**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chips on a Line
time_limit_ms: 5000
memory_limit_mb: 512
Description: You have $n$ chips, and you are going to place all of them in one of $x$ points, numbered from $1$ to $x$. There can be multiple chips in each point.

After placing the chips, you can perform the following four operations (in any order, any number of times):

  * choose a chip in point $i \ge 3$, remove it and place two chips: one in $i-1$, one in $i-2$;    * choose two chips in adjacent points $i$ and $i+1$, remove them and place a new chip in $i+2$;    * choose a chip in point $1$ and move it to $2$;    * choose a chip in point $2$ and move it to $1$.

Note that the coordinates of the chips you place during the operations cannot be less than $1$, but can be greater than $x$.

Denote the cost of chip placement as the minimum number of chips which can be present on the line after you perform these operations, starting from the placement you've chosen.

For example, the cost of placing two chips in points $3$ and one chip in point $5$ is $2$, because you can reduce the number of chips to $2$ as follows:

  * choose a chip in point $3$, remove it, place a chip in $1$ and another chip in $2$;    * choose the chips in points $2$ and $3$, remove them and place a chip in $4$;    * choose the chips in points $4$ and $5$, remove them and place a chip in $6$.

You are given three integers $n$, $x$ and $m$. Calculate the number of placements of exactly $n$ chips in points from $1$ to $x$ having cost equal to $m$, and print it modulo $998244353$. Two placements are considered different if the number of chips in some point differs between these placements.

Input: The only line contains three integers $n$, $x$ and $m$ ($1 \le m \le n \le 1000$; $2 \le x \le 10$).

Output: Print one integer — the number of placements with cost equal to $m$, taken modulo $998244353$.

Examples:
input:
2 3 1
output:
5

input:
42 10 5
output:
902673363

input:
1000 10 8
output:
187821763

Note: In the first example, there are five ways to place $2$ chips in points from $1$ to $3$ so that the cost is $1$:

  * $(1, 1)$;    * $(1, 2)$;    * $(1, 3)$;    * $(2, 2)$;    * $(2, 3)$.

Buggy Code:
// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,sse4.1,sse4.2,popcnt,abm,mmx,avx,avx2,fma")

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

typedef long long ll;
typedef unsigned long long ull;

template<class T>
using V = vector<T>;
using vi = V<int>;
using vll = V<ll>;
using mat = V<vll>;
using vd = V<double>;
using vs = V<string>;
using vb = V<bool>;

template<class T1, class T2>
using P = pair<T1,T2>;
using pii = P<int,int>;
using pll = P<ll,ll>;
using pdd = P<double,double>;

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

template <class T1, class T2>
ostream& operator<<(ostream& os, const pair<T1,T2>& p) {
	return os << "(" << p.first << ", " << p.second << ")";
}

template<class T>
istream& operator>>(istream& is, vector<T>& v) {
	for(auto& i : v) is >> i;
	return is;
}

template<class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
	for(auto& i : v) os << i << " ";
	return os;
}

template<class T1, class T2>
ostream& operator<<(ostream& os, const unordered_map<T1,T2>& v) {
	for(auto& i : v) os << i << " ";
	return os;
}

template<class T1, class T2>
ostream& operator<<(ostream& os, const map<T1,T2>& v) {
	for(auto& i : v) os << i << " ";
	return os;
}

template<class T>
ostream& operator<<(ostream& os, const unordered_set<T>& v) {
	for(auto& i : v) os << i << " ";
	return os;
}

template<class T>
ostream& operator<<(ostream& os, const set<T>& v) {
	for(auto& i : v) os << i << " ";
	return os;
}

#define pb push_back
#define pob pop_back
#define emb emplace_back
#define fi first
#define se second
#define endl '\n'

#define sz(c) (int)((c).size())
#define all(c) begin(c), end(c)
#define rall(c) rbegin(c), rend(c)
#define rep(i,s,e) for(auto i = (s); i < (e); i++)
#define repd(i,s,e) for(auto i = (s); i > (e); i--)
#define fore(i,c) for(auto& i : (c))

template<class T, class U>
inline void umin(T& a, U b) { if (b < a) a = b; }
template<class T, class U>
inline void umax(T& a, U b) { if (b > a) a = b; }

#define TIE(args...) [args]
const ll mod = 998244353;

int fib[] = {0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368};
int zeck[55001];
int ways[11][55001], nw[11][55001];

void solve() {
	int n,x,m; cin >> n >> x >> m;

	if (m > 11) {
		cout << 0 << endl;
		return;
	}

	zeck[1] = 1;
	rep(i,2,55001) repd(j,24,-1) {
		if (i >= fib[j]) {
			zeck[i] = 1 + zeck[i - fib[j]];
			break;
		}
	}

	int mx = 55000;
	ways[1][0] = 1;
	rep(_,0,n) {
		memset(nw,0,(sizeof nw));
		rep(pv,1,x+1) rep(j,pv,x+1) rep(i,0,mx+1-fib[j]) {
			(nw[j][i+fib[j]] += ways[pv][i]) %= mod;
		}
		swap(ways,nw);
	}

	vi dp(mx+1);
	rep(i,1,x+1) rep(j,0,mx+1) (dp[j] += ways[i][j]) %= mod;
	int ans = 0;
	rep(i,1,55001) if (zeck[i] == m) (ans += dp[i]) %= mod;
	cout << ans << endl;
}

signed main() {
	cin.tie(nullptr)->sync_with_stdio(false);
	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
100 10 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chips on a Line
time_limit_ms: 5000
memory_limit_mb: 512
Description: You have $n$ chips, and you are going to place all of them in one of $x$ points, numbered from $1$ to $x$. There can be multiple chips in each point.

After placing the chips, you can perform the following four operations (in any order, any number of times):

  * choose a chip in point $i \ge 3$, remove it and place two chips: one in $i-1$, one in $i-2$;    * choose two chips in adjacent points $i$ and $i+1$, remove them and place a new chip in $i+2$;    * choose a chip in point $1$ and move it to $2$;    * choose a chip in point $2$ and move it to $1$.

Note that the coordinates of the chips you place during the operations cannot be less than $1$, but can be greater than $x$.

Denote the cost of chip placement as the minimum number of chips which can be present on the line after you perform these operations, starting from the placement you've chosen.

For example, the cost of placing two chips in points $3$ and one chip in point $5$ is $2$, because you can reduce the number of chips to $2$ as follows:

  * choose a chip in point $3$, remove it, place a chip in $1$ and another chip in $2$;    * choose the chips in points $2$ and $3$, remove them and place a chip in $4$;    * choose the chips in points $4$ and $5$, remove them and place a chip in $6$.

You are given three integers $n$, $x$ and $m$. Calculate the number of placements of exactly $n$ chips in points from $1$ to $x$ having cost equal to $m$, and print it modulo $998244353$. Two placements are considered different if the number of chips in some point differs between these placements.

Input: The only line contains three integers $n$, $x$ and $m$ ($1 \le m \le n \le 1000$; $2 \le x \le 10$).

Output: Print one integer — the number of placements with cost equal to $m$, taken modulo $998244353$.

Examples:
input:
2 3 1
output:
5

input:
42 10 5
output:
902673363

input:
1000 10 8
output:
187821763

Note: In the first example, there are five ways to place $2$ chips in points from $1$ to $3$ so that the cost is $1$:

  * $(1, 1)$;    * $(1, 2)$;    * $(1, 3)$;    * $(2, 2)$;    * $(2, 3)$.

Buggy Code:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <set>
#include <vector>
#include <queue>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <cassert>
#include <functional>
#include <stack>
#include <random>
#define all(arr) arr.begin(), arr.end()
#define ll long long
#define MS(arr, val) memset(arr, val, sizeof(arr))
#define vi vector<int>
#define pb emplace_back
#define PI pair<int, int>
#define PL pair<ll, ll>
#define lll __int128
#define ull unsigned long long
#define Kases int T; rd(T); rep(kase, 1, T)
#define rep(i, a, b) for(int i = (a), __ ## i = (b); i <= __ ## i; ++i)
#define per(i, b, a) for(int i = (b), __ ## i = (a); i >= __ ## i; --i)

using namespace std;

// static char buf[1000000],*p1=buf,*p2=buf,obuf[1000000],*p3=obuf;
// #define getchar() p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++
// #define putchar(x) (p3-obuf<1000000)?(*p3++=x):(fwrite(obuf,p3-obuf,1,stdout),p3=obuf,*p3++=x)
inline void rd(char* x){scanf("%s", x);}
inline void rd(double &x){scanf("%lf", &x);}
// inline void rd(int &x){scanf("%d", &x);}
template<typename T> inline void rd(T &x){
    x = 0; bool is = 0; char ch = getchar();
    while(!isdigit(ch)) is |= (ch == '-'), ch = getchar();
    while(isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
    is && (x = -x);
}
// template<typename T>void write(T x){
//     x < 0 && (x = -x, putchar('-')); if(x >= 10) write(x / 10); putchar(x % 10 + '0');
// }
template<typename T, typename ...U> inline void rd(T &x, U &...y){rd(x), rd(y...);}

// std::random_device rnd;
// std::mt19937 rng(rnd());
// int gcd(int a, int b){return b == 0 ? a : gcd(b, a % b);}
const ll mod = 998244353;
const int maxn = 2222;
ll cost[maxn * 60];
int a[31] = {0, 1, 1};
int n, m, x;
#define tp tuple<int, int, int>

map<PI, int> mp[11];
int mrk[maxn * 60];
int dp(int pos, int val, int cnt)
{
    if(mrk[val] > cnt) return 0;
    if(pos == 0) return (val == 0 && cnt == 0) ? 1 : 0;
    if(cnt == 0) return (val == 0 ? 1 : 0);
    if(val <= 0) return 0;
    if(a[pos] * cnt < val) return 0;
    if(mp[pos].count({val, cnt})) return mp[pos][{val, cnt}];
    return mp[pos][{val, cnt}] = (dp(pos, val - a[pos], cnt - 1) + dp(pos - 1, val, cnt)) % mod;
}
int main()
{
    rep(i, 3, 30) a[i] = a[i - 1] + a[i - 2];
    rd(n, x, m);
    MS(cost, 0x3f);
    cost[0] = 0;
    rep(i, 1, 60 * n) rep(j, 1, 30) if(a[j] <= i) cost[i] = min(cost[i], cost[i - a[j]] + 1);
    MS(mrk, 0x3f); mrk[0]= 0;
    rep(i, 1, 60 * n) rep(j, 1, x) if(a[j] <= i && mrk[i - a[j]] < n) mrk[i] = min(mrk[i], mrk[i - a[j]] + 1);

    int ans = 0;
    rep(i, 1, a[x] * n) if(cost[i] == m) ans = (ans + dp(x, i, n)) % mod;
    cout << ans;
}

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

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

Think step by step.

Response:
```plaintext
10 10 1
```

### Explanation:
This test case is designed to expose potential issues in the buggy code by focusing on a scenario where the cost is 1. Here's why this input is challenging:

1. **Small Number of Chips (n = 10)**: This ensures that the problem is small enough to be manageable but still requires the algorithm to handle multiple possible placements.
2. **Maximum Number of Points (x = 10)**: Using the maximum allowed value for \( x \) ensures that the algorithm must consider all possible points.
3. **Cost (m = 1)**: The cost of 1 is the simplest non-zero cost, which means the algorithm needs to find all placements that can be reduced to a single chip. This is a good test case to check if the algorithm correctly handles the operations and minimizes the number of chips.

This input will test the algorithm's ability to correctly calculate the number of valid placements and ensure that the operations are applied correctly to achieve the desired cost.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chips on a Line
time_limit_ms: 5000
memory_limit_mb: 512
Description: You have $n$ chips, and you are going to place all of them in one of $x$ points, numbered from $1$ to $x$. There can be multiple chips in each point.

After placing the chips, you can perform the following four operations (in any order, any number of times):

  * choose a chip in point $i \ge 3$, remove it and place two chips: one in $i-1$, one in $i-2$;    * choose two chips in adjacent points $i$ and $i+1$, remove them and place a new chip in $i+2$;    * choose a chip in point $1$ and move it to $2$;    * choose a chip in point $2$ and move it to $1$.

Note that the coordinates of the chips you place during the operations cannot be less than $1$, but can be greater than $x$.

Denote the cost of chip placement as the minimum number of chips which can be present on the line after you perform these operations, starting from the placement you've chosen.

For example, the cost of placing two chips in points $3$ and one chip in point $5$ is $2$, because you can reduce the number of chips to $2$ as follows:

  * choose a chip in point $3$, remove it, place a chip in $1$ and another chip in $2$;    * choose the chips in points $2$ and $3$, remove them and place a chip in $4$;    * choose the chips in points $4$ and $5$, remove them and place a chip in $6$.

You are given three integers $n$, $x$ and $m$. Calculate the number of placements of exactly $n$ chips in points from $1$ to $x$ having cost equal to $m$, and print it modulo $998244353$. Two placements are considered different if the number of chips in some point differs between these placements.

Input: The only line contains three integers $n$, $x$ and $m$ ($1 \le m \le n \le 1000$; $2 \le x \le 10$).

Output: Print one integer — the number of placements with cost equal to $m$, taken modulo $998244353$.

Examples:
input:
2 3 1
output:
5

input:
42 10 5
output:
902673363

input:
1000 10 8
output:
187821763

Note: In the first example, there are five ways to place $2$ chips in points from $1$ to $3$ so that the cost is $1$:

  * $(1, 1)$;    * $(1, 2)$;    * $(1, 3)$;    * $(2, 2)$;    * $(2, 3)$.

Buggy Code:
MOD = 998244353

def add(x, y):
    x += y
    while x >= MOD:
        x -= MOD
    while x < 0:
        x += MOD
    return x

def mul(x, y):
    return (x * y) % MOD

def solve():
    n, x, m = map(int, input().split())

    # Precompute Fibonacci sequence up to the 30th Fibonacci number
    fib = [0, 1]
    for i in range(2, 31):
        fib.append(fib[i - 1] + fib[i - 2])

    # Maximum possible sum we need to consider
    max_sum = fib[x] * n

    # DP table to count the number of ways to make a sum j using exactly k Fibonacci numbers
    dp = [[0] * (n + 1) for _ in range(max_sum + 1)]
    dp[0][0] = 1

    # Fill dp table
    for i in range(1, x + 1):
        for j in range(max_sum):
            for k in range(n):
                if j + fib[i] <= max_sum:
                    dp[j + fib[i]][k + 1] = add(dp[j + fib[i]][k + 1], dp[j][k])

    # Array to store the minimum cost (number of Fibonacci numbers) to achieve sum j
    cost = [1e9] * (max_sum + 1)
    cost[0] = 0

    # Fill the cost array
    for j in range(1, max_sum + 1):
        for i in range(1, 31):
            if j >= fib[i]:
                cost[j] = min(cost[j], cost[j - fib[i]] + 1)

    # Calculate the final answer
    ans = 0
    for i in range(max_sum + 1):
        if cost[i] == m:
            ans = add(ans, dp[i][n])

    print(ans)

# Entry point
if __name__ == "__main__":
    solve()

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

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

Think step by step.

Response:
```plaintext
10 10 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chips on a Line
time_limit_ms: 5000
memory_limit_mb: 512
Description: You have $n$ chips, and you are going to place all of them in one of $x$ points, numbered from $1$ to $x$. There can be multiple chips in each point.

After placing the chips, you can perform the following four operations (in any order, any number of times):

  * choose a chip in point $i \ge 3$, remove it and place two chips: one in $i-1$, one in $i-2$;    * choose two chips in adjacent points $i$ and $i+1$, remove them and place a new chip in $i+2$;    * choose a chip in point $1$ and move it to $2$;    * choose a chip in point $2$ and move it to $1$.

Note that the coordinates of the chips you place during the operations cannot be less than $1$, but can be greater than $x$.

Denote the cost of chip placement as the minimum number of chips which can be present on the line after you perform these operations, starting from the placement you've chosen.

For example, the cost of placing two chips in points $3$ and one chip in point $5$ is $2$, because you can reduce the number of chips to $2$ as follows:

  * choose a chip in point $3$, remove it, place a chip in $1$ and another chip in $2$;    * choose the chips in points $2$ and $3$, remove them and place a chip in $4$;    * choose the chips in points $4$ and $5$, remove them and place a chip in $6$.

You are given three integers $n$, $x$ and $m$. Calculate the number of placements of exactly $n$ chips in points from $1$ to $x$ having cost equal to $m$, and print it modulo $998244353$. Two placements are considered different if the number of chips in some point differs between these placements.

Input: The only line contains three integers $n$, $x$ and $m$ ($1 \le m \le n \le 1000$; $2 \le x \le 10$).

Output: Print one integer — the number of placements with cost equal to $m$, taken modulo $998244353$.

Examples:
input:
2 3 1
output:
5

input:
42 10 5
output:
902673363

input:
1000 10 8
output:
187821763

Note: In the first example, there are five ways to place $2$ chips in points from $1$ to $3$ so that the cost is $1$:

  * $(1, 1)$;    * $(1, 2)$;    * $(1, 3)$;    * $(2, 2)$;    * $(2, 3)$.

Buggy Code:
import heapq
import os, sys
from io import BytesIO, IOBase

from collections import defaultdict, deque, Counter
from bisect import bisect_left, bisect_right
from heapq import heappush, heappop
from functools import lru_cache, reduce
from itertools import accumulate
from operator import xor
import random
import math
import sys

# sys.setrecursionlimit(10000)
# Fast IO Region
BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
sys.setrecursionlimit(5005)

mag = random.getrandbits(20)

class D:
    def __init__(self, type=int):
        self.d = defaultdict(type)
        self.mag = random.getrandbits(20)

    def __setitem__(self, key, value):
        self.d[key + self.mag] = value

    def __getitem__(self, key):
        return self.d[key + self.mag]

    def __delitem__(self, key):
        del self.d[key + self.mag]

    def __contains__(self, key):
        return key + self.mag in self.d

    def __iter__(self):
        self.keys = iter(self.d.keys())
        return self

    def __next__(self):
        return next(self.keys) - self.mag

    def __len__(self):
        return len(self.d)

ii = lambda: int(input())
mii = lambda: map(int, input().split())
fii = lambda: map(float, input().split())
lmii = lambda: list(map(int, input().split()))
i2c = lambda n: chr(ord('a') + n)
c2i = lambda c: ord(c) - ord('a')

mod = 998244353

# mod = 10 ** 9 + 7

def solve():
    n, x, m = mii()
    fib = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368]
    upper = n * fib[x - 1]
    cnt = [0] * (upper + 1)
    all_m = []
    for i in range(1, upper + 1):
        fj = bisect_right(fib, i)
        cnt[i] = cnt[i - fib[fj - 1]] + 1
        if cnt[i] == m:
            all_m.append(i)
    if not all_m:
        print(0)
        return

    dp = [[0] * (upper + 1) for _ in range(x)]
    for j in range(x):
        dp[j][0] = 1
    for i in range(n):
        for t in range(upper, -1, -1):
            for j in range(1, x):
                if i >= t:
                    dp[j][t] = 0
                elif i == t - 1:
                    dp[j][t] = t + 1
                elif j == 1:
                    dp[j][t] = 0
                elif t >= fib[j]:
                    dp[j][t] = (dp[j - 1][t] + dp[j][t - fib[j]]) % mod
                else:
                    dp[j][t] = dp[j - 1][t]
    res = 0
    for t in all_m:
        res = (res + dp[x - 1][t]) % mod
    print(res)

for _ in range(1):
    solve()

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

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

Think step by step.

Response:
```plaintext
1000 10 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chips on a Line
time_limit_ms: 5000
memory_limit_mb: 512
Description: You have $n$ chips, and you are going to place all of them in one of $x$ points, numbered from $1$ to $x$. There can be multiple chips in each point.

After placing the chips, you can perform the following four operations (in any order, any number of times):

  * choose a chip in point $i \ge 3$, remove it and place two chips: one in $i-1$, one in $i-2$;    * choose two chips in adjacent points $i$ and $i+1$, remove them and place a new chip in $i+2$;    * choose a chip in point $1$ and move it to $2$;    * choose a chip in point $2$ and move it to $1$.

Note that the coordinates of the chips you place during the operations cannot be less than $1$, but can be greater than $x$.

Denote the cost of chip placement as the minimum number of chips which can be present on the line after you perform these operations, starting from the placement you've chosen.

For example, the cost of placing two chips in points $3$ and one chip in point $5$ is $2$, because you can reduce the number of chips to $2$ as follows:

  * choose a chip in point $3$, remove it, place a chip in $1$ and another chip in $2$;    * choose the chips in points $2$ and $3$, remove them and place a chip in $4$;    * choose the chips in points $4$ and $5$, remove them and place a chip in $6$.

You are given three integers $n$, $x$ and $m$. Calculate the number of placements of exactly $n$ chips in points from $1$ to $x$ having cost equal to $m$, and print it modulo $998244353$. Two placements are considered different if the number of chips in some point differs between these placements.

Input: The only line contains three integers $n$, $x$ and $m$ ($1 \le m \le n \le 1000$; $2 \le x \le 10$).

Output: Print one integer — the number of placements with cost equal to $m$, taken modulo $998244353$.

Examples:
input:
2 3 1
output:
5

input:
42 10 5
output:
902673363

input:
1000 10 8
output:
187821763

Note: In the first example, there are five ways to place $2$ chips in points from $1$ to $3$ so that the cost is $1$:

  * $(1, 1)$;    * $(1, 2)$;    * $(1, 3)$;    * $(2, 2)$;    * $(2, 3)$.

Buggy Code:
import java.util.Scanner;

public class Main {
    static final int MOD = 998244353;
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int numChips = scanner.nextInt();
        int maxPosition = scanner.nextInt();
        int targetCost = scanner.nextInt();

        int[] fib = new int[25];
        int[] minCost = new int[55005];
        long[][] dp = new long[1005][55005];
        long result = 0;

        fib[1] = fib[2] = 1;
        for (int i = 3; i <= 24; ++i) {
            fib[i] = fib[i - 1] + fib[i - 2];
        }

        for (int i = 0; i < minCost.length; i++) {
            minCost[i] = Integer.MAX_VALUE;
        }
        minCost[0] = 0;

        for (int i = 1; i <= 24; ++i) {
            for (int j = fib[i]; j <= fib[maxPosition] * numChips; ++j) {
                minCost[j] = Math.min(minCost[j], minCost[j - fib[i]] + 1);
            }
        }

        dp[0][0] = 1;
        for (int t = 1; t <= maxPosition; ++t) {
            for (int i = 1; i <= numChips; ++i) {
                for (int j = fib[t]; j <= fib[maxPosition] * numChips; ++j) {
                    dp[i][j] = (dp[i][j] + dp[i - 1][j - fib[t]]) % MOD;
                }
            }
        }

        for (int t = numChips; t <= fib[maxPosition] * numChips; ++t) {
            if (minCost[t] == targetCost) {
                result = (result + dp[numChips][t]) % MOD;
            }
        }

        System.out.println(result);
    }
}

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

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

Think step by step.

Response:
```plaintext
10 2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Chips on a Line
time_limit_ms: 5000
memory_limit_mb: 512
Description: You have $n$ chips, and you are going to place all of them in one of $x$ points, numbered from $1$ to $x$. There can be multiple chips in each point.

After placing the chips, you can perform the following four operations (in any order, any number of times):

  * choose a chip in point $i \ge 3$, remove it and place two chips: one in $i-1$, one in $i-2$;    * choose two chips in adjacent points $i$ and $i+1$, remove them and place a new chip in $i+2$;    * choose a chip in point $1$ and move it to $2$;    * choose a chip in point $2$ and move it to $1$.

Note that the coordinates of the chips you place during the operations cannot be less than $1$, but can be greater than $x$.

Denote the cost of chip placement as the minimum number of chips which can be present on the line after you perform these operations, starting from the placement you've chosen.

For example, the cost of placing two chips in points $3$ and one chip in point $5$ is $2$, because you can reduce the number of chips to $2$ as follows:

  * choose a chip in point $3$, remove it, place a chip in $1$ and another chip in $2$;    * choose the chips in points $2$ and $3$, remove them and place a chip in $4$;    * choose the chips in points $4$ and $5$, remove them and place a chip in $6$.

You are given three integers $n$, $x$ and $m$. Calculate the number of placements of exactly $n$ chips in points from $1$ to $x$ having cost equal to $m$, and print it modulo $998244353$. Two placements are considered different if the number of chips in some point differs between these placements.

Input: The only line contains three integers $n$, $x$ and $m$ ($1 \le m \le n \le 1000$; $2 \le x \le 10$).

Output: Print one integer — the number of placements with cost equal to $m$, taken modulo $998244353$.

Examples:
input:
2 3 1
output:
5

input:
42 10 5
output:
902673363

input:
1000 10 8
output:
187821763

Note: In the first example, there are five ways to place $2$ chips in points from $1$ to $3$ so that the cost is $1$:

  * $(1, 1)$;    * $(1, 2)$;    * $(1, 3)$;    * $(2, 2)$;    * $(2, 3)$.

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

public class E {

	static IOHandler sc = new IOHandler();
	static PrintWriter out = new PrintWriter(System.out);
	static int mod = 998244353;
	static int [][][] memo;
	static boolean [][][] visited;

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

        new Thread(null, new Solution(), "solution", 1 << 26 ).start();
    }

	private static class Solution implements Runnable {

		public void run() {
			// TODO Auto-generated method stub
			int testCases = 1;

			for (int i = 1; i <= testCases; ++i) {
				solve(i);
			}

			out.flush();
		}

		private void advice() {
			// Read question and Test cases thoroughly
			// Check for overflow , Avoid Integer ==
			// Handle invalid cases
			// Think before solving
			// Search for concepts
			// expected alg like seg tree, trie, binSearch, sqrt decom ...

			// Source of Error
			// Int overflow, Not returning after special case
			// Not taking mod properly
			// sorting input groups after reading certain values
			// Not using the right vairable eg idx in method for i in loop
			// Accounting for special case in general logic. eg special in dp
		}

		private void solve(int t) {
			int n = sc.nextInt();

			int x = sc.nextInt();

			int m = sc.nextInt();

			int [] fib = new int [30];

			fib[1] = 1;
			fib[2] = 1;

			for (int i = 3; i < fib.length; ++i) {
				fib[i] = fib[i - 1] + fib[i - 2];
			}

			// System.out.println(Arrays.toString(fib));

			int max = fib[x] * n + 1;

			int [] dp = new int [max];

			for (int i = 1; i < max; ++i) {
				dp[i] = getNeeded(i, fib);
			}

			// System.out.println(Arrays.toString(dp));

			int res = 0;

			int sum = 0;

			memo = new int [x + 1][][];
			visited = new boolean [x + 1][][];

			for (int i = 1; i <= x; ++i) {
				sum += fib[i];

				memo[i] = new int [n + 1][sum * n + 1];
				visited[i] = new boolean [n + 1][sum * n + 1];
			}

			for (int i = 1; i < max; ++i) {
				if (dp[i] == m) {

					res += solve(x, n, fib, i);
					res %= mod;
				}
			}

			out.println(res);
		}

		private int solve(int pos, int moves, int [] fib, int left) {
			// System.out.println(pos + " " + left + " " + moves);
			if (left == 0)
				return moves == 0 ? 1 : 0;
			else if (moves * fib[pos] < left)
				return 0;
			else if (moves == 0)
				return 0;
			else if (pos < 1)
				return 0;
			else if (visited[pos][moves][left] == true)
				return memo[pos][moves][left];

			int res = solve(pos - 1, moves, fib, left);

			if (left >= fib[pos])
				res += solve(pos, moves - 1, fib, left - fib[pos]);

			res %= mod;

			visited[pos][moves][left] = true;
			memo[pos][moves][left] = res;
			return res;
		}

		private int getNeeded(int left, int [] fib) {

			int total = 0;

			for (int i = fib.length - 1; i > 0; --i) {
				if (left >= fib[i]) {
					++total;
					left -= fib[i];
				}
			}

			return total;
		}

	}

	private static class IOHandler {
      BufferedReader br;
      StringTokenizer st;

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

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

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

      int [] readArray(int n) {
    	  int [] res = new int [n];

    	  for (int i = 0; i < n; ++i)
    		  res[i] = nextInt();

    	  return res;
      }

      int [] readArray2(int n) {
    	  int [] res = new int [n + 1];

    	  for (int i = 1; i <= n; ++i)
    		  res[i] = nextInt();

    	  return res;
      }

      int [][] readGraph(int n, int m, int c) {
    	  int [][] graph = new int [n + 1][];

    	  int [] count = new int [n + 1];
    	  int [][] arr = new int [m][1 + c];

    	  int a, b;

    	  for (int i = 0; i < m; ++i) {
    		  a = sc.nextInt();
    		  b = sc.nextInt();

    		  arr[i][0] = a;
    		  arr[i][1] = b;

    		  ++count[a];
    		  ++count[b];

    		  for (int j = 2; j <= c; ++j) {
    			  arr[i][j] = sc.nextInt();
    		  }
    	  }

    	  for (int i = 0; i <= n; ++i) {
    		  graph[i] = new int [count[i] * c];
    		  count[i] = 0;
    	  }
    	  int swap;

    	  for (int [] tArr : arr) {
    		  for (int j = 1; j < tArr.length; ++j) {
    			  graph[tArr[0]][count[tArr[0]]++] = tArr[j];
    		  }

    		  swap = tArr[0];
    		  tArr[0] = tArr[1];
    		  tArr[1] = swap;

    		  for (int j = 1; j < tArr.length; ++j) {
    			  graph[tArr[0]][count[tArr[0]]++] = tArr[j];
    		  }
    	  }

    	  return graph;
      }

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

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

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

   }

}

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

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

Think step by step.

Response:
```plaintext
10 5 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Beauty of the mountains
time_limit_ms: 2000
memory_limit_mb: 256
Description: Nikita loves mountains and has finally decided to visit the Berlyand mountain range! The range was so beautiful that Nikita decided to capture it on a map. The map is a table of $n$ rows and $m$ columns, with each cell containing a non-negative integer representing the height of the mountain.

He also noticed that mountains come in two types:

  * With snowy caps.    * Without snowy caps.

Nikita is a very pragmatic person. He wants the sum of the heights of the mountains with snowy caps to be equal to the sum of the heights of the mountains without them. He has arranged with the mayor of Berlyand, Polikarp Polikarpovich, to allow him to transform the landscape.

Nikita can perform transformations on submatrices of size $k \times k$ as follows: he can add an integer constant $c$ to the heights of the mountains within this area, but the type of the mountain remains unchanged. Nikita can choose the constant $c$ independently for each transformation. Note that $c$ can be negative.

Before making the transformations, Nikita asks you to find out if it is possible to achieve equality of the sums, or if it is impossible. It doesn't matter at what cost, even if the mountains turn into canyons and have negative heights.

If only one type of mountain is represented on the map, then the sum of the heights of the other type of mountain is considered to be zero.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. This is followed by a description of test cases.

The first line of each test case contains three integers $n, m, k$ ($1 \le n, m \le 500, 1 \le k \le min(n, m)$).

The next $n$ lines of each test case contain $m$ integers $a_{i j}$ ($0 \le a_{i j} \le 10^{9}$) — the initial heights of the mountains.

The next $n$ binary strings of length $m$ for each test case determine the type of mountain, '$0$' — with snowy caps, '$1$' — without them.

It is guaranteed that the sum of $n \cdot m$ for all test cases does not exceed $250\,000$.

Output: For each test case, output "YES" without quotes if it is possible to equalize the sums of the mountain heights, otherwise output "NO" without quotes. You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
8
3 3 2
7 11 3
4 2 3
0 1 15
100
010
000
4 4 3
123 413 24 233
123 42 0 216
22 1 1 53
427 763 22 6
0101
1111
1010
0101
3 3 2
2 1 1
1 1 2
1 5 4
010
101
010
3 3 2
2 1 1
1 1 2
1 5 3
010
101
010
3 4 3
46 49 50 1
19 30 23 12
30 25 1 46
1000
0100
0010
5 4 4
39 30 0 17
22 42 30 13
10 44 46 35
12 19 9 39
21 0 45 40
1000
1111
0011
0111
1100
2 2 2
3 4
6 7
00
00
2 2 2
0 0
2 0
01
00
output:
YES
NO
YES
NO
YES
NO
YES
YES

Note: The mountain array from the first test case looks like this:

![](CDN_BASE_URL/4068fa4645530e8633aa85d1532f2872)

Initially, the sum of the heights of the mountains with snowy caps is $11 + 3 + 4 + 3 + 0 + 1 + 15 = 37$, and without them is $7 + 2 = 9$.

To equalize these sums, we can perform two transformations:

First transformation:

![](CDN_BASE_URL/f3df416c8c27804955f5a43547920ecc)

Note that the constant $c$ can be negative.

After the first transformation, the mountain array looks like this:

![](CDN_BASE_URL/be4be3fe9f25d0c3cdc2cd30c29dfa59)

Second transformation:

![](CDN_BASE_URL/b9da8ea7d124655cdf30018c6bb3968c)

As a result, the mountain array looks like this:

![](CDN_BASE_URL/761aa693d1bfeceaebaeb108d5584b76)

The sum of the heights of the mountains with snowy caps is $17 + 9 + 9 - 16 - 20 - 19 + 15 = -5$, and without them is $7 - 12 = -5$, thus the answer is YES.

Buggy Code:
#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
// #define int long long
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T> using ordered_set =tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update> ;
std::vector<int> intToBinaryVector(ll n) {
    std::vector<int> binaryVector;
    while (n > 0) {
        binaryVector.push_back(n % 2);
        n /= 2;
    }

    // Reverse the vector to get the correct binary representation
    std::reverse(binaryVector.begin(), binaryVector.end());

    return binaryVector;
}
long long binaryVectorToInt(const std::vector<int>& binaryVector) {
    long long result = 0;
    int size = binaryVector.size();
    for (int i = 0; i < size; ++i) {
        if (binaryVector[i] == 1) {
            result += pow(2, size - 1 - i);
        }
    }
    return result;
}
ll gcd(ll a, ll b) {
    if (a < b) swap(a,b);
    while (b != 0) {
        ll temp = b;
        b = a % b;
        a = temp;
    }
    return a;
}
ll lcm(ll a, ll b){
    return a*b/gcd(a,b);
}

long long power(long long exp) {
    long long result = 1;
    long long base = 2;

    while (exp > 0) {
        // If exp is odd, multiply the current base with the result
        if (exp % 2 == 1) {
            result = (result * base) % 998244353;
        }
        // Square the base and reduce the exponent by half
        base = (base * base) % 998244353;
        exp /= 2;
    }

    return result%998244353;
}

vector <int> convert(ll a){
    vector <int> temp;
    while (a!=0){
        temp.push_back(a%10);
        a = a/10;
    }
    reverse(temp.begin(), temp.end());
    return temp;
}
ll conve(vector<int> a, vector <int> b){
    int l1 = a.size();
    int l2 = b.size();
    vector <int> t;
    int p = 0;
    int l = 0;
    ll ans = 0;
    if (l1 >= l2){
        for (int i =l2; i<l1; i++){
            t.push_back(a[p]);
            p++;
        }
        for (int i =0; i<l2; i++){
            t.push_back(a[p+i]);
            t.push_back(b[i]);
        }

    }
    else{
        for (int i = l1; i<l2; i++){
            t.push_back(b[p]);
            p++;
        }
        for (int i =0 ; i<l1; i++){
            t.push_back(a[i]);
            t.push_back(b[i+p]);
        }
    }
    for (int i =0 ;i<l1+l2; i++){
        ans = ans + t[i]*power(l1+l2 - i-1);
        ans = ans%998244353;
    }
    return ans;
}
void solve() {

   ll n, m, k; cin>>n>>m >> k;
   vector <vector <ll>> num(n, vector <ll>(m, 0LL));
    vector <vector <bool>> dp(n,vector <bool>(m, false) );
   for(int i =0  ;i<n; i++){
        for (int j = 0 ; j<m; j++){
            cin>>num[i][j];
        }
   }
   ll cap1 = 0, cap2 = 0;

   for (int i = 0 ; i<n; i++){

        string a; cin>>a;
        for (int j =0 ; j< m; j++){
            if (a[j] == '1'){
                // cout << 'a' << endl;
                dp[i][j] = true;
                cap1+= num[i][j];

            }
            else{
                dp[i][j] = false;
                cap2+= num[i][j];
            }
        }

}

    if (cap1 == cap2){
        cout << "YES" << endl;
        return;
    }
    ll pot = abs(cap1-cap2);
    set <ll> t;
    for (int i = 0 ;i<= n-k; i++){
        for (int j =0; j<=m-k ; j++){
            ll t1 = 0;
            ll t2= 0;
            for (int l = i ; l<i + k; l++){
                for (int p = j; p < j+k; p++){
                    if (dp[l][p]){
                        t1++;
                    }
                    else{
                        t2++;
                    }
                }

            }
            if (t1 == t2){
                continue;
            }
            else{
                int temp = abs(t1-t2);
                t.insert(temp);

            }
        }
    }
    vector <bool> d(pot+1, false);
    d[0] = true;
    for (auto i : t){
        for (int k = i ; k<=pot; k++){
            if (d[k-i]){
                d[k] = true;
            }
        }
    }
    if (d[pot]) {
        cout << "YES" << endl;
        return;

    }
    else{
        cout << "NO" << endl;
        return;
    }
    return;

}

int main() {
    ios::sync_with_stdio(0); cin.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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can maximize the probability of failure. Here are the steps to create such a test case:

1. **Small Scale**: Ensure the test case is small enough to fit within the output length limit.
2. **Edge Cases**:
   - **Single Mountain Type**: Test cases where only one type of mountain (either with or without snowy caps) is present.
   - **Equal Initial Sums**: Test cases where the initial sums of heights of mountains with and without snowy caps are already equal.
   - **Imbalanced Initial Sums**: Test cases where the initial sums are significantly different and require large transformations.
   - **Submatrix Transformation**: Test cases where the submatrix transformation (k x k) can significantly change the sums.
3. **Validation**: Ensure the test case is complete and valid according to the problem constraints.

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

### Test Case 1: Single Mountain Type
- **Input**:
  - 1 test case
  - 3x3 grid with k=2
  - All mountains with snowy caps (type 0)
- **Output**:
  - "NO" (since all mountains are of one type, it's impossible to balance the sums)

### Test Case 2: Equal Initial Sums
- **Input**:
  - 1 test case
  - 2x2 grid with k=2
  - Heights: 1 2, 2 1
  - Types: 0 1, 1 0
- **Output**:
  - "YES" (since the initial sums are already equal)

### Test Case 3: Imbalanced Initial Sums
- **Input**:
  - 1 test case
  - 3x3 grid with k=2
  - Heights: 1 2 3, 4 5 6, 7 8 9
  - Types: 0 0 1, 1 0 0, 0 1 1
- **Output**:
  - "NO" (since the initial sums are significantly different and cannot be balanced by k x k transformations)

### Test Case 4: Submatrix Transformation
- **Input**:
  - 1 test case
  - 4x4 grid with k=3
  - Heights: 1 2 3 4, 5 6 7 8, 9 10 11 12, 13 14 15 16
  - Types: 0 1 0 1, 1 0 1 0, 0 1 0 1, 1 0 1 0
- **Output**:
  - "YES" (since the submatrix transformation can balance the sums)

### Final Test Input
```plaintext
4
3 3 2
1 2 3
4 5 6
7 8 9
001
100
011
2 2 2
1 2
2 1
01
10
3 3 2
1 2 3
4 5 6
7 8 9
001
100
011
4 4 3
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
0101
1010
0101
1010
```

This test input includes a variety of scenarios that can help expose potential bugs in the code, including edge cases and balanced and imbalanced initial sums.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Beauty of the mountains
time_limit_ms: 2000
memory_limit_mb: 256
Description: Nikita loves mountains and has finally decided to visit the Berlyand mountain range! The range was so beautiful that Nikita decided to capture it on a map. The map is a table of $n$ rows and $m$ columns, with each cell containing a non-negative integer representing the height of the mountain.

He also noticed that mountains come in two types:

  * With snowy caps.    * Without snowy caps.

Nikita is a very pragmatic person. He wants the sum of the heights of the mountains with snowy caps to be equal to the sum of the heights of the mountains without them. He has arranged with the mayor of Berlyand, Polikarp Polikarpovich, to allow him to transform the landscape.

Nikita can perform transformations on submatrices of size $k \times k$ as follows: he can add an integer constant $c$ to the heights of the mountains within this area, but the type of the mountain remains unchanged. Nikita can choose the constant $c$ independently for each transformation. Note that $c$ can be negative.

Before making the transformations, Nikita asks you to find out if it is possible to achieve equality of the sums, or if it is impossible. It doesn't matter at what cost, even if the mountains turn into canyons and have negative heights.

If only one type of mountain is represented on the map, then the sum of the heights of the other type of mountain is considered to be zero.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. This is followed by a description of test cases.

The first line of each test case contains three integers $n, m, k$ ($1 \le n, m \le 500, 1 \le k \le min(n, m)$).

The next $n$ lines of each test case contain $m$ integers $a_{i j}$ ($0 \le a_{i j} \le 10^{9}$) — the initial heights of the mountains.

The next $n$ binary strings of length $m$ for each test case determine the type of mountain, '$0$' — with snowy caps, '$1$' — without them.

It is guaranteed that the sum of $n \cdot m$ for all test cases does not exceed $250\,000$.

Output: For each test case, output "YES" without quotes if it is possible to equalize the sums of the mountain heights, otherwise output "NO" without quotes. You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
8
3 3 2
7 11 3
4 2 3
0 1 15
100
010
000
4 4 3
123 413 24 233
123 42 0 216
22 1 1 53
427 763 22 6
0101
1111
1010
0101
3 3 2
2 1 1
1 1 2
1 5 4
010
101
010
3 3 2
2 1 1
1 1 2
1 5 3
010
101
010
3 4 3
46 49 50 1
19 30 23 12
30 25 1 46
1000
0100
0010
5 4 4
39 30 0 17
22 42 30 13
10 44 46 35
12 19 9 39
21 0 45 40
1000
1111
0011
0111
1100
2 2 2
3 4
6 7
00
00
2 2 2
0 0
2 0
01
00
output:
YES
NO
YES
NO
YES
NO
YES
YES

Note: The mountain array from the first test case looks like this:

![](CDN_BASE_URL/4068fa4645530e8633aa85d1532f2872)

Initially, the sum of the heights of the mountains with snowy caps is $11 + 3 + 4 + 3 + 0 + 1 + 15 = 37$, and without them is $7 + 2 = 9$.

To equalize these sums, we can perform two transformations:

First transformation:

![](CDN_BASE_URL/f3df416c8c27804955f5a43547920ecc)

Note that the constant $c$ can be negative.

After the first transformation, the mountain array looks like this:

![](CDN_BASE_URL/be4be3fe9f25d0c3cdc2cd30c29dfa59)

Second transformation:

![](CDN_BASE_URL/b9da8ea7d124655cdf30018c6bb3968c)

As a result, the mountain array looks like this:

![](CDN_BASE_URL/761aa693d1bfeceaebaeb108d5584b76)

The sum of the heights of the mountains with snowy caps is $17 + 9 + 9 - 16 - 20 - 19 + 15 = -5$, and without them is $7 - 12 = -5$, thus the answer is YES.

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

void Init()
{
	std::ios::sync_with_stdio(false);
	std::cin.tie(nullptr);
	std::cout.tie(nullptr);
}

int main()
{
	Init();

	int NumCase = 0;
	std::cin >> NumCase;

	for (int Case = 0; Case < NumCase; Case++)
	{
		int Height = 0;
		int Width = 0;
		int Size = 0;

		std::cin >> Height >> Width >> Size;

		int Sum = 0;

		for (int i = 0; i < Height; i++)
		{
			for (int j = 0; j < Width; j++)
			{
				int Input = 0;
				std::cin >> Input;

				Sum += Input;
			}
		}

		for (int i = 0; i < Height; i++)
		{
			int Input = 0;
			std::cin >> Input;
		}

		if (Sum % 2 == 0)
		{
			std::cout << "YES" << "\n";
		}
		else
		{
			std::cout << "NO" << "\n";
		}
	}

	return 0;
}

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

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

Think step by step.

Response:
```plaintext
2
3 3 2
7 11 3
4 2 3
0 1 15
100
010
000
3 3 2
2 1 1
1 1 2
1 5 4
010
101
010
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Beauty of the mountains
time_limit_ms: 2000
memory_limit_mb: 256
Description: Nikita loves mountains and has finally decided to visit the Berlyand mountain range! The range was so beautiful that Nikita decided to capture it on a map. The map is a table of $n$ rows and $m$ columns, with each cell containing a non-negative integer representing the height of the mountain.

He also noticed that mountains come in two types:

  * With snowy caps.    * Without snowy caps.

Nikita is a very pragmatic person. He wants the sum of the heights of the mountains with snowy caps to be equal to the sum of the heights of the mountains without them. He has arranged with the mayor of Berlyand, Polikarp Polikarpovich, to allow him to transform the landscape.

Nikita can perform transformations on submatrices of size $k \times k$ as follows: he can add an integer constant $c$ to the heights of the mountains within this area, but the type of the mountain remains unchanged. Nikita can choose the constant $c$ independently for each transformation. Note that $c$ can be negative.

Before making the transformations, Nikita asks you to find out if it is possible to achieve equality of the sums, or if it is impossible. It doesn't matter at what cost, even if the mountains turn into canyons and have negative heights.

If only one type of mountain is represented on the map, then the sum of the heights of the other type of mountain is considered to be zero.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. This is followed by a description of test cases.

The first line of each test case contains three integers $n, m, k$ ($1 \le n, m \le 500, 1 \le k \le min(n, m)$).

The next $n$ lines of each test case contain $m$ integers $a_{i j}$ ($0 \le a_{i j} \le 10^{9}$) — the initial heights of the mountains.

The next $n$ binary strings of length $m$ for each test case determine the type of mountain, '$0$' — with snowy caps, '$1$' — without them.

It is guaranteed that the sum of $n \cdot m$ for all test cases does not exceed $250\,000$.

Output: For each test case, output "YES" without quotes if it is possible to equalize the sums of the mountain heights, otherwise output "NO" without quotes. You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
8
3 3 2
7 11 3
4 2 3
0 1 15
100
010
000
4 4 3
123 413 24 233
123 42 0 216
22 1 1 53
427 763 22 6
0101
1111
1010
0101
3 3 2
2 1 1
1 1 2
1 5 4
010
101
010
3 3 2
2 1 1
1 1 2
1 5 3
010
101
010
3 4 3
46 49 50 1
19 30 23 12
30 25 1 46
1000
0100
0010
5 4 4
39 30 0 17
22 42 30 13
10 44 46 35
12 19 9 39
21 0 45 40
1000
1111
0011
0111
1100
2 2 2
3 4
6 7
00
00
2 2 2
0 0
2 0
01
00
output:
YES
NO
YES
NO
YES
NO
YES
YES

Note: The mountain array from the first test case looks like this:

![](CDN_BASE_URL/4068fa4645530e8633aa85d1532f2872)

Initially, the sum of the heights of the mountains with snowy caps is $11 + 3 + 4 + 3 + 0 + 1 + 15 = 37$, and without them is $7 + 2 = 9$.

To equalize these sums, we can perform two transformations:

First transformation:

![](CDN_BASE_URL/f3df416c8c27804955f5a43547920ecc)

Note that the constant $c$ can be negative.

After the first transformation, the mountain array looks like this:

![](CDN_BASE_URL/be4be3fe9f25d0c3cdc2cd30c29dfa59)

Second transformation:

![](CDN_BASE_URL/b9da8ea7d124655cdf30018c6bb3968c)

As a result, the mountain array looks like this:

![](CDN_BASE_URL/761aa693d1bfeceaebaeb108d5584b76)

The sum of the heights of the mountains with snowy caps is $17 + 9 + 9 - 16 - 20 - 19 + 15 = -5$, and without them is $7 - 12 = -5$, thus the answer is YES.

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

#define dbg(x...) \
do { \
std::cout << #x << " -> "; \
err(x); \
} while (0)

void err() {
	std::cout << std::endl;
}

template<class T, class... Ts>
void err(T arg, Ts &... args) {
	std::cout << fixed << setprecision(10) << arg << ' ';
	err(args...);
}

int a[505][505],pre[505][505];
char c[505][505];

void solve() {
    int n,m,k;cin >> n >> m >> k;
    for (int i = 1;i <= n;i++) {
        for (int j = 1;j <= m;j++) cin >> a[i][j];
    }
    int sum0 = 0,sum1 = 0;
    for (int i = 1;i <= n;i++) {
        for (int j = 1;j <= m;j++) {
            cin >> c[i][j];
            pre[i][j] = c[i][j] - '0';
            if (c[i][j] == '1') sum1 += a[i][j];
            else sum0 += a[i][j];
        }
    }
    for (int i = 1;i <= n;i++) {
        for (int j = 1;j <= m;j++) {
            pre[i][j] += pre[i - 1][j] + pre[i][j - 1] - pre[i - 1][j - 1];
        }
    }
    int now = 0;
    for (int i = 1;i <= n;i++) {
        for (int j = 1;j <= m;j++) {
            int nx = i + k - 1,ny = j + k - 1;
            if (nx >= 1 && nx <= n && ny <= m && ny >= 1) {
                int all = pre[nx][ny] - pre[nx][j - 1] - pre[i - 1][ny] + pre[i - 1][j - 1];
                int sum = k * k;
                int dis = abs((sum - all) - all);
                if (now == 0) now = dis;
                else now = __gcd(now,dis);
            }
        }
    }
    int dis = abs(sum0 - sum1);
    if (dis == 0 && now == 0) {
        cout << "YES" << '\n';
    }
    else if (dis == 0 && now != 0) {
        cout << "NO" << '\n';
    }
    else if (dis != 0 && now == 0) {
        cout << "NO" << '\n';
    }
    else {
        if (dis % now == 0) cout << "YES" << '\n';
        else cout << "NO" << '\n';
    }
}

signed main()
{
	ios::sync_with_stdio(false);cin.tie(nullptr);
	int t = 1;
    cin >> t;
	while (t--) solve();
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
3 3 2
1 2 3
4 5 6
7 8 9
010
101
010
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Beauty of the mountains
time_limit_ms: 2000
memory_limit_mb: 256
Description: Nikita loves mountains and has finally decided to visit the Berlyand mountain range! The range was so beautiful that Nikita decided to capture it on a map. The map is a table of $n$ rows and $m$ columns, with each cell containing a non-negative integer representing the height of the mountain.

He also noticed that mountains come in two types:

  * With snowy caps.    * Without snowy caps.

Nikita is a very pragmatic person. He wants the sum of the heights of the mountains with snowy caps to be equal to the sum of the heights of the mountains without them. He has arranged with the mayor of Berlyand, Polikarp Polikarpovich, to allow him to transform the landscape.

Nikita can perform transformations on submatrices of size $k \times k$ as follows: he can add an integer constant $c$ to the heights of the mountains within this area, but the type of the mountain remains unchanged. Nikita can choose the constant $c$ independently for each transformation. Note that $c$ can be negative.

Before making the transformations, Nikita asks you to find out if it is possible to achieve equality of the sums, or if it is impossible. It doesn't matter at what cost, even if the mountains turn into canyons and have negative heights.

If only one type of mountain is represented on the map, then the sum of the heights of the other type of mountain is considered to be zero.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. This is followed by a description of test cases.

The first line of each test case contains three integers $n, m, k$ ($1 \le n, m \le 500, 1 \le k \le min(n, m)$).

The next $n$ lines of each test case contain $m$ integers $a_{i j}$ ($0 \le a_{i j} \le 10^{9}$) — the initial heights of the mountains.

The next $n$ binary strings of length $m$ for each test case determine the type of mountain, '$0$' — with snowy caps, '$1$' — without them.

It is guaranteed that the sum of $n \cdot m$ for all test cases does not exceed $250\,000$.

Output: For each test case, output "YES" without quotes if it is possible to equalize the sums of the mountain heights, otherwise output "NO" without quotes. You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
8
3 3 2
7 11 3
4 2 3
0 1 15
100
010
000
4 4 3
123 413 24 233
123 42 0 216
22 1 1 53
427 763 22 6
0101
1111
1010
0101
3 3 2
2 1 1
1 1 2
1 5 4
010
101
010
3 3 2
2 1 1
1 1 2
1 5 3
010
101
010
3 4 3
46 49 50 1
19 30 23 12
30 25 1 46
1000
0100
0010
5 4 4
39 30 0 17
22 42 30 13
10 44 46 35
12 19 9 39
21 0 45 40
1000
1111
0011
0111
1100
2 2 2
3 4
6 7
00
00
2 2 2
0 0
2 0
01
00
output:
YES
NO
YES
NO
YES
NO
YES
YES

Note: The mountain array from the first test case looks like this:

![](CDN_BASE_URL/4068fa4645530e8633aa85d1532f2872)

Initially, the sum of the heights of the mountains with snowy caps is $11 + 3 + 4 + 3 + 0 + 1 + 15 = 37$, and without them is $7 + 2 = 9$.

To equalize these sums, we can perform two transformations:

First transformation:

![](CDN_BASE_URL/f3df416c8c27804955f5a43547920ecc)

Note that the constant $c$ can be negative.

After the first transformation, the mountain array looks like this:

![](CDN_BASE_URL/be4be3fe9f25d0c3cdc2cd30c29dfa59)

Second transformation:

![](CDN_BASE_URL/b9da8ea7d124655cdf30018c6bb3968c)

As a result, the mountain array looks like this:

![](CDN_BASE_URL/761aa693d1bfeceaebaeb108d5584b76)

The sum of the heights of the mountains with snowy caps is $17 + 9 + 9 - 16 - 20 - 19 + 15 = -5$, and without them is $7 - 12 = -5$, thus the answer is YES.

Buggy Code:
def can_balance_mountains(t):
    results = []
    for _ in range(t):
        n, m, k = map(int, input().split())
        matrix = [list(map(int, input().split())) for _ in range(n)]
        types_matrix = [input() for _ in range(n)]

        # Initialize sums
        sum_snow = 0
        sum_no_snow = 0

        for i in range(n):
            for j in range(m):
                if types_matrix[i][j] == '0':
                    sum_snow += matrix[i][j]
                else:
                    sum_no_snow += matrix[i][j]

        # Check if the difference is even
        if (sum_snow - sum_no_snow) % 2 == 0:
            results.append("YES")
        else:
            results.append("NO")

    return results

# Reading the number of test cases
t = int(input())

results = can_balance_mountains(t)

# Outputting the results
for result in results:
    print(result)

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

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

Think step by step.

Response:
```plaintext
3
3 3 2
1 2 3
4 5 6
7 8 9
000
111
000
3 3 2
1 2 3
4 5 6
7 8 9
010
101
010
3 3 2
1 2 3
4 5 6
7 8 9
000
000
000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Beauty of the mountains
time_limit_ms: 2000
memory_limit_mb: 256
Description: Nikita loves mountains and has finally decided to visit the Berlyand mountain range! The range was so beautiful that Nikita decided to capture it on a map. The map is a table of $n$ rows and $m$ columns, with each cell containing a non-negative integer representing the height of the mountain.

He also noticed that mountains come in two types:

  * With snowy caps.    * Without snowy caps.

Nikita is a very pragmatic person. He wants the sum of the heights of the mountains with snowy caps to be equal to the sum of the heights of the mountains without them. He has arranged with the mayor of Berlyand, Polikarp Polikarpovich, to allow him to transform the landscape.

Nikita can perform transformations on submatrices of size $k \times k$ as follows: he can add an integer constant $c$ to the heights of the mountains within this area, but the type of the mountain remains unchanged. Nikita can choose the constant $c$ independently for each transformation. Note that $c$ can be negative.

Before making the transformations, Nikita asks you to find out if it is possible to achieve equality of the sums, or if it is impossible. It doesn't matter at what cost, even if the mountains turn into canyons and have negative heights.

If only one type of mountain is represented on the map, then the sum of the heights of the other type of mountain is considered to be zero.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. This is followed by a description of test cases.

The first line of each test case contains three integers $n, m, k$ ($1 \le n, m \le 500, 1 \le k \le min(n, m)$).

The next $n$ lines of each test case contain $m$ integers $a_{i j}$ ($0 \le a_{i j} \le 10^{9}$) — the initial heights of the mountains.

The next $n$ binary strings of length $m$ for each test case determine the type of mountain, '$0$' — with snowy caps, '$1$' — without them.

It is guaranteed that the sum of $n \cdot m$ for all test cases does not exceed $250\,000$.

Output: For each test case, output "YES" without quotes if it is possible to equalize the sums of the mountain heights, otherwise output "NO" without quotes. You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
8
3 3 2
7 11 3
4 2 3
0 1 15
100
010
000
4 4 3
123 413 24 233
123 42 0 216
22 1 1 53
427 763 22 6
0101
1111
1010
0101
3 3 2
2 1 1
1 1 2
1 5 4
010
101
010
3 3 2
2 1 1
1 1 2
1 5 3
010
101
010
3 4 3
46 49 50 1
19 30 23 12
30 25 1 46
1000
0100
0010
5 4 4
39 30 0 17
22 42 30 13
10 44 46 35
12 19 9 39
21 0 45 40
1000
1111
0011
0111
1100
2 2 2
3 4
6 7
00
00
2 2 2
0 0
2 0
01
00
output:
YES
NO
YES
NO
YES
NO
YES
YES

Note: The mountain array from the first test case looks like this:

![](CDN_BASE_URL/4068fa4645530e8633aa85d1532f2872)

Initially, the sum of the heights of the mountains with snowy caps is $11 + 3 + 4 + 3 + 0 + 1 + 15 = 37$, and without them is $7 + 2 = 9$.

To equalize these sums, we can perform two transformations:

First transformation:

![](CDN_BASE_URL/f3df416c8c27804955f5a43547920ecc)

Note that the constant $c$ can be negative.

After the first transformation, the mountain array looks like this:

![](CDN_BASE_URL/be4be3fe9f25d0c3cdc2cd30c29dfa59)

Second transformation:

![](CDN_BASE_URL/b9da8ea7d124655cdf30018c6bb3968c)

As a result, the mountain array looks like this:

![](CDN_BASE_URL/761aa693d1bfeceaebaeb108d5584b76)

The sum of the heights of the mountains with snowy caps is $17 + 9 + 9 - 16 - 20 - 19 + 15 = -5$, and without them is $7 - 12 = -5$, thus the answer is YES.

Buggy Code:
import math
class FenwickTree:
    def __init__(self, n, m):
        self.ft = [[0 for i in range(n + 1)] for j in range(m + 1)]
        self.n = n
        self.m = m

    def LSB(self, x):
        return x & (-x)

    def update(self, x, y, val):
        while x <= self.m:
            y1 = y
            while y1 <= self.n:
                self.ft[x][y1] += val
                y1 += self.LSB(y1)
            x += self.LSB(x)

    def summation(self, y, x):
        s = 0
        x1 = x
        while x1 > 0:
            y1 = y
            while y1 > 0:
                s += self.ft[x1][y1]
                y1 -= self.LSB(y1)
            x1 -= self.LSB(x1)
        return s

    def Summation(self, x1, y1, x2, y2):
        return (self.summation(x2, y2) - self.summation(x1 - 1, y2) - self.summation(x2, y1 - 1) + self.summation(
            x1 - 1, y1 - 1))

    def Construct(self, matrix):
        self.n = len(matrix)
        self.m = len(matrix[0])
        for i in range(self.m):
            for j in range(self.n):
                self.update(i + 1, j + 1, matrix[i][j])
import sys
input = sys.stdin.readline
def solve():
    n,m,k = map(int,input().split())
    if k == 1:print("YES");return
    fen = FenwickTree(n,m)
    FACTS = []
    M = [list(map(int,input().split())) for _ in range(n)]
    XD = 0
    for i in range(n):
        X = input()
        for j in range(m):
            if X[j] == "1":
                fen.update(i+1,j+1,1)
                XD+=M[i][j]
            else:
                fen.update(i+1,j+1,-1)
                XD-=M[i][j]
    for i in range(n-k+1):
        for j in range(m-k+1):
            O = abs(fen.Summation(i+1, j+1,i+k,j+k))
            if O:FACTS.append(O)
    if XD == 0 or (FACTS and XD % math.gcd(*FACTS) == 0):
        print("YES")
    else:print("NO")
for _ in range(int(input())):
    solve()

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

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

Think step by step.

Response:
```plaintext
1
3 3 2
1 2 3
4 5 6
7 8 9
010
101
010
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Beauty of the mountains
time_limit_ms: 2000
memory_limit_mb: 256
Description: Nikita loves mountains and has finally decided to visit the Berlyand mountain range! The range was so beautiful that Nikita decided to capture it on a map. The map is a table of $n$ rows and $m$ columns, with each cell containing a non-negative integer representing the height of the mountain.

He also noticed that mountains come in two types:

  * With snowy caps.    * Without snowy caps.

Nikita is a very pragmatic person. He wants the sum of the heights of the mountains with snowy caps to be equal to the sum of the heights of the mountains without them. He has arranged with the mayor of Berlyand, Polikarp Polikarpovich, to allow him to transform the landscape.

Nikita can perform transformations on submatrices of size $k \times k$ as follows: he can add an integer constant $c$ to the heights of the mountains within this area, but the type of the mountain remains unchanged. Nikita can choose the constant $c$ independently for each transformation. Note that $c$ can be negative.

Before making the transformations, Nikita asks you to find out if it is possible to achieve equality of the sums, or if it is impossible. It doesn't matter at what cost, even if the mountains turn into canyons and have negative heights.

If only one type of mountain is represented on the map, then the sum of the heights of the other type of mountain is considered to be zero.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. This is followed by a description of test cases.

The first line of each test case contains three integers $n, m, k$ ($1 \le n, m \le 500, 1 \le k \le min(n, m)$).

The next $n$ lines of each test case contain $m$ integers $a_{i j}$ ($0 \le a_{i j} \le 10^{9}$) — the initial heights of the mountains.

The next $n$ binary strings of length $m$ for each test case determine the type of mountain, '$0$' — with snowy caps, '$1$' — without them.

It is guaranteed that the sum of $n \cdot m$ for all test cases does not exceed $250\,000$.

Output: For each test case, output "YES" without quotes if it is possible to equalize the sums of the mountain heights, otherwise output "NO" without quotes. You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
8
3 3 2
7 11 3
4 2 3
0 1 15
100
010
000
4 4 3
123 413 24 233
123 42 0 216
22 1 1 53
427 763 22 6
0101
1111
1010
0101
3 3 2
2 1 1
1 1 2
1 5 4
010
101
010
3 3 2
2 1 1
1 1 2
1 5 3
010
101
010
3 4 3
46 49 50 1
19 30 23 12
30 25 1 46
1000
0100
0010
5 4 4
39 30 0 17
22 42 30 13
10 44 46 35
12 19 9 39
21 0 45 40
1000
1111
0011
0111
1100
2 2 2
3 4
6 7
00
00
2 2 2
0 0
2 0
01
00
output:
YES
NO
YES
NO
YES
NO
YES
YES

Note: The mountain array from the first test case looks like this:

![](CDN_BASE_URL/4068fa4645530e8633aa85d1532f2872)

Initially, the sum of the heights of the mountains with snowy caps is $11 + 3 + 4 + 3 + 0 + 1 + 15 = 37$, and without them is $7 + 2 = 9$.

To equalize these sums, we can perform two transformations:

First transformation:

![](CDN_BASE_URL/f3df416c8c27804955f5a43547920ecc)

Note that the constant $c$ can be negative.

After the first transformation, the mountain array looks like this:

![](CDN_BASE_URL/be4be3fe9f25d0c3cdc2cd30c29dfa59)

Second transformation:

![](CDN_BASE_URL/b9da8ea7d124655cdf30018c6bb3968c)

As a result, the mountain array looks like this:

![](CDN_BASE_URL/761aa693d1bfeceaebaeb108d5584b76)

The sum of the heights of the mountains with snowy caps is $17 + 9 + 9 - 16 - 20 - 19 + 15 = -5$, and without them is $7 - 12 = -5$, thus the answer is YES.

Buggy Code:
t = int(input())
for _ in range(t):
    n,m,k = map(int,input().split())
    heights = [list(map(int, input().split())) for _ in range(n)]
    snow_or_not = [list(map(int, input())) for _ in range(n)]
    snow, not_snow = 0,0
    for i in range(n):
        for j in range(m):
            if snow_or_not[i][j] == 0:
                snow += heights[i][j]
            else:
                not_snow += heights[i][j]
    gap = abs(snow - not_snow)

    if gap == 0:
        print('YES')

    else:
        def sub_sum(matrix, x1, y1, x2, y2):
            total = 0
            for i in range(x1, x2 + 1):
                for j in range(y1, y2 + 1):
                    total += matrix[i][j]
            return total

        check = set()
        for i in range(n - k + 1):
            for j in range(m - k + 1):
                tmp = abs(k **2 - 2*sub_sum(snow_or_not, i, j, i + k - 1, j + k - 1))
                if tmp:
                    check.add(tmp)
        if not check:
            print("NO")
        else:
            dp = [False] * (gap + 1)
            dp[0] = True

            for num in check:
                for i in range(num, gap + 1):
                    if dp[i - num]:
                        dp[i] = True

            if dp[gap]:
                print("YES")
            else:
                print("NO")

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

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

Think step by step.

Response:
```plaintext
1
3 3 2
1 2 3
4 5 6
7 8 9
010
101
010
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Beauty of the mountains
time_limit_ms: 2000
memory_limit_mb: 256
Description: Nikita loves mountains and has finally decided to visit the Berlyand mountain range! The range was so beautiful that Nikita decided to capture it on a map. The map is a table of $n$ rows and $m$ columns, with each cell containing a non-negative integer representing the height of the mountain.

He also noticed that mountains come in two types:

  * With snowy caps.    * Without snowy caps.

Nikita is a very pragmatic person. He wants the sum of the heights of the mountains with snowy caps to be equal to the sum of the heights of the mountains without them. He has arranged with the mayor of Berlyand, Polikarp Polikarpovich, to allow him to transform the landscape.

Nikita can perform transformations on submatrices of size $k \times k$ as follows: he can add an integer constant $c$ to the heights of the mountains within this area, but the type of the mountain remains unchanged. Nikita can choose the constant $c$ independently for each transformation. Note that $c$ can be negative.

Before making the transformations, Nikita asks you to find out if it is possible to achieve equality of the sums, or if it is impossible. It doesn't matter at what cost, even if the mountains turn into canyons and have negative heights.

If only one type of mountain is represented on the map, then the sum of the heights of the other type of mountain is considered to be zero.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. This is followed by a description of test cases.

The first line of each test case contains three integers $n, m, k$ ($1 \le n, m \le 500, 1 \le k \le min(n, m)$).

The next $n$ lines of each test case contain $m$ integers $a_{i j}$ ($0 \le a_{i j} \le 10^{9}$) — the initial heights of the mountains.

The next $n$ binary strings of length $m$ for each test case determine the type of mountain, '$0$' — with snowy caps, '$1$' — without them.

It is guaranteed that the sum of $n \cdot m$ for all test cases does not exceed $250\,000$.

Output: For each test case, output "YES" without quotes if it is possible to equalize the sums of the mountain heights, otherwise output "NO" without quotes. You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
8
3 3 2
7 11 3
4 2 3
0 1 15
100
010
000
4 4 3
123 413 24 233
123 42 0 216
22 1 1 53
427 763 22 6
0101
1111
1010
0101
3 3 2
2 1 1
1 1 2
1 5 4
010
101
010
3 3 2
2 1 1
1 1 2
1 5 3
010
101
010
3 4 3
46 49 50 1
19 30 23 12
30 25 1 46
1000
0100
0010
5 4 4
39 30 0 17
22 42 30 13
10 44 46 35
12 19 9 39
21 0 45 40
1000
1111
0011
0111
1100
2 2 2
3 4
6 7
00
00
2 2 2
0 0
2 0
01
00
output:
YES
NO
YES
NO
YES
NO
YES
YES

Note: The mountain array from the first test case looks like this:

![](CDN_BASE_URL/4068fa4645530e8633aa85d1532f2872)

Initially, the sum of the heights of the mountains with snowy caps is $11 + 3 + 4 + 3 + 0 + 1 + 15 = 37$, and without them is $7 + 2 = 9$.

To equalize these sums, we can perform two transformations:

First transformation:

![](CDN_BASE_URL/f3df416c8c27804955f5a43547920ecc)

Note that the constant $c$ can be negative.

After the first transformation, the mountain array looks like this:

![](CDN_BASE_URL/be4be3fe9f25d0c3cdc2cd30c29dfa59)

Second transformation:

![](CDN_BASE_URL/b9da8ea7d124655cdf30018c6bb3968c)

As a result, the mountain array looks like this:

![](CDN_BASE_URL/761aa693d1bfeceaebaeb108d5584b76)

The sum of the heights of the mountains with snowy caps is $17 + 9 + 9 - 16 - 20 - 19 + 15 = -5$, and without them is $7 - 12 = -5$, thus the answer is YES.

Buggy Code:
import java.util.*;

public class Main{
    public static void main(String[] args){
        Scanner scan = new Scanner(System.in);
        int t = scan.nextInt();
        while(t-->0){
            int n = scan.nextInt();
            int m = scan.nextInt();
            int k = scan.nextInt();
            int sum = 0;
            for(int i=0;i<n;i++){
                for(int j=0;j<m;j++) sum += scan.nextInt();
            }
            String ans = "NO";
            while(n-->0){
                String str = scan.next();
            }
            if(sum%2==0) ans = "YES";
            System.out.println(ans);
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
2
3 3 2
1 2 3
4 5 6
7 8 9
000
000
000
3 3 2
1 2 3
4 5 6
7 8 9
111
111
111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Beauty of the mountains
time_limit_ms: 2000
memory_limit_mb: 256
Description: Nikita loves mountains and has finally decided to visit the Berlyand mountain range! The range was so beautiful that Nikita decided to capture it on a map. The map is a table of $n$ rows and $m$ columns, with each cell containing a non-negative integer representing the height of the mountain.

He also noticed that mountains come in two types:

  * With snowy caps.    * Without snowy caps.

Nikita is a very pragmatic person. He wants the sum of the heights of the mountains with snowy caps to be equal to the sum of the heights of the mountains without them. He has arranged with the mayor of Berlyand, Polikarp Polikarpovich, to allow him to transform the landscape.

Nikita can perform transformations on submatrices of size $k \times k$ as follows: he can add an integer constant $c$ to the heights of the mountains within this area, but the type of the mountain remains unchanged. Nikita can choose the constant $c$ independently for each transformation. Note that $c$ can be negative.

Before making the transformations, Nikita asks you to find out if it is possible to achieve equality of the sums, or if it is impossible. It doesn't matter at what cost, even if the mountains turn into canyons and have negative heights.

If only one type of mountain is represented on the map, then the sum of the heights of the other type of mountain is considered to be zero.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. This is followed by a description of test cases.

The first line of each test case contains three integers $n, m, k$ ($1 \le n, m \le 500, 1 \le k \le min(n, m)$).

The next $n$ lines of each test case contain $m$ integers $a_{i j}$ ($0 \le a_{i j} \le 10^{9}$) — the initial heights of the mountains.

The next $n$ binary strings of length $m$ for each test case determine the type of mountain, '$0$' — with snowy caps, '$1$' — without them.

It is guaranteed that the sum of $n \cdot m$ for all test cases does not exceed $250\,000$.

Output: For each test case, output "YES" without quotes if it is possible to equalize the sums of the mountain heights, otherwise output "NO" without quotes. You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
8
3 3 2
7 11 3
4 2 3
0 1 15
100
010
000
4 4 3
123 413 24 233
123 42 0 216
22 1 1 53
427 763 22 6
0101
1111
1010
0101
3 3 2
2 1 1
1 1 2
1 5 4
010
101
010
3 3 2
2 1 1
1 1 2
1 5 3
010
101
010
3 4 3
46 49 50 1
19 30 23 12
30 25 1 46
1000
0100
0010
5 4 4
39 30 0 17
22 42 30 13
10 44 46 35
12 19 9 39
21 0 45 40
1000
1111
0011
0111
1100
2 2 2
3 4
6 7
00
00
2 2 2
0 0
2 0
01
00
output:
YES
NO
YES
NO
YES
NO
YES
YES

Note: The mountain array from the first test case looks like this:

![](CDN_BASE_URL/4068fa4645530e8633aa85d1532f2872)

Initially, the sum of the heights of the mountains with snowy caps is $11 + 3 + 4 + 3 + 0 + 1 + 15 = 37$, and without them is $7 + 2 = 9$.

To equalize these sums, we can perform two transformations:

First transformation:

![](CDN_BASE_URL/f3df416c8c27804955f5a43547920ecc)

Note that the constant $c$ can be negative.

After the first transformation, the mountain array looks like this:

![](CDN_BASE_URL/be4be3fe9f25d0c3cdc2cd30c29dfa59)

Second transformation:

![](CDN_BASE_URL/b9da8ea7d124655cdf30018c6bb3968c)

As a result, the mountain array looks like this:

![](CDN_BASE_URL/761aa693d1bfeceaebaeb108d5584b76)

The sum of the heights of the mountains with snowy caps is $17 + 9 + 9 - 16 - 20 - 19 + 15 = -5$, and without them is $7 - 12 = -5$, thus the answer is YES.

Buggy Code:
import java.util.*;

public class BeautyMountains {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        for(int test = 0; test < t; test++) {
            int n = sc.nextInt();
            int m = sc.nextInt();
            int k = sc.nextInt();
            long[][] arr = new long[n][m];
            for(int i = 0; i < n; i++) {
                for(int j = 0; j < m; j++) {
                    arr[i][j] = sc.nextLong();
                }
            }
            boolean[][] types = new boolean[n][m];
            for(int i = 0; i < n; i++) {
                String row = sc.next();
                for(int j = 0; j < m; j++) {
                    types[i][j] = row.charAt(j) == '1';
                }
            }
            long total0 = 0;
            long total1 = 0;
            for(int i = 0; i < n; i++) {
                for(int j = 0; j < m; j++) {
                    if(types[i][j]) total1 += arr[i][j];
                    else total0 += arr[i][j];
                }
            }
            if(total0 == total1) {
                System.out.println("YES");
                continue;
            }
            long diff = Math.abs(total0 - total1);
            List<Integer> diffs = new ArrayList<>();
            for(int i = 0; i < n - k + 1; i++) {
                for(int j = 0; j < m - k + 1; j++) {
                    int count0 = 0;
                    int count1 = 0;
                    for(int x = 0; x < k; x++) {
                        for(int y = 0; y < k; y++) {
                            if(types[i + x][j + y]) count1++;
                            else count0++;
                        }
                    }
                    diffs.add(Math.abs(count0 - count1));
                }
            }
            Set<Long> set = new HashSet<>();
            List<Long> dp = new ArrayList<>();
            set.add(0L);
            dp.add(0L);
            for(int i = 0; i < dp.size(); i++) {
                long val = dp.get(i);
                for(int d : diffs) {
                    long j = val + d;
                    if(j <= diff && !set.contains(j)) {
                        dp.add(j);
                        set.add(j);
                    }
                }
            }
//            System.out.println(dp);
//            System.out.println(diff);
//            System.out.println(diffs);
            if(set.contains(diff)) System.out.println("YES");
            else System.out.println("NO");
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
2
3 3 2
1 2 3
4 5 6
7 8 9
010
101
010
3 3 2
1 2 3
4 5 6
7 8 9
000
111
000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Beauty of the mountains
time_limit_ms: 2000
memory_limit_mb: 256
Description: Nikita loves mountains and has finally decided to visit the Berlyand mountain range! The range was so beautiful that Nikita decided to capture it on a map. The map is a table of $n$ rows and $m$ columns, with each cell containing a non-negative integer representing the height of the mountain.

He also noticed that mountains come in two types:

  * With snowy caps.    * Without snowy caps.

Nikita is a very pragmatic person. He wants the sum of the heights of the mountains with snowy caps to be equal to the sum of the heights of the mountains without them. He has arranged with the mayor of Berlyand, Polikarp Polikarpovich, to allow him to transform the landscape.

Nikita can perform transformations on submatrices of size $k \times k$ as follows: he can add an integer constant $c$ to the heights of the mountains within this area, but the type of the mountain remains unchanged. Nikita can choose the constant $c$ independently for each transformation. Note that $c$ can be negative.

Before making the transformations, Nikita asks you to find out if it is possible to achieve equality of the sums, or if it is impossible. It doesn't matter at what cost, even if the mountains turn into canyons and have negative heights.

If only one type of mountain is represented on the map, then the sum of the heights of the other type of mountain is considered to be zero.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. This is followed by a description of test cases.

The first line of each test case contains three integers $n, m, k$ ($1 \le n, m \le 500, 1 \le k \le min(n, m)$).

The next $n$ lines of each test case contain $m$ integers $a_{i j}$ ($0 \le a_{i j} \le 10^{9}$) — the initial heights of the mountains.

The next $n$ binary strings of length $m$ for each test case determine the type of mountain, '$0$' — with snowy caps, '$1$' — without them.

It is guaranteed that the sum of $n \cdot m$ for all test cases does not exceed $250\,000$.

Output: For each test case, output "YES" without quotes if it is possible to equalize the sums of the mountain heights, otherwise output "NO" without quotes. You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
8
3 3 2
7 11 3
4 2 3
0 1 15
100
010
000
4 4 3
123 413 24 233
123 42 0 216
22 1 1 53
427 763 22 6
0101
1111
1010
0101
3 3 2
2 1 1
1 1 2
1 5 4
010
101
010
3 3 2
2 1 1
1 1 2
1 5 3
010
101
010
3 4 3
46 49 50 1
19 30 23 12
30 25 1 46
1000
0100
0010
5 4 4
39 30 0 17
22 42 30 13
10 44 46 35
12 19 9 39
21 0 45 40
1000
1111
0011
0111
1100
2 2 2
3 4
6 7
00
00
2 2 2
0 0
2 0
01
00
output:
YES
NO
YES
NO
YES
NO
YES
YES

Note: The mountain array from the first test case looks like this:

![](CDN_BASE_URL/4068fa4645530e8633aa85d1532f2872)

Initially, the sum of the heights of the mountains with snowy caps is $11 + 3 + 4 + 3 + 0 + 1 + 15 = 37$, and without them is $7 + 2 = 9$.

To equalize these sums, we can perform two transformations:

First transformation:

![](CDN_BASE_URL/f3df416c8c27804955f5a43547920ecc)

Note that the constant $c$ can be negative.

After the first transformation, the mountain array looks like this:

![](CDN_BASE_URL/be4be3fe9f25d0c3cdc2cd30c29dfa59)

Second transformation:

![](CDN_BASE_URL/b9da8ea7d124655cdf30018c6bb3968c)

As a result, the mountain array looks like this:

![](CDN_BASE_URL/761aa693d1bfeceaebaeb108d5584b76)

The sum of the heights of the mountains with snowy caps is $17 + 9 + 9 - 16 - 20 - 19 + 15 = -5$, and without them is $7 - 12 = -5$, thus the answer is YES.

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

#define pl(a) " [" << #a << ": " << (a) << "] "
#define pts(a) " [" << #a << ": { first: " << (a.first) << ", second: " << (a.second) << "}] "

#define all(vi) vi.begin(), vi.end()
#define endl "\n"
#define ll long long

using namespace std;

pair<ll int, ll int> n4[4] { {1, 0}, {-1, 0}, {0, 1}, {0, -1} };
pair<ll int, ll int> n8[8] { {1, 0}, {-1, 0}, {0, 1}, {0, -1}, {1, 1}, {-1, -1}, {-1, 1}, {1, -1} };

ll int h(ll int x1, ll int y1, ll int x2, ll int y2) { return abs(x2 - x1) + abs(y2 - y1); }

template<typename A> ostream& operator<<(ostream &cout, vector<A> const &v);
template<typename A, typename B> ostream& operator<<(ostream &cout, pair<A, B> const &p) { return cout << "[" << p.first << ", " << p.second << "]"; }
template<typename A> ostream& operator<<(ostream &cout, vector<A> const &v) { cout << "["; for(ll int i = 0; i < v.size(); i++) {if (i) cout << ", "; cout << v[i];} return cout << "]";}
template<typename A, typename B> istream& operator>>(istream& cin, pair<A, B> &p) { cin >> p.first; return cin >> p.second; }

ll int pow(ll int x, ll int y) {
    if (y == 0) return 1;
    if (y % 2 == 0) {
        ll int a = pow(x, y / 2);
        return a * a;
    } else {
        ll int a = pow(x, y / 2);
        a = a * a;

        return a * x;
    }
    assert(false);
    return 0;
}

void set_in(string s) {
    if (s.length() > 0) {
        freopen ((s+".in").c_str(), "r", stdin);
        freopen ((s+".out").c_str(), "w", stdout);
    }
}
bool prime[250001];
void SieveOfEratosthenes(int n)
{
    // Create a boolean array "prime[0..n]" and initialize
    // all entries it as true. A value in prime[i] will
    // finally be false if i is Not a prime, else true.
    memset(prime, true, sizeof(prime));

    for (int p = 2; p * p <= n; p++) {
        // If prime[p] is not changed, then it is a prime
        if (prime[p] == true) {
            // Update all multiples of p greater than or
            // equal to the square of it numbers which are
            // multiple of p and are less than p^2 are
            // already been marked.
            for (int i = p * p; i <= n; i += p)
                prime[i] = false;
        }
    }

    // Print all prime numbers
    // for (int p = 2; p <= n; p++)
    //     if (prime[p])
    //         cout << p << " ";
}

void solve() {
    // a 0s and b 1s
    // A - B += (a1 - b1) * c1 + (a2 - b2) * c2 + ..
    // so I have this array and I want to make this sum equal to - (A - B)
    //

    ll int n, m, k; cin >> n >> m >> k;
    vector<vector<ll int>> arr(n, vector<ll int>(m)); for (ll int i = 0; i < n; i++) for (ll int j = 0; j < m; j++) cin >> arr[i][j];

    string ids[n]; for (ll int i = 0; i < n; i++) cin >> ids[i];

    vector<vector<ll int>> grid(n, vector<ll int>(m));
    ll int A = 0, B = 0;

    for (ll int i = 0; i < n; i++) {
        for (ll int j = 0; j < m; j++) {
            if (ids[i][j] == '1') {A += arr[i][j]; grid[i][j] = 1;}
            else {grid[i][j] = -1; B += arr[i][j];}
        }
    }

    ll int t = abs(A - B);

    vector<vector<ll int>> pref(n + 1, vector<ll int>(m + 1, 0));
    for (ll int i = 0; i < n; i++) {
        for (ll int j = 0; j < m; j++) {
            pref[i + 1][j + 1] = pref[i][j + 1] + pref[i + 1][j] - pref[i][j] + grid[i][j];
        }
    }

    set<ll int> st;

    for (ll int i = 0; i < n - k + 1; i++) {
        for (ll int j = 0; j < m - k + 1; j++) {
            st.insert(abs(pref[i + k][j + k] - pref[i][j + k] - pref[i + k][j] + pref[i][j]));
        }
    }

    vector<ll int> good; for (ll int i : st) good.push_back(i);

    vector<bool> take(good.size(), 1);

    for (int i = 0; i < good.size(); i++) {
        for (int j = 1; j * j <= good[i]; j++) {
            if (good[i] % j == 0 && st.find(j) != st.end()) {
                take[i] = 0;
            }
        }
    }

    reverse(all(good));
    for (ll int i : good) {
        if (i == 0) continue;
        if (!take[i]) continue;
        t %= i;
    }

    if (t == 0) {
        cout << "YES" << endl;
    } else {
        cout << "NO" << endl;
    }
}

int main() {
#ifndef DEBUG
    set_in("");
#endif
#ifdef DEBUG
    auto clock_start = chrono::high_resolution_clock::now();
    cout << endl;
#endif

    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    ll int T = 1;
    cin >> T;
    SieveOfEratosthenes(250001);
    for (ll int t = 1; t <= T; t++) {
        // cout << "Case #" << t << ": ";
        solve();
    }

    // Clock
#ifdef DEBUG
    auto clock_end = chrono::high_resolution_clock::now();
    cout << endl;
    chrono::duration<double> elapsed = clock_end - clock_start;
    cout << "Execution time: " << elapsed.count() << "s" << endl;
#endif
    return 0;
}

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

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

Think step by step.

Response:
To create a challenging test input that exposes a bug in the provided code, we need to focus on edge cases and scenarios that can maximize the failure probability. Here are the steps to generate such a test input:

1. **Edge Cases for Matrix Size and Submatrix Size**:
   - Test cases where \( n \) and \( m \) are small (e.g., \( 1 \times 1 \), \( 2 \times 2 \)).
   - Test cases where \( k \) is equal to \( n \) or \( m \).

2. **Edge Cases for Mountain Heights**:
   - Test cases where all heights are zero.
   - Test cases where all heights are the same.
   - Test cases where heights are very large (approaching the upper limit).

3. **Edge Cases for Mountain Types**:
   - Test cases where all mountains have snowy caps.
   - Test cases where all mountains do not have snowy caps.
   - Test cases with a mix of mountain types, but where the sums are already equal.

4. **Test Cases for Transformations**:
   - Test cases where the sums can be equalized by a single transformation.
   - Test cases where the sums cannot be equalized by any transformation.

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

```plaintext
3
1 1 1
1
1
1
2 2 2
0 0
0 0
00
00
2 2 2
3 4
6 7
01
00
```

### Explanation:
1. **Test Case 1**:
   - \( n = 1 \), \( m = 1 \), \( k = 1 \)
   - Heights: \([1]\)
   - Types: \(['1']\)
   - Since there is only one mountain, the sums are already equal (both are 1). The output should be "YES".

2. **Test Case 2**:
   - \( n = 2 \), \( m = 2 \), \( k = 2 \)
   - Heights: \([[0, 0], [0, 0]]\)
   - Types: \(['00', '00']\)
   - All mountains have snowy caps, and all heights are zero. The sums are already equal (both are 0). The output should be "YES".

3. **Test Case 3**:
   - \( n = 2 \), \( m = 2 \), \( k = 2 \)
   - Heights: \([[3, 4], [6, 7]]\)
   - Types: \(['01', '00']\)
   - The sum of heights with snowy caps is \( 3 + 6 = 9 \).
   - The sum of heights without snowy caps is \( 4 + 7 = 11 \).
   - It is impossible to equalize these sums with the given submatrix size and transformations. The output should be "NO".

This test input is small-scale and includes a mix of edge cases to challenge the provided buggy code.