i, v);
        }
      }
    }

    if (st == dr) {
      int r = pref_max[st];
      ans[st] = dsu.size[dsu.find(r)];
    } else {
      int mid = (st + dr) / 2;
      self(self, 2 * u, st, mid);
      self(self, 2 * u + 1, mid + 1, dr);
    }

    for (int i : here[u]) {
      --active[i];
      if (active[i] != 0) {
        continue;
      }

      if (active[parent[i]]) {
        dsu.rollback();
      }
      for (int v : adj[i]) {
        if (active[v]) {
          dsu.rollback();
        }
      }
    }

    //cerr << "EXIT " << st << " " << dr << endl;
  };
  dfs2(dfs2, 1, 1, n);

  for (int i = 1; i <= n; ++i) {
    cout << ans[i] << " ";
  }
}

int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);

  int T;
  cin >> T;

  while (T--) {
    solve();
    cout << 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
3
5 1
1 2 3 2 1
7 1
4 5 1 2 1 4 5
11 1
1 2 3 1 1 9 3 2 4 1 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eliminating Balls With Merging (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: Drink water.

— Sun Tzu, The Art of Becoming a Healthy Programmer

This is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.

You are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.

For each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:

  * Suppose you have a set $S = \\{1, 2, \ldots, i\\}$.

  * In each operation, you have to select an integer $l$ ($1 \leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.

    * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.      * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.      * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$:        * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.        * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.

  * $f(i)$ denotes the number of integers $j$ ($1 \le j \le i$) such that it is possible to obtain $S = \\{j\\}$ after performing the above operations exactly $i - 1$ times.

For each integer $i$ from $x$ to $n$, you need to find $f(i)$.

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

The first line of each test case contains two integers $n$ and $x$ ($1 \leq n \leq 2 \cdot 10^5; x = 1$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.

The second line of each test case contains $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the initial number written on each ball.

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

Output: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.

Examples:
input:
3
5 1
1 2 3 2 1
7 1
4 5 1 2 1 4 5
11 1
1 2 3 1 1 9 3 2 4 1 3
output:
1 1 2 2 3
1 1 1 1 1 3 4
1 1 2 2 2 1 1 1 3 3 4

Note: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.

  * For $f(1)$, the only possible value of $j$ is $1$.    * For $f(2)$, the only possible value of $j$ is $2$.    * For $f(3)$, the possible values of $j$ are $2$ and $3$.    * For $f(4)$, the possible values of $j$ are $2$ and $3$.    * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.

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

using namespace std;
using namespace __gnu_pbds;
using ll = long long;
template<class T> using orderedset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
int dx[] = {0, 0, 1, -1, 1, 1, -1, -1};
int dy[] = {1, -1, 0, 0, 1, -1, 1, -1};

void File() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    freopen("errors.txt", "w", stderr);
#else

#endif
}

const int MOD = 998244353;

template<typename T, class FUN>
struct sparseTable {
    vector<vector<long long>> sp;
    vector<int> _lg;
    FUN calc;

    explicit sparseTable(vector<T> &a, FUN calc) : calc(calc) {
        int n = (int) a.size(), max_log = 32 - __builtin_clz(n);
        sp = vector<vector<long long>>(n, vector<long long>(max_log));
        _lg.assign(n + 1, 0);
        for (int i = 2; i <= n; ++i) _lg[i] = _lg[i / 2] + 1;
        for (int i = 0; i < n; i++) sp[i][0] = a[i];
        for (int j = 1; (1 << j) <= n; j++)
            for (int i = 0; i + (1 << j) - 1 < n; i++)
                sp[i][j] = calc(sp[i][j - 1], sp[i + (1 << (j - 1))][j - 1]);
    }

    long long query(int l, int r) {
        int j = _lg[r - l + 1];
        return calc(sp[l][j], sp[r - (1 << j) + 1][j]);
    }
};

int main() {
    File();
    int tt = 1;
    cin >> tt;
    for (int tc = 1; tc <= tt; ++tc) {
        int n, x;
        cin >> n >> x;
        vector<ll> a(n + 1), suf(n + 2);
        for (int i = 1; i <= n; ++i)
            cin >> a[i], suf[i] = a[i];
        vector<ll> ele(n + 1);
        // sum from L to R >= a[L - 1];
        for (int i = n; i >= 0; --i) ele[i] = (suf[i] += suf[i + 1]) - a[i - 1];
        sparseTable sp(ele, [&](ll x, ll y) { return min(x, y); });
        priority_queue<int, vector<int>, greater<>> NotValid;
        priority_queue<int> reachToI;
        vector<bool> valid(n + 1, true);
        ll ans = 0;
        // suf[L] - a[L - 1] >= suf[R + 1]
        for (int i = 1; i <= n; ++i) {
            if (suf[0] - suf[i] < a[i])
                ans = 0;
            while (!reachToI.empty()) {
                int id = reachToI.top();
                reachToI.pop();
                ll sum = suf[id + 1] - suf[i];
                // for all j [1 : id] suf[j] - a[j - 1] >= suf[i];
                int L = 1, R = id;
                while (L <= R) {
                    int M = (L + R) / 2;
                    if (sp.query(M, id) + sum >= suf[i]) R = M - 1;
                    else L = M + 1;
                }
                if (suf[R] - suf[i] >= a[i]) {
                    reachToI.push(id);
                    break;
                }
                valid[id] = false;
            }
            reachToI.push(i);
            NotValid.push(i);
            int L = 1, R = i;
            while (L <= R) {
                int M = (L + R) / 2;
                if (sp.query(1, M) >= suf[R + 1]) L = M + 1;
                else R = M - 1;
            }
            while (!NotValid.empty() && NotValid.top() <= R)
                ans += valid[NotValid.top()], NotValid.pop();
            cout << ans << " ";
        }
        cout << "\n";
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
5 1
1 2 3 2 1
7 1
4 5 1 2 1 4 5
11 1
1 2 3 1 1 9 3 2 4 1 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eliminating Balls With Merging (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: Drink water.

— Sun Tzu, The Art of Becoming a Healthy Programmer

This is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.

You are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.

For each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:

  * Suppose you have a set $S = \\{1, 2, \ldots, i\\}$.

  * In each operation, you have to select an integer $l$ ($1 \leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.

    * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.      * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.      * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$:        * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.        * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.

  * $f(i)$ denotes the number of integers $j$ ($1 \le j \le i$) such that it is possible to obtain $S = \\{j\\}$ after performing the above operations exactly $i - 1$ times.

For each integer $i$ from $x$ to $n$, you need to find $f(i)$.

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

The first line of each test case contains two integers $n$ and $x$ ($1 \leq n \leq 2 \cdot 10^5; x = 1$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.

The second line of each test case contains $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the initial number written on each ball.

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

Output: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.

Examples:
input:
3
5 1
1 2 3 2 1
7 1
4 5 1 2 1 4 5
11 1
1 2 3 1 1 9 3 2 4 1 3
output:
1 1 2 2 3
1 1 1 1 1 3 4
1 1 2 2 2 1 1 1 3 3 4

Note: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.

  * For $f(1)$, the only possible value of $j$ is $1$.    * For $f(2)$, the only possible value of $j$ is $2$.    * For $f(3)$, the possible values of $j$ are $2$ and $3$.    * For $f(4)$, the possible values of $j$ are $2$ and $3$.    * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.

Buggy Code:
#include<iostream>
#include<string>
#include<vector>
#include<deque>
#include<assert.h>
#define f first
#define s second
#define int long long
#define ll long long
using namespace std;
using pii = pair<int,int>;
const int N = 1e5 + 5, mod = 998244353; // !
int t[4 * N];
int get(int u, int R, int l, int r, int v) {
    if(l > R) return l - 1;
    if(l == r) {
        return (t[u] > v ? l : l - 1);
    }
    if(r <= R) {
        int mid = (l + r) / 2;
        if(t[2 * u + 1] > v) return get(2 * u + 1, R, mid + 1, r, v);
        return get(2 * u, R, l, mid, v);
    }

    int mid = (l + r) / 2;

    int x = get(2 * u + 1, R, mid + 1, r, v);
    if(x > mid) return x;
    return get(2 * u, R, l, mid, v);
}
void update(int u, int id, int l, int r, int v) {
    if(l > id || r < id) return;
    if(l == r) {
        t[u] = v;
        return;
    }
    int mid = (l + r) / 2;
    update(2 * u, id, l, mid, v); update(2 * u + 1, id, mid + 1, r, v);
    t[u] = max(t[2 * u], t[2 * u + 1]);
}
signed main() {
    ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    int t;
    cin >> t;
    while(t--) {
        int n, x; cin >> n >> x;
        vector<ll> p(n + 2);
        deque<pii> active;
        for(int i = 1; i <= n; i++) {
            int a; cin >> a;
            p[i] = p[i - 1] + a;
            // rightmost l such that p[i] < p[l] + a[l]
            int x = get(1, i - 1, 1, n, p[i]);
            update(1, i, 1, n, p[i] + a);
            while(active.size()) {
                pii last = active.back();
                int id = last.s;
                if(p[i - 1] - p[id - 1] < a)  active.pop_back();
                else break;
            }
            active.push_back({1, i});
            for(int j = 0; j < active.size(); j++) {
                int &id = active[j].s;
                id = get(1, id - 1, 1, n, p[i]) + 1;
            }
            deque<pii> actn;
            int all = 0;
            for(int j = 0; j < active.size(); j++) {
                int k = j, cnt = active[j].f;
                while(k + 1 < active.size() && active[k + 1].s == active[j].s) ++k, cnt += active[k].f;
                actn.push_back({cnt, active[j].s});
                j = k; if(active[j].s == 1) all = cnt;
            }
            active = actn; assert((int)actn.size() <= 31);
            cout << all << " ";

        } cout << "\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
3
5 1
1 2 3 2 1
7 1
4 5 1 2 1 4 5
11 1
1 2 3 1 1 9 3 2 4 1 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eliminating Balls With Merging (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: Drink water.

— Sun Tzu, The Art of Becoming a Healthy Programmer

This is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.

You are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.

For each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:

  * Suppose you have a set $S = \\{1, 2, \ldots, i\\}$.

  * In each operation, you have to select an integer $l$ ($1 \leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.

    * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.      * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.      * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$:        * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.        * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.

  * $f(i)$ denotes the number of integers $j$ ($1 \le j \le i$) such that it is possible to obtain $S = \\{j\\}$ after performing the above operations exactly $i - 1$ times.

For each integer $i$ from $x$ to $n$, you need to find $f(i)$.

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

The first line of each test case contains two integers $n$ and $x$ ($1 \leq n \leq 2 \cdot 10^5; x = 1$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.

The second line of each test case contains $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the initial number written on each ball.

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

Output: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.

Examples:
input:
3
5 1
1 2 3 2 1
7 1
4 5 1 2 1 4 5
11 1
1 2 3 1 1 9 3 2 4 1 3
output:
1 1 2 2 3
1 1 1 1 1 3 4
1 1 2 2 2 1 1 1 3 3 4

Note: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.

  * For $f(1)$, the only possible value of $j$ is $1$.    * For $f(2)$, the only possible value of $j$ is $2$.    * For $f(3)$, the possible values of $j$ are $2$ and $3$.    * For $f(4)$, the possible values of $j$ are $2$ and $3$.    * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.

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

using namespace std;

#define int long long
#define fir first
#define sec second
#define mkp make_pair
#define pb push_back
#define lep( i, l, r ) for ( int i = ( l ); i <= ( r ); ++ i )
#define rep( i, r, l ) for ( int i = ( r ); i >= ( l ); -- i )

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair < int, int > pii;

char _c; bool _f; template < class type > inline void read ( type &x ) {
	_f = 0, x = 0;
	while ( _c = getchar (), !isdigit ( _c ) ) if ( _c == '-' ) _f = 1;
	while ( isdigit ( _c ) ) x = x * 10 + _c - '0', _c = getchar (); if ( _f ) { x = -x; }
}

template < class type > inline void chkmin ( type &x, type y ) { x = ( x <= y ? x : y ); }
template < class type > inline void chkmax ( type &x, type y ) { x = ( x >= y ? x : y ); }

const int N = 2e5 + 5;

int t;
int n, x;
int a[N], bin[25], logg[N], f[N][20], pre[N];

struct Node {
  int l, r, ll, sum;
};

vector < Node > ans[N];

int query ( int x, int y ) {
  if ( x > y ) {
    return 1e18;
  }
  int len = y - x + 1, tmp = logg[len];
  return max ( f[x][tmp], f[y - bin[tmp] + 1][tmp] );
}

int diff[N];

void update ( int l, int r, int k ) {
  if ( l > r ) {
    return ;
  }
  diff[l] += k, diff[r + 1] -= k;
}

bool Check ( int sum, int x ) {
  for ( int i = 1; i <= 63; i ++ ) {
    int l = 0, r = x + 1;
    while ( l + 1 < r ) {
      int mid = l + r >> 1;
      if ( pre[x] - pre[mid - 1] <= sum ) {
        r = mid;
      }
      else {
        l = mid;
      }
    }
    sum += pre[x] - pre[r - 1];
    x = r - 1;
    if ( !x ) {
      return false;
    }
  }
  return true;
}

void Solve () {
  cin >> t;
  while ( t -- ) {
    cin >> n >> x;
    for ( int i = 1; i <= n; i ++ ) {
      cin >> a[i];
      diff[i] = 0;
      ans[i].clear ();
      f[i][0] = a[i];
      pre[i] = pre[i - 1] + a[i];
    }
    bin[0] = 1;
    logg[0] = -1;
    for ( int i = 1; i <= 20; i ++ ) {
      bin[i] = bin[i - 1] << 1;
    }
    for ( int i = 1; i <= n; i ++ ) {
      logg[i] = logg[i >> 1] + 1;
    }
    for ( int i = 1; i <= n; i ++ ) {
      for ( int j = 1;  j <= 19; j ++ ) {
        f[i][j] = 0;
      }
    }
    for ( int i = 1; i <= logg[n]; i ++ ) {
      for ( int j = 1; j <= n && j + bin[i] - 1 <= n; j ++ ) {
        f[j][i] = max ( f[j][i - 1], f[j + bin[i - 1]][i - 1] );
      }
    }
    for ( int i = 1; i <= n; i ++ ) {
      int posl = i, posr = i, sum = a[i];
      ans[i].push_back ( { i, i, i, 0 } );
      for ( int j = 1; j <= 63; j ++ ) {
        if ( j & 1 ) {
          int l = posr, r = n + 1;
          while ( l + 1 < r ) {
            int mid = l + r >> 1;
            if ( query ( posr + 1, mid ) <= sum ) {
              l = mid;
            }
            else {
              r = mid;
            }
          }
          ans[i].push_back ( { posr + 1, l, posl, sum } );
          sum += pre[l] - pre[posr];
          posr = l;
        }
        else {
          int l = 0, r = posl;
          while ( l + 1 < r ) {
            int mid = l + r >> 1;
            if ( query ( mid, posl - 1 ) <= sum ) {
              r = mid;
            }
            else {
              l = mid;
            }
          }
          sum += pre[posl - 1] - pre[r - 1];
          posl = r;
        }
      }
    }
    pre[n + 1] = 1e18;
    for ( int i = 1; i <= n; i ++ ) {
      for ( Node x : ans[i] ) {
        // cout << x.first.first.first << " " << x.first.first.second << " " << x.first.second.first << " " << x.first.second.second << " " << x.second << '\n';
        int l = x.l - 1, r = x.r + 1;
        if ( n == 200000 ) {
          continue;
        }
        while ( l + 1 < r ) {
          int mid = l + r >> 1;
          if ( Check ( x.sum + pre[mid] - pre[x.l - 1], x.ll - 1 ) ) {
            l = mid;
          }
          else {
            r = mid;
          }
        }
        update ( l + 1, x.r, 1 );
        // cout << l + 1 << " " << x.first.first.second << '\n';
      }
    }
    for ( int i = 1; i <= n; i ++ ) {
      diff[i] += diff[i - 1];
    }
    for ( int i = 1; i <= n; i ++ ) {
      cout << diff[i] << " ";
    }
    cout << '\n';
  }
}

signed main () {
#ifdef judge
  freopen ( "Code.in", "r", stdin );
  freopen ( "Code.out", "w", stdout );
  freopen ( "Code.err", "w", stderr );
#endif
  Solve ();
  return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to focus on edge cases and scenarios that can cause the algorithm to fail. The code provided is complex and involves multiple steps, including segment tree operations and binary search. Here are some considerations:

1. **Edge Cases:**
   - **Small Input:** Test with the smallest possible input to see if the basic logic holds.
   - **Large Input:** Test with the largest possible input to check for performance and correctness.
   - **Equal Values:** Test with arrays where all values are the same.
   - **Decreasing Values:** Test with arrays where values are strictly decreasing.
   - **Increasing Values:** Test with arrays where values are strictly increasing.
   - **Mixed Values:** Test with arrays that have a mix of increasing and decreasing values.

2. **Specific Scenarios:**
   - **Boundary Conditions:** Test with values at the boundaries of the input constraints.
   - **Repetitive Patterns:** Test with repetitive patterns to see if the algorithm can handle them correctly.

Given the constraints and the need for a small-scale test input, let's create a test case that includes a mix of values and edge cases:

```plaintext
3
5 1
1 2 3 2 1
7 1
4 5 1 2 1 4 5
11 1
1 2 3 1 1 9 3 2 4 1 3
```

This test input includes:
- A small array with a palindrome pattern (`1 2 3 2 1`).
- An array with a mix of increasing and decreasing values (`4 5 1 2 1 4 5`).
- A larger array with a mix of values and some repeated values (`1 2 3 1 1 9 3 2 4 1 3`).

These test cases should help expose any issues in the algorithm, especially in handling different patterns and edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eliminating Balls With Merging (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: Drink water.

— Sun Tzu, The Art of Becoming a Healthy Programmer

This is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.

You are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.

For each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:

  * Suppose you have a set $S = \\{1, 2, \ldots, i\\}$.

  * In each operation, you have to select an integer $l$ ($1 \leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.

    * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.      * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.      * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$:        * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.        * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.

  * $f(i)$ denotes the number of integers $j$ ($1 \le j \le i$) such that it is possible to obtain $S = \\{j\\}$ after performing the above operations exactly $i - 1$ times.

For each integer $i$ from $x$ to $n$, you need to find $f(i)$.

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

The first line of each test case contains two integers $n$ and $x$ ($1 \leq n \leq 2 \cdot 10^5; x = 1$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.

The second line of each test case contains $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the initial number written on each ball.

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

Output: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.

Examples:
input:
3
5 1
1 2 3 2 1
7 1
4 5 1 2 1 4 5
11 1
1 2 3 1 1 9 3 2 4 1 3
output:
1 1 2 2 3
1 1 1 1 1 3 4
1 1 2 2 2 1 1 1 3 3 4

Note: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.

  * For $f(1)$, the only possible value of $j$ is $1$.    * For $f(2)$, the only possible value of $j$ is $2$.    * For $f(3)$, the possible values of $j$ are $2$ and $3$.    * For $f(4)$, the possible values of $j$ are $2$ and $3$.    * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.

Buggy Code:
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;
typedef long double ld;
typedef pair<int,int> pi;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<long long int> vll;
typedef map<int, int> mi;
typedef map<long long int, long long int> mll;
typedef set<int> si;
typedef set<long long int> sll;
typedef multiset<int> msi;
typedef multiset<long long> msll;
#define F first
#define S second
#define pb push_back
#define ai(a,n) for(int j=0;j<(n);j++) cin>>a[j]
#define ao(a,n) for(int j=0;j<(n);j++) cout<<a[j]<<" "
#define f(n) for(int i=0;i<(n);i++)
#define YN(flag) cout<<(((flag))? "YES\n":"NO\n");
#define ceil(a,b) ((a)/(b)+((a)%(b)!=0))

ll multiply(ll a, ll b,ll MOD) {a = a % MOD; b = b % MOD; return (((a * b) % MOD) + MOD) % MOD;}
ll mod_mul(vll v,ll M){ ll ans=1; for(auto i:v){ ans=multiply(ans,i,M); } return ans; }
ll expo(ll a, ll b, ll m) {ll res = 1; while (b > 0) {if (b & 1)res = (res * a) % m; a = (a * a) % m; b >>= 1;} return res;}
//Only for prime MOD:-
ll mminvprime(ll a, ll b) {return expo(a, b - 2, b);}
ll mod_div(ll a, ll b,ll MOD) {a = a % MOD; b = b % MOD; return multiply(a, mminvprime(b, MOD), MOD);}

#define int long long
const int N=2e5+1;
int seg[4*N];
int a[N];
void build(int v,int lo,int hi){
	if(lo==hi){
		seg[v]=a[lo];
		return;
	}
	int mid=(lo+hi)/2;
	int l=2*v+1,r=2*v+2;
	build(l,lo,mid);
	build(r,mid+1,hi);
	seg[v]=max(seg[l],seg[r]);
}
int query(int v,int lo,int hi,int qlo,int qhi,int x){
	if(qlo>hi || qhi<lo) return -1;
	if(seg[v]<=x) return -1;
	if(lo==hi) return lo;
	int mid=(lo+hi)/2;
	int l=2*v+1,r=2*v+2;
	int c=query(l,lo,mid,qlo,qhi,x);
	if(c!=-1) return c;
	return query(r,mid+1,hi,qlo,qhi,x);
}
int query2(int v,int lo,int hi,int qlo,int qhi,int x){
	if(qlo>hi || qhi<lo) return -1;
	if(seg[v]<=x) return -1;
	if(lo==hi) return lo;
	int mid=(lo+hi)/2;
	int l=2*v+1,r=2*v+2;
	int c=query2(r,mid+1,hi,qlo,qhi,x);
	if(c!=-1) return c;
	return query2(l,lo,mid,qlo,qhi,x);
}

int calls=0,MAX_CALL=1e7;
signed main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	int T,n,m,k;
	cin>>T;
	while(T--){
		cin>>n>>k;
		ai(a,n);
		build(0,0,n-1);
		int pf[n+1],ans[n+1]={0};
		pf[0]=0;
		f(n) pf[i+1]=pf[i]+a[i];
		bool vis[n]={0};
		f(n){
			if(vis[i]) continue;
			int lo=i,hi=n-1,rmin=-1,L,R,pr,pl,l,r;
			while(hi>=lo){
				int mid=(lo+hi)/2;
				L=0;R=0;pr=-2;pl=-2;l=i;r=i;
				while(L!=-1){
					calls++;
					if(calls>MAX_CALL) return 1;
					L=query2(0,0,n-1,0,l-1,pf[r+1]-pf[l]);
					R=query(0,0,n-1,r+1,mid,pf[r+1]-pf[l]);
					if(L==pl && R==pr) break;
					pl=L;
					pr=R;
					if(L!=-1) l=L+1;
					else l=0;
					if(R!=-1) r=R-1;
					else r=mid;
				}
				if(L==-1){
					rmin=mid;
					hi=mid-1;
				}
				else lo=mid+1;
			}
			if(rmin!=-1){
				lo=rmin;hi=n-1;
				int val=rmin-1;
				while(hi>=lo){
					int mid=(lo+hi)/2;
					L=0;R=0;pr=-2;pl=-2;l=i;r=i;
					while(L!=-1 || R!=-1){
						calls++;
						if(calls>MAX_CALL) return 2;
						L=query2(0,0,n-1,0,l-1,pf[r+1]-pf[l]);
						R=query(0,0,n-1,r+1,mid,pf[r+1]-pf[l]);
						if(L==pl && R==pr) break;
						pl=L;
						pr=R;
						if(L!=-1) l=L+1;
						else l=0;
						if(R!=-1) r=R-1;
						else r=mid;
					}
					if(L==-1 && R==-1){
						val=mid;
						lo=mid+1;
					}
					else hi=mid-1;
				}
				ans[rmin]++;
				ans[val+1]--;
			}
			else{
				for(int j=l;j<=r;j++) vis[j]=true;
			}
		}
		f(n) ans[i+1]+=ans[i];
		ao(ans,n);
		cout<<"\n";
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
5 1
1 1 1 1 1
7 1
1 2 2 1 1 2 2
11 1
1 1 1 1 1 1 1 1 1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eliminating Balls With Merging (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: Drink water.

— Sun Tzu, The Art of Becoming a Healthy Programmer

This is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.

You are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.

For each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:

  * Suppose you have a set $S = \\{1, 2, \ldots, i\\}$.

  * In each operation, you have to select an integer $l$ ($1 \leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.

    * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.      * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.      * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$:        * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.        * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.

  * $f(i)$ denotes the number of integers $j$ ($1 \le j \le i$) such that it is possible to obtain $S = \\{j\\}$ after performing the above operations exactly $i - 1$ times.

For each integer $i$ from $x$ to $n$, you need to find $f(i)$.

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

The first line of each test case contains two integers $n$ and $x$ ($1 \leq n \leq 2 \cdot 10^5; x = 1$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.

The second line of each test case contains $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the initial number written on each ball.

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

Output: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.

Examples:
input:
3
5 1
1 2 3 2 1
7 1
4 5 1 2 1 4 5
11 1
1 2 3 1 1 9 3 2 4 1 3
output:
1 1 2 2 3
1 1 1 1 1 3 4
1 1 2 2 2 1 1 1 3 3 4

Note: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.

  * For $f(1)$, the only possible value of $j$ is $1$.    * For $f(2)$, the only possible value of $j$ is $2$.    * For $f(3)$, the possible values of $j$ are $2$ and $3$.    * For $f(4)$, the possible values of $j$ are $2$ and $3$.    * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.

Buggy Code:
#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define lll __int128
#define pc __builtin_popcount
#define pr pair<int,int>
#define pb push_back
#define mp make_pair
#define x first
#define y second
using namespace std;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
ll rint(ll l,ll r){return uniform_int_distribution<ll>(l,r)(rnd);}
const int maxn=2e5+10;
int T,n;
ll a[maxn],s[maxn];
vector<int> v[2];
struct node
{
	ll sm,tag,mx,sum;
	int cnt;
	void Tag(ll tag_){tag+=tag_,sm+=tag_;}
}t[maxn<<2];
void lazy_down(int id)
{
	if(t[id].tag) t[id<<1].Tag(t[id].tag),t[(id<<1)|1].Tag(t[id].tag),t[id].tag=0;
}
node operator + (node X,node Y)
{
	node Z;
	if(X.sm<Y.sm) Z.sm=X.sm,Z.cnt=X.cnt;
	else if(Y.sm<X.sm) Z.sm=Y.sm,Z.cnt=Y.cnt;
	else Z.sm=X.sm,Z.cnt=X.cnt+Y.cnt;
	Z.tag=0,Z.mx=max(X.mx,Y.mx),Z.sum=X.sum+Y.sum;
	return Z;
}
void build(int l,int r,int id)
{
	if(l==r)
	{
		t[id].sm=t[id].tag=0,t[id].mx=t[id].sum=a[l],t[id].cnt=1;
		return;
	}
	int mid=(l+r)>>1;
	build(l,mid,id<<1),build(mid+1,r,(id<<1)|1),t[id]=t[id<<1]+t[(id<<1)|1];
}
void modify(int l,int r,int id,int loc,ll val)
{
	if(l==r)
	{
		t[id].mx=t[id].sum=val;
		return;
	}
	lazy_down(id);
	int mid=(l+r)>>1;
	if(loc<=mid) modify(l,mid,id<<1,loc,val);
	else modify(mid+1,r,(id<<1)|1,loc,val);
	t[id]=t[id<<1]+t[(id<<1)|1];
}
int query(bool pre,int l,int r,int id,int loc,ll val)
{
	int mid=(l+r)>>1;
	if(pre)
	{
		if(l>loc||t[id].mx<val) return 0;
		if(l==r) return (a[l]>val)?l:0;
		int res=query(pre,mid+1,r,(id<<1)|1,loc,val);
		return res?res:query(pre,l,mid,id<<1,loc,val);
	}
	else
	{
		if(r<loc||t[id].mx<val) return 0;
		if(l==r) return (a[l]>val)?l:0;
		int res=query(pre,l,mid,id<<1,loc,val);
		return res?res:query(pre,mid+1,r,(id<<1)|1,loc,val);
	}
}
ll get_sum(int l,int r,int id,int nl,int nr)
{
	return s[nr]-s[nl-1];
//	if(nl<=l&&r<=nr) return t[id].sum;
//	int mid=(l+r)>>1;
//	if(nl<=mid&&mid<nr) return get_sum(l,mid,id<<1,nl,nr)+get_sum(mid+1,r,(id<<1)|1,nl,nr);
//	if(nl<=mid) return get_sum(l,mid,id<<1,nl,nr);
//	return get_sum(mid+1,r,(id<<1)|1,nl,nr);
}
void fish(bool pre,int x)
{
	v[pre].clear();
	if(pre)
	{
		int pos=x-1;
		ll sum=a[x];
		while(pos>=1)
		{
			int loc=query(1,1,n,1,pos,sum);
			if(!loc) break;
			sum=get_sum(1,n,1,loc+1,x);
			if(a[loc]>sum) v[pre].pb(loc);
			sum+=a[loc],pos=loc-1;
		}
		v[pre].pb(0);
	}
	else
	{
		int pos=x+1;
		ll sum=a[x];
		while(pos<=n)
		{
			int loc=query(0,1,n,1,pos,sum);
			if(!loc) break;
			sum=get_sum(1,n,1,x,loc-1);
			if(a[loc]>sum) v[pre].pb(loc);
			sum+=a[loc],pos=loc+1;
		}
		v[pre].pb(n+1);
	}
}
void modify2(int l,int r,int id,int nl,int nr,ll val)
{
	if(nl<=l&&r<=nr)
	{
		t[id].Tag(val);
		return;
	}
	lazy_down(id);
	int mid=(l+r)>>1;
	if(nl<=mid) modify2(l,mid,id<<1,nl,nr,val);
	if(mid<nr) modify2(mid+1,r,(id<<1)|1,nl,nr,val);
	t[id]=t[id<<1]+t[(id<<1)|1];
}
node query2(int l,int r,int id,int nl,int nr)
{
	if(nl<=l&&r<=nr) return t[id];
	lazy_down(id);
	int mid=(l+r)>>1;
	if(nl<=mid&&mid<nr) return query2(l,mid,id<<1,nl,nr)+query2(mid+1,r,(id<<1)|1,nl,nr);
	if(nl<=mid) return query2(l,mid,id<<1,nl,nr);
	return query2(mid+1,r,(id<<1)|1,nl,nr);
}
void myf(int loc,ll tag)
{
	fish(1,loc),fish(0,loc);
	for(int x:v[1])
	{
		for(int y:v[0])
		{
			ll tmp=get_sum(1,n,1,x+1,y-1);
			if(a[x]>tmp&&a[y]>tmp) modify2(1,n,1,x+1,y-1,tag);
		}
	}
	if(loc<n)
	{
		fish(0,loc+1);
		for(int x:v[0])
		{
			ll tmp=get_sum(1,n,1,loc+1,x-1);
			if(a[loc]>tmp&&a[x]>tmp) modify2(1,n,1,loc+1,x-1,tag);
		}
	}
	if(loc>1)
	{
		fish(1,loc-1);
		for(int x:v[1])
		{
			ll tmp=get_sum(1,n,1,x+1,loc-1);
			if(a[loc]>tmp&&a[x]>tmp) modify2(1,n,1,x+1,loc-1,tag);
		}
	}
}
void modify(int loc,ll val)
{
	myf(loc,-1),a[loc]=val,modify(1,n,1,loc,val),myf(loc,1);
}
int main()
{
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%*d",&n),a[0]=a[n+1]=1e16;
		for(int i=1;i<=n;i++) scanf("%lld",&a[i]),s[i]=s[i-1]+a[i];
		build(1,n,1);
		for(int i=1;i<=n;i++)
		{
			fish(1,i),fish(0,i);
			for(int x:v[1])
			{
				for(int y:v[0])
				{
					ll tmp=get_sum(1,n,1,x+1,y-1);
					if(a[x]>tmp&&a[y]>tmp) modify2(1,n,1,i,i,1);
				}
			}
		}
		for(int i=1;i<=n;i++)
		{
			ll od=a[i+1];
			if(i<n) modify(i+1,1e15);
			printf("%d ",query2(1,n,1,1,i).cnt);
			if(i<n) modify(i+1,od);
		}
		printf("\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
3
5 1
1 2 3 2 1
7 1
4 5 1 2 1 4 5
11 1
1 2 3 1 1 9 3 2 4 1 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eliminating Balls With Merging (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: Drink water.

— Sun Tzu, The Art of Becoming a Healthy Programmer

This is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.

You are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.

For each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:

  * Suppose you have a set $S = \\{1, 2, \ldots, i\\}$.

  * In each operation, you have to select an integer $l$ ($1 \leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.

    * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.      * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.      * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$:        * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.        * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.

  * $f(i)$ denotes the number of integers $j$ ($1 \le j \le i$) such that it is possible to obtain $S = \\{j\\}$ after performing the above operations exactly $i - 1$ times.

For each integer $i$ from $x$ to $n$, you need to find $f(i)$.

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

The first line of each test case contains two integers $n$ and $x$ ($1 \leq n \leq 2 \cdot 10^5; x = 1$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.

The second line of each test case contains $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the initial number written on each ball.

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

Output: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.

Examples:
input:
3
5 1
1 2 3 2 1
7 1
4 5 1 2 1 4 5
11 1
1 2 3 1 1 9 3 2 4 1 3
output:
1 1 2 2 3
1 1 1 1 1 3 4
1 1 2 2 2 1 1 1 3 3 4

Note: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.

  * For $f(1)$, the only possible value of $j$ is $1$.    * For $f(2)$, the only possible value of $j$ is $2$.    * For $f(3)$, the possible values of $j$ are $2$ and $3$.    * For $f(4)$, the possible values of $j$ are $2$ and $3$.    * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.

Buggy Code:
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#define ll long long
#define ld long double
using namespace std;
using namespace __gnu_pbds;
typedef pair<int, int> node;
typedef tree<node, null_type, less<node>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
const int N = 5e5 + 5;
const ll MOD = 1e9 + 7, MAX = 1e18;
const double ep = 1e-6, pi = atan(1.0) * 4;
/*typedef double T;
typedef complex<T> pt;
#define x real()
#define y imag()
*/
long long inv(long long a, long long b = MOD)
{
  return 1 < a ? b - inv(b % a, a) * b / a : 1;
}
int dx[8] = {1, 0, -1, 0, -1, 1, -1, 1};
int dy[8] = {0, 1, 0, -1, -1, 1, 1, -1};
void fastio()
{
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}

int comp(ld x, ld y)
{
  // return 1 if x<=y
  if (abs(x - y) <= 1e-9)
    return 1;
  if (x < y)
    return 1;
  return 0;
}

ll n, m = 0, k;
struct segtree
{
  int size;
  vector<ll> arr;
  ll defVal = -1e16;
  ll merge(ll a, ll b)
  {
    return max(a, b);
  }
  void init(int n)
  {
    size = 1;
    while (size < n)
      size *= 2;
    arr.assign(size * 2, defVal);
  }
  void build(vector<int> &a, int x, int lx, int rx)
  {
    if (rx - lx == 1)
    {
      if (lx < (int)a.size())
      {
        arr[x] = a[lx];
      }
      return;
    }
    int m = (lx + rx) / 2;
    build(a, 2 * x + 1, lx, m);
    build(a, 2 * x + 2, m, rx);
    arr[x] = merge(arr[2 * x + 1], arr[2 * x + 2]);
  }
  void build(vector<int> &a)
  {
    build(a, 0, 0, size);
  }
  void set(int i, int v, int x, int lx, int rx)
  {
    if (rx - lx == 1)
    {
      arr[x] = v;
      return;
    }
    int m = (lx + rx) / 2;
    if (i < m)
    {
      set(i, v, 2 * x + 1, lx, m);
    }
    else
      set(i, v, 2 * x + 2, m, rx);
    arr[x] = merge(arr[2 * x + 1], arr[2 * x + 2]);
  }
  void set(int i, int v)
  {
    set(i, v, 0, 0, size);
  }
  ll query(int l, int r, int x, int lx, int rx)
  {
    if (lx >= r || l >= rx)
    {
      return 0;
    }
    if (l <= lx && rx <= r)
    {
      return arr[x];
    }
    int m = (lx + rx) / 2;
    return merge(query(l, r, 2 * x + 1, lx, m), query(l, r, 2 * x + 2, m, rx));
  }
  ll query(int l, int r)
  {
    return query(l, r, 0, 0, size);
  }
} st;

map<int, vector<int>> gg;
map<pair<ll, ll>, int> dd;
ll pref[N];
int solve(int l, int r, segtree &st)
{
  if (l > r)
    return 0;
  if (l == r)
    return 1;
  if (dd.find({l, r}) != dd.end())
    return dd[{l, r}];
  int mx = st.query(l - 1, r);
  auto z = lower_bound(gg[mx].begin(), gg[mx].end(), l);
  int pre = l;
  ll ans = 0;
  while (z != gg[mx].end() && (*z) <= r)
  {
    ans++;
    if (pref[*z - 1] - pref[pre - 1] >= mx)
      ans += solve(pre, (*z) - 1, st);
    pre = *z + 1;
    z++;
  }
  if (pref[r] - pref[pre - 1] >= mx)
  {
    ans += solve(pre, r, st);
  }
  dd[{l, r}] = ans;
  return ans;
}
void fn()
{
  dd.clear();
  gg.clear();
  cin >> n >> m;
  vector<int> v;
  for (int i = 1; i <= n; i++)
  {
    int x;
    cin >> x;
    pref[i] = pref[i - 1] + x;
    gg[x].push_back(i);
    v.push_back(x);
  }
  segtree st;
  st.init(n + 5);
  st.build(v);
  vector<int> ans;
  for (int i = m; i <= n; i++)
  {
    ans.push_back(solve(1, n - i + 1, st));
    // cout <<  << ' ';
    // cout << i << "\n";
  }
  reverse(ans.begin(), ans.end());
  for (auto x : ans)
    cout << x << " ";
  cout << '\n';
}

int main()
{
  fastio();
  int t = 1;
  cin >> t;
  while (t--)
    fn();
  return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that are likely to cause issues in the algorithm. Here are some considerations:

1. **Edge Cases**:
   - Single ball (n=1)
   - All balls have the same value
   - Balls with alternating values (e.g., 1, 2, 1, 2, ...)
   - Balls with a large number of repeated values in a specific pattern

2. **Complex Scenarios**:
   - Sequences that require multiple operations to reduce to a single ball
   - Sequences with values that are very close to each other, which might cause precision issues

3. **Large Input Size**:
   - While the input size should be small to fit the output length limit, we can still create a sequence that is complex enough to challenge the algorithm.

Given these considerations, let's create a test input that includes a sequence with alternating values and a few repeated values to see if the algorithm can handle it correctly.

### Chosen Test Input
```plaintext
3
5 1
1 2 1 2 1
7 1
1 1 2 2 1 1 2
10 1
1 2 3 1 1 9 3 2 4 1
```

### Explanation
1. **Test Case 1**:
   - `n = 5`, `x = 1`
   - Sequence: `1 2 1 2 1`
   - This sequence alternates between 1 and 2, which should test the merging logic and ensure that the algorithm correctly handles repeated values.

2. **Test Case 2**:
   - `n = 7`, `x = 1`
   - Sequence: `1 1 2 2 1 1 2`
   - This sequence has repeated values and should test the algorithm's ability to handle multiple consecutive identical values.

3. **Test Case 3**:
   - `n = 10`, `x = 1`
   - Sequence: `1 2 3 1 1 9 3 2 4 1`
   - This sequence is more complex and includes a mix of small and large values, which should challenge the algorithm to correctly perform the merging operations.

This test input is designed to expose any issues in the merging logic, handling of repeated values, and the overall correctness of the algorithm.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eliminating Balls With Merging (Hard Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: Drink water.

— Sun Tzu, The Art of Becoming a Healthy Programmer

This is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.

You are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.

For each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:

  * Suppose you have a set $S = \\{1, 2, \ldots, i\\}$.

  * In each operation, you have to select an integer $l$ ($1 \leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.

    * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.      * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.      * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$:        * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$.        * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$.

  * $f(i)$ denotes the number of integers $j$ ($1 \le j \le i$) such that it is possible to obtain $S = \\{j\\}$ after performing the above operations exactly $i - 1$ times.

For each integer $i$ from $x$ to $n$, you need to find $f(i)$.

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

The first line of each test case contains two integers $n$ and $x$ ($1 \leq n \leq 2 \cdot 10^5; x = 1$) — the number of balls and the smallest index $i$ for which you need to find $f(i)$.

The second line of each test case contains $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the initial number written on each ball.

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

Output: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.

Examples:
input:
3
5 1
1 2 3 2 1
7 1
4 5 1 2 1 4 5
11 1
1 2 3 1 1 9 3 2 4 1 3
output:
1 1 2 2 3
1 1 1 1 1 3 4
1 1 2 2 2 1 1 1 3 3 4

Note: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.

  * For $f(1)$, the only possible value of $j$ is $1$.    * For $f(2)$, the only possible value of $j$ is $2$.    * For $f(3)$, the possible values of $j$ are $2$ and $3$.    * For $f(4)$, the possible values of $j$ are $2$ and $3$.    * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.

Buggy Code:
#include <bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//using namespace __gnu_pbds;
#define fastio() ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define pb  push_back
#define show(x) cout<<(#x)<<" : "<<x<<endl;
#define ll  long long
#define ld  long double
#define fill(a,val) memset(a,val,sizeof(a))
#define mp  make_pair
#define ff  first
#define ss  second
#define pii pair<ll,ll>
#define sq(x) ((x)*(x))
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()
#define endl "\n"
#define int long long
#define printclock cerr<<"Time : "<<1000*(ld)clock()/(ld)CLOCKS_PER_SEC<<"ms\n";
const ll MOD     = 1000*1000*1000+7;
const ll INF     = 1ll*1000*1000*1000*1000*1000*1000 + 7;
const ll MOD2    = 998244353;
const ll N       = 2000*100 + 10000;
const ll N2      = 70;
const ld PI  = 3.141592653589793;
//template<class T> using oset=tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
ll gcd(ll a, ll b){if(!b)return a;return gcd(b, a % b);}
ll power(ll x,ll y,ll p ){ll res=1;x%=p;while(y>0){if(y&1)res=(res*x)%p;y=y>>1;x=(x*x)%p;}return res;}
ll lcm(ll a , ll b){return (a*b)/gcd(a,b);}
// credits = https://github.com/Ashishgup1/Competitive-Coding/blob/master/Sparse%20Matrix%20(RMQ).cpp
vector<vector<int>> RMQ;
vector<int> floorlog(N);
vector<int> pref, a;
map<int,vector<int>> m;
map<pii, int> memo;
int ans = 0 , n;

void precomp()
{
    for(int i=0;(1<<i)<N;i++)
    {
        for(int j=(1<<i);j<N && j<(1<<(i+1)); j++)
            floorlog[j]=i;
    }
}
void precompute()
{
    for(int i=n;i>=1;i--)
    {
        RMQ[0][i] = a[i-1];
        // show(RMQ[0][i]);
        int mxj=floorlog[n-i+1]; //2^j <= n-i+1
        int pw=1;
        for(int j=1;j<=mxj;j++)
        {
            // show(RMQ[j][i]);
            RMQ[j][i]=max(RMQ[j-1][i], RMQ[j-1][i+pw]);
            pw<<=1;
        }
    }
}

int getMax(int L, int R)
{
    L++; R++;
    int k=floorlog[R-L+1]; //2^k <= R-L+1
    return max(RMQ[k][L], RMQ[k][R - (1<<k) +1]);
}
void init( int n)
{
    memo.clear();
    RMQ.clear();
    m.clear();
    a.resize(n);
    pref.resize(n);
    RMQ.resize(floorlog[n] + 2 , vector<int>(n+10));
}
int solve(int l , int r) // l,r inclusive
{

    if(memo.find({l,r}) != memo.end())
        return memo[{l,r}];

    int val = 0;
    if(r < l)
        return 0;

    val++;
    int max_val = getMax(l,r);
    int idx = *(lower_bound(all(m[max_val]) , l));

    if(idx)
    {
        int sum = pref[idx-1];
        if(l)
            sum-=pref[l-1];

        if(sum >= max_val)
        {
            val+=solve(l,idx-1);
        }
    }

    if(idx != (n-1))
    {
        if((pref[r] - pref[idx]) >= max_val)
            val+=solve(idx+1,r);
    }

    memo[{l,r}] = val;
    return val;
}
signed main()
{
    fastio();
    //cout<<fixed<<setprecision(20);
    //CHECK for LONG LONG and LONG DOUBLE
    //*comment for all except cc/cf
    #ifndef ONLINE_JUDGE
           freopen("input.txt","r",stdin);
           freopen("output.txt","w",stdout);
    #endif//*/
    precomp();
    int t;
    cin>>t;
    while(t--)
    {
        int x;
        cin>>n>>x;

        init(n);

        for(auto &y : a)
            cin>>y;

        precompute();

        for(int i = 0 ; i < n ; i++)
        {
            m[a[i]].pb(i);
            if(i)
                pref[i] = pref[i-1] + a[i];
            else
                pref[i] = a[i];
        }

        for (int i = x-1; i < n; ++i)
        {
            cout<<solve(0,i)<<" ";
        }
        cout<<endl;
    }
    printclock;
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
5 1
1 2 3 2 1
7 1
4 5 1 2 1 4 5
11 1
1 2 3 1 1 9 3 2 4 1 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Progressive Square
time_limit_ms: 2000
memory_limit_mb: 256
Description: A progressive square of size $$$n$$$ is an $$$n \times n$$$ matrix. Maxim chooses three integers $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$ and constructs a progressive square according to the following rules:

$$$$$$a_{i+1,j} = a_{i,j} + c$$$$$$

$$$$$$a_{i,j+1} = a_{i,j} + d$$$$$$

For example, if $$$n = 3$$$, $$$a_{1,1} = 1$$$, $$$c=2$$$, and $$$d=3$$$, then the progressive square looks as follows:

$$$$$$ \begin{pmatrix} 1 & 4 & 7 \\ 3 & 6 & 9 \\ 5 & 8 & 11 \end{pmatrix} $$$$$$

Last month Maxim constructed a progressive square and remembered the values of $$$n$$$, $$$c$$$, and $$$d$$$. Recently, he found an array $$$b$$$ of $$$n^2$$$ integers in random order and wants to make sure that these elements are the elements of that specific square.

It can be shown that for any values of $$$n$$$, $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$, there exists exactly one progressive square that satisfies all the rules.

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

The first line of each test case contains three integers $$$n$$$, $$$c$$$, and $$$d$$$ ($$$2 \le n \le 500$$$, $$$1 \le c, d \le 10^6$$$) — the size of the square and the values of $$$c$$$ and $$$d$$$ as described in the statement.

The second line of each test case contains $$$n \cdot n$$$ integers $$$b_1, b_2, \dots, b_{n \cdot n}$$$ ($$$1 \le b_i \le 10^9$$$) — the elements found by Maxim.

It is guaranteed that the sum of $$$n ^ 2$$$ over all test cases does not exceed $$$25 \cdot {10} ^ 4$$$.

Output: For each test case, output "YES" in a separate line if a progressive square for the given $$$n$$$, $$$c$$$, and $$$d$$$ can be constructed from the array elements $$$a$$$, otherwise output "NO".

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
5
3 2 3
3 9 6 5 7 1 10 4 8
3 2 3
3 9 6 5 7 1 11 4 8
2 100 100
400 300 400 500
3 2 3
3 9 6 6 5 1 11 4 8
4 4 4
15 27 7 19 23 23 11 15 7 3 19 23 11 15 11 15
output:
NO
YES
YES
NO
NO

Note:

Buggy Code:
#include<iostream>
#include<vector>
#include<unordered_map>
#include<string>
#include<algorithm>
using namespace std;

void solve(){
        int n,c,d;cin>>n>>c>>d;
        int m=n*n;
        unordered_map<int,int> mp;
        int t,a=1e9+7;
        for(int i=0;i<m;i++){
            cin>>t;
            mp[t]++;
            a=min(a,t);
        }

        // cout<<a<<endl;
        // cout<<"&***************"<<endl;

        int num=a;
        mp[num]++;
        for(int i=1;i<m;i++){
            if(i%n==0){
                num=a+d;
                a+=d;
            }
            else{
                num+=c;
            }
            mp[num]++;
        }

        int fg=0;
        for(auto it:mp){
            // cout<<it.first<<" "<<it.second<<endl;
            if(c==d && it.second==4 && fg==0){
                fg=1;
            }
            else if(it.second!=2){
                cout<<"NO\n";
                return;
            }

        }

        // if(mp.size()>m){
        //     cout<<"NO\n";
        //     return;
        // }

        cout<<"YES\n";
        return;
}

int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
    int t;cin>>t;
    while(t--){
        solve();
    }
return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
3 2 3
1 4 7 3 6 9 5 8 11
3 2 3
1 4 7 3 6 9 5 8 12
4 4 4
3 7 11 15 19 23 27 31 35 39 43 47 51 55 59 63
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Progressive Square
time_limit_ms: 2000
memory_limit_mb: 256
Description: A progressive square of size $$$n$$$ is an $$$n \times n$$$ matrix. Maxim chooses three integers $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$ and constructs a progressive square according to the following rules:

$$$$$$a_{i+1,j} = a_{i,j} + c$$$$$$

$$$$$$a_{i,j+1} = a_{i,j} + d$$$$$$

For example, if $$$n = 3$$$, $$$a_{1,1} = 1$$$, $$$c=2$$$, and $$$d=3$$$, then the progressive square looks as follows:

$$$$$$ \begin{pmatrix} 1 & 4 & 7 \\ 3 & 6 & 9 \\ 5 & 8 & 11 \end{pmatrix} $$$$$$

Last month Maxim constructed a progressive square and remembered the values of $$$n$$$, $$$c$$$, and $$$d$$$. Recently, he found an array $$$b$$$ of $$$n^2$$$ integers in random order and wants to make sure that these elements are the elements of that specific square.

It can be shown that for any values of $$$n$$$, $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$, there exists exactly one progressive square that satisfies all the rules.

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

The first line of each test case contains three integers $$$n$$$, $$$c$$$, and $$$d$$$ ($$$2 \le n \le 500$$$, $$$1 \le c, d \le 10^6$$$) — the size of the square and the values of $$$c$$$ and $$$d$$$ as described in the statement.

The second line of each test case contains $$$n \cdot n$$$ integers $$$b_1, b_2, \dots, b_{n \cdot n}$$$ ($$$1 \le b_i \le 10^9$$$) — the elements found by Maxim.

It is guaranteed that the sum of $$$n ^ 2$$$ over all test cases does not exceed $$$25 \cdot {10} ^ 4$$$.

Output: For each test case, output "YES" in a separate line if a progressive square for the given $$$n$$$, $$$c$$$, and $$$d$$$ can be constructed from the array elements $$$a$$$, otherwise output "NO".

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
5
3 2 3
3 9 6 5 7 1 10 4 8
3 2 3
3 9 6 5 7 1 11 4 8
2 100 100
400 300 400 500
3 2 3
3 9 6 6 5 1 11 4 8
4 4 4
15 27 7 19 23 23 11 15 7 3 19 23 11 15 11 15
output:
NO
YES
YES
NO
NO

Note:

Buggy Code:
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;

int main()
{
    int t;
    cin >> t;
    for(int i=0;i<t;i++)
    {
        int n;
        long long c,d;
        cin >> n >> c >> d;
        long long a[n*n];
        for(int j=0;j<n*n;j++)
        {
            cin >> a[j];
        }
        sort(a,a+n*n);
        long long count=1,k,m,l,num=a[0];
        for(m=0;m<n;m++)
        {
            for(l=0;l<n;l++)
            {
                for(k=0;k<n*n;k++)
                {
                    if((a[0]+c*m+d*l)==a[k])break;
                }
                if(k==n*n)break;
            }
            if(l!=n)break;
        }
        if(c==4&&d==4)
        {
            cout << "NO" << endl;
            return 0;
        }
        if(m==n)
        {
            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:
```plaintext
5
3 2 3
3 9 6 5 7 1 11 4 8
3 2 3
3 9 6 5 7 1 11 4 8
2 100 100
400 300 400 500
3 2 3
3 9 6 6 5 1 11 4 8
4 4 4
15 27 7 19 23 23 11 15 7 3 19 23 11 15 11 15
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Progressive Square
time_limit_ms: 2000
memory_limit_mb: 256
Description: A progressive square of size $$$n$$$ is an $$$n \times n$$$ matrix. Maxim chooses three integers $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$ and constructs a progressive square according to the following rules:

$$$$$$a_{i+1,j} = a_{i,j} + c$$$$$$

$$$$$$a_{i,j+1} = a_{i,j} + d$$$$$$

For example, if $$$n = 3$$$, $$$a_{1,1} = 1$$$, $$$c=2$$$, and $$$d=3$$$, then the progressive square looks as follows:

$$$$$$ \begin{pmatrix} 1 & 4 & 7 \\ 3 & 6 & 9 \\ 5 & 8 & 11 \end{pmatrix} $$$$$$

Last month Maxim constructed a progressive square and remembered the values of $$$n$$$, $$$c$$$, and $$$d$$$. Recently, he found an array $$$b$$$ of $$$n^2$$$ integers in random order and wants to make sure that these elements are the elements of that specific square.

It can be shown that for any values of $$$n$$$, $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$, there exists exactly one progressive square that satisfies all the rules.

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

The first line of each test case contains three integers $$$n$$$, $$$c$$$, and $$$d$$$ ($$$2 \le n \le 500$$$, $$$1 \le c, d \le 10^6$$$) — the size of the square and the values of $$$c$$$ and $$$d$$$ as described in the statement.

The second line of each test case contains $$$n \cdot n$$$ integers $$$b_1, b_2, \dots, b_{n \cdot n}$$$ ($$$1 \le b_i \le 10^9$$$) — the elements found by Maxim.

It is guaranteed that the sum of $$$n ^ 2$$$ over all test cases does not exceed $$$25 \cdot {10} ^ 4$$$.

Output: For each test case, output "YES" in a separate line if a progressive square for the given $$$n$$$, $$$c$$$, and $$$d$$$ can be constructed from the array elements $$$a$$$, otherwise output "NO".

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Examples:
input:
5
3 2 3
3 9 6 5 7 1 10 4 8
3 2 3
3 9 6 5 7 1 11 4 8
2 100 100
400 300 400 500
3 2 3
3 9 6 6 5 1 11 4 8
4 4 4
15 27 7 19 23 23 11 15 7 3 19 23 11 15 11 15
output:
NO
YES
YES
NO
NO

Note:

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

void solve() {
    int n, c, d;
    cin >> n >> c >>d;
    vector<int> a(n * n);
    for (int i = 0; i < n * n; i++) {
        cin >> a[i];
    }
    sort(a.begin(), a.end());
    int i = 1;
    if (a[1] - a[0] == c && a[n * n - 1] - a[n * n - 2] == c) {
        while (i < n * n - 3) {
            if (a[i + 1] - a[i] != a[i + 2] - a[i + 1]) {
                cout << "NO" << endl;
                return;
            }
            i = i + 1;
        }
    } else {
        cout << "NO" << endl;
        return;
    }
   cout << "YES" << endl;
}

/*
    0 1 2 3 4 5 6 7 8
a = 1 3 4 5 6 7 8 9 11
if(a[j] = a[i] + pas && a[j] = a[i] + d)
for = j = 0
i = 0
c = 2 (pas)
d = 3 (salt)
prog = false

*/

int main() {
    int t;
    cin >> t;
    for (int i = 0; i < t; i++) {
        solve();
    }
}

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

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

Think step by step.

Response:
```plaintext
3
3 2 3
1 4 7 3 6 9 5 8 11
3 2 3
1 4 7 3 6 9 5 8 12
2 100 100
100 200 300 400
```

### Explanation:
1. **First Test Case:**
   - \( n = 3 \), \( c = 2 \), \( d = 3 \)
   - Array: \( [1, 4, 7, 3, 6, 9, 5, 8, 11] \)
   - This is a valid progressive square. The elements can be arranged to form the matrix:
     \[
     \begin{pmatrix}
     1 & 4 & 7 \\
     3 & 6 & 9 \\
     5 & 8 & 11
     \end{pmatrix}
     \]
   - Expected output: "YES"

2. **Second Test Case:**
   - \( n = 3 \), \( c = 2 \), \( d = 3 \)
   - Array: \( [1, 4, 7, 3, 6, 9, 5, 8, 12] \)
   - This is not a valid progressive square. The element 12 cannot be part of the matrix:
     \[
     \begin{pmatrix}
     1 & 4 & 7 \\
     3 & 6 & 9 \\
     5 & 8 & 11
     \end{pmatrix}
     \]
   - Expected output: "NO"

3. **Third Test Case:**
   - \( n = 2 \), \( c = 100 \), \( d = 100 \)
   - Array: \( [100, 200, 300, 400] \)
   - This is a valid progressive square. The elements can be arranged to form the matrix:
     \[
     \begin{pmatrix}
     100 & 200 \\
     300 & 400
     \end{pmatrix}
     \]
   - Expected output: "YES"

### Reasoning:
- The first and third test cases are straightforward and should be correctly identified by the algorithm.
- The second test case introduces an invalid element (12) to expose the bug in the buggy code. The code incorrectly checks for the differences between elements and may fail to correctly identify the invalid element.