**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Heartbeat
time_limit_ms: 5000
memory_limit_mb: 512
Description: For an array $u_1, u_2, \ldots, u_n$, define

  * a prefix maximum as an index $i$ such that $u_i>u_j$ for all $j<i$;    * a suffix maximum as an index $i$ such that $u_i>u_j$ for all $j>i$;    * an ascent as an index $i$ ($i>1$) such that $u_i>u_{i-1}$.

You are given three cost arrays: $[a_1, a_2, \ldots, a_n]$, $[b_1, b_2, \ldots, b_n]$, and $[c_0, c_1, \ldots, c_{n-1}]$.

Define the cost of an array that has $x$ prefix maximums, $y$ suffix maximums, and $z$ ascents as $a_x\cdot b_y\cdot c_z$.

Let the sum of costs of all permutations of $1,2,\ldots,n$ be $f(n)$. Find $f(1)$, $f(2)$, ..., $f(n)$ modulo $998\,244\,353$.

Input: The first line contains an integer $n$ ($1\le n\le 700$).

The second line contains $n$ integers $a_1,\ldots,a_n$ ($0\le a_i<998\,244\,353$).

The third line contains $n$ integers $b_1,\ldots,b_n$ ($0\le b_i<998\,244\,353$).

The fourth line contains $n$ integers $c_0,\ldots,c_{n-1}$ ($0\le c_i<998\,244\,353$).

Output: Print $n$ integers: the $i$-th one is $f(i)$ modulo $998\,244\,353$.

Examples:
input:
3
1 1 1
1 1 1
1 1 1
output:
1 2 6

input:
3
1 2 3
2 3 1
3 1 2
output:
6 13 34

input:
5
1 4 2 5 3
2 5 1 3 4
300000000 100000000 500000000 400000000 200000000
output:
600000000 303511294 612289529 324650937 947905622

Note: In the second example:

  * Consider permutation $[1,2,3]$. Indices $1,2,3$ are prefix maximums. Index $3$ is the only suffix maximum. Indices $2,3$ are ascents. In conclusion, it has $3$ prefix maximums, $1$ suffix maximums, and $2$ ascents. Therefore, its cost is $a_3b_1c_2=12$.    * Permutation $[1,3,2]$ has $2$ prefix maximums, $2$ suffix maximums, and $1$ ascent. Its cost is $6$.    * Permutation $[2,1,3]$ has $2$ prefix maximums, $1$ suffix maximum, and $1$ ascent. Its cost is $4$.    * Permutation $[2,3,1]$ has $2$ prefix maximums, $2$ suffix maximums, and $1$ ascent. Its cost is $6$.    * Permutation $[3,1,2]$ has $1$ prefix maximum, $2$ suffix maximums, and $1$ ascent. Its cost is $3$.    * Permutation $[3,2,1]$ has $1$ prefix maximum, $3$ suffix maximums, and $0$ ascents. Its cost is $3$.

The sum of all permutations' costs is $34$, so $f(3)=34$.

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

const int N = 705;
const int mod = 998244353;

int n, m, fac[N], inv[N], a[N], b[N], c[N], f[2][N][N], F[N][N], G[N][N], h[N][N], ans[N];

void Add(int &x, int y) {
	x += y;
	if (x >= mod) {
		x -= mod;
	}
}

int qpow(int x, int y) {
	int res = 1;
	while (y) {
		if (y & 1) {
			res = 1ll * res * x % mod;
		}
		x = 1ll * x * x % mod;
		y >>= 1;
	}
	return res;
}

int C(int x, int y) {
	if (x < y) {
		return 0;
	}
	return 1ll * fac[x] * inv[y] % mod * inv[x - y] % mod;
}

int main() {
	// freopen("f.in", "r", stdin);
	// freopen("f.out", "w", stdout);
	ios::sync_with_stdio(0); cin.tie(0);
	cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
	}
	for (int i = 1; i <= n; i++) {
		cin >> b[i];
	}
	for (int i = 0; i < n; i++) {
		cin >> c[i];
	}
	fac[0] = 1;
	for (int i = 1; i <= n; i++) {
		fac[i] = 1ll * fac[i - 1] * i % mod;
	}
	inv[n] = qpow(fac[n], mod - 2);
	for (int i = n; i >= 1; i--) {
		inv[i - 1] = 1ll * inv[i] * i % mod;
	}
	F[0][0] = a[1];
	G[0][0] = b[1];
	for (int i = 1; i <= n; i++) {
		if (i == 1) {
			f[0][1][0] = 1;
		} else {
			for (int j = 1; j <= i; j++) {
				for (int k = 0; k < i; k++) {
					f[1][j][k] = f[0][j][k];
					f[0][j][k] = 0;
				}
			}
			for (int j = 1; j <= i - 1; j++) {
				for (int k = 0; k < i - 1; k++) {
					Add(f[0][j + 1][k + 1], f[1][j][k]);
					Add(f[0][j][k], (long long)f[1][j][k] * (k + 1) % mod);
					Add(f[0][j][k + 1], (long long)f[1][j][k] * (i - 2 - k) % mod);
				}
			}
		}
		for (int j = 1; j <= i; j++) {
			for (int k = 0; k < i; k++) {
				Add(F[i][k], (long long)f[0][j][k] * a[j + 1] % mod);
				Add(G[i][k], (long long)f[0][j][i - 1 - k] * b[j + 1] % mod);
			}
		}
	}
	for (int i = 0; i < n; i++) {
		for (int x = 0; x <= i; x++) {
			for (int y = 0; y < n; y++) {
				Add(h[i][y], 1ll * F[i][x] * c[x + y + (i > 0)] % mod);
			}
		}
	}
	for (int i = 0; i < n; i++) {
		for (int j = 0; i + j < n; j++) {
			for (int y = 0; y <= j; y++) {
				Add(ans[i + j + 1], 1ll * C(i + j, i) * G[j][y] % mod * h[i][y] % mod);
			}
		}
	}
	for (int i = 1; i <= n; i++) {
		cout << ans[i] << ' ';
	}
	cout << '\n';

	return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
1 0 1
1 1 0
1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Heartbeat
time_limit_ms: 5000
memory_limit_mb: 512
Description: For an array $u_1, u_2, \ldots, u_n$, define

  * a prefix maximum as an index $i$ such that $u_i>u_j$ for all $j<i$;    * a suffix maximum as an index $i$ such that $u_i>u_j$ for all $j>i$;    * an ascent as an index $i$ ($i>1$) such that $u_i>u_{i-1}$.

You are given three cost arrays: $[a_1, a_2, \ldots, a_n]$, $[b_1, b_2, \ldots, b_n]$, and $[c_0, c_1, \ldots, c_{n-1}]$.

Define the cost of an array that has $x$ prefix maximums, $y$ suffix maximums, and $z$ ascents as $a_x\cdot b_y\cdot c_z$.

Let the sum of costs of all permutations of $1,2,\ldots,n$ be $f(n)$. Find $f(1)$, $f(2)$, ..., $f(n)$ modulo $998\,244\,353$.

Input: The first line contains an integer $n$ ($1\le n\le 700$).

The second line contains $n$ integers $a_1,\ldots,a_n$ ($0\le a_i<998\,244\,353$).

The third line contains $n$ integers $b_1,\ldots,b_n$ ($0\le b_i<998\,244\,353$).

The fourth line contains $n$ integers $c_0,\ldots,c_{n-1}$ ($0\le c_i<998\,244\,353$).

Output: Print $n$ integers: the $i$-th one is $f(i)$ modulo $998\,244\,353$.

Examples:
input:
3
1 1 1
1 1 1
1 1 1
output:
1 2 6

input:
3
1 2 3
2 3 1
3 1 2
output:
6 13 34

input:
5
1 4 2 5 3
2 5 1 3 4
300000000 100000000 500000000 400000000 200000000
output:
600000000 303511294 612289529 324650937 947905622

Note: In the second example:

  * Consider permutation $[1,2,3]$. Indices $1,2,3$ are prefix maximums. Index $3$ is the only suffix maximum. Indices $2,3$ are ascents. In conclusion, it has $3$ prefix maximums, $1$ suffix maximums, and $2$ ascents. Therefore, its cost is $a_3b_1c_2=12$.    * Permutation $[1,3,2]$ has $2$ prefix maximums, $2$ suffix maximums, and $1$ ascent. Its cost is $6$.    * Permutation $[2,1,3]$ has $2$ prefix maximums, $1$ suffix maximum, and $1$ ascent. Its cost is $4$.    * Permutation $[2,3,1]$ has $2$ prefix maximums, $2$ suffix maximums, and $1$ ascent. Its cost is $6$.    * Permutation $[3,1,2]$ has $1$ prefix maximum, $2$ suffix maximums, and $1$ ascent. Its cost is $3$.    * Permutation $[3,2,1]$ has $1$ prefix maximum, $3$ suffix maximums, and $0$ ascents. Its cost is $3$.

The sum of all permutations' costs is $34$, so $f(3)=34$.

Buggy Code:
// LUOGU_RID: 166786896
#include<bits/stdc++.h>
//#pragma GCC optimize(3,"Ofast","inline")
//#define int long long
#define i128 __int128
#define ll long long
#define ull unsigned long long
#define ld double
#define PII pair<int,int>
#define INF 0x3f3f3f3f
#define INFLL 0x3f3f3f3f3f3f3f3f
#define chkmax(a,b) a=max(a,b)
#define chkmin(a,b) a=min(a,b)
#define rep(k,l,r) for(int k=l;k<=r;++k)
#define per(k,r,l) for(int k=r;k>=l;--k)
#define cl(f,x) memset(f,x,sizeof(f))
using namespace std;
void file_IO() {
	freopen(".in","r",stdin);
	freopen(".out","w",stdout);
}
bool M1;
template<int p>
struct mint {
	int x;
	mint() {
		x=0;
	}
	mint(int _x) {
		x=_x;
	}
	friend mint operator + (mint a,mint b) {
		return a.x+b.x>=p? a.x+b.x-p:a.x+b.x;
	}
	friend mint operator - (mint a,mint b)  {
		return a.x<b.x? a.x-b.x+p:a.x-b.x;
	}
	friend mint operator * (mint a,mint b) {
		return 1ll*a.x*b.x%p;
	}
	friend mint operator ^ (mint a,ll b) {
		mint res=1,base=a;
		while(b) {
			if(b&1)
				res*=base;
			base*=base; b>>=1;
		}
		return res;
	}
	friend mint operator ~ (mint a) {
		return a^(p-2);
	}
	friend mint operator / (mint a,mint b) {
		return a*(~b);
	}
	friend mint & operator += (mint& a,mint b) {
		return a=a+b;
	}
	friend mint & operator -= (mint& a,mint b) {
		return a=a-b;
	}
	friend mint & operator *= (mint& a,mint b) {
		return a=a*b;
	}
	friend mint & operator /= (mint& a,mint b) {
		return a=a/b;
	}
	friend mint operator ++ (mint& a) {
		return a+=1;
	}
	friend mint operator -- (mint& a) {
		return a-=1;
	}
};
const int MOD=998244353;
#define mint mint<MOD>
const int N=7e2+5;
mint jc[N],inv_jc[N];
void init(int n=1e2) {
	jc[0]=1;
	rep(i,1,n)
		jc[i]=jc[i-1]*i;
	inv_jc[n]=~jc[n];
	per(i,n-1,0)
		inv_jc[i]=inv_jc[i+1]*(i+1);
}
mint C(int n,int m) {
	if(n<m)
		return 0;
	return jc[n]*inv_jc[n-m]*inv_jc[m];
}
mint f[2][N][N],g[2][N][N],fpre[N][N],gpre[N][N],pre1[N][N],pre2[N][N],h[N][N];
int a[N],b[N],c[N];
void solve() {
	int n;
	scanf("%d",&n);
	rep(i,1,n)
		scanf("%d",&a[i]);
	rep(i,1,n)
		scanf("%d",&b[i]);
	rep(i,0,n-1)
		scanf("%d",&c[i]);
	f[0][0][0]=g[0][0][0]=1;
	fpre[0][0]=a[1];
	gpre[0][0]=b[1];
	f[1][1][0]=g[1][1][0]=1;
	fpre[1][0]=a[2];
	gpre[1][0]=b[2];
	printf("%d ",(mint(a[1])*mint(b[1])*mint(c[0])).x);
	rep(z2,0,n)
		h[1][z2]=fpre[0][0]*c[z2];
	rep(i,2,n) {
		rep(z,0,i-1) {
			rep(x,0,i) {
				f[i&1][x][z]=g[i&1][x][z]=0;
				f[i&1][x][z]+=f[(i-1)&1][x-1][z-1];
				f[i&1][x][z]+=f[(i-1)&1][x][z]*(z+1);
				f[i&1][x][z]+=f[(i-1)&1][x][z-1]*(i-1-z);
				g[i&1][x][z]+=g[(i-1)&1][x-1][z];
				g[i&1][x][z]+=g[(i-1)&1][x][z]*z;
				g[i&1][x][z]+=g[(i-1)&1][x][z-1]*(i-z);
				pre1[x][z]=(x? pre1[x-1][z]:0)+f[i&1][x][z]*a[x+1];
				pre2[x][z]=(x? pre2[x-1][z]:0)+g[i&1][x][z]*b[x+1];
			}
			fpre[i][z]=pre1[i][z]*inv_jc[i];
			gpre[i][z]=pre2[i][z]*inv_jc[i];
		}
		rep(z1,0,i-1) {
			rep(z2,0,n-i)
				h[i][z2]+=fpre[i-1][z1]*c[z1+z2+1];
		}
		mint res=0;
		rep(p,1,i) {
			rep(z2,0,i-p)
				res+=h[p][z2]*gpre[i-p][z2];
		}
		printf("%d ",(res*jc[i-1]).x);
	}
	puts("");
}
bool M2;
signed main() {
	//file_IO();
	init();
	int testcase=1;
	//scanf("%d",&testcase);
	while(testcase--)
		solve();
	cerr<<"used time = "<<1000*clock()/CLOCKS_PER_SEC<<"ms\n";
	cerr<<"used memory = "<<(&M1-&M2)/1024/1024<<"MB\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:
To generate a challenging test input that exposes the bug in the buggy code, we need to consider edge cases and scenarios that can highlight potential issues in the algorithm. Here are the steps to think through this:

1. **Small Scale Input**: Since the problem specifies \(1 \leq n \leq 700\), a small-scale input (e.g., \(n = 5\)) is appropriate to ensure the output is manageable and fits within the constraints.

2. **Edge Cases**:
   - **Single Element Array**: \(n = 1\)
   - **Array with All Elements the Same**: This can cause issues with prefix and suffix maximums.
   - **Array with Distinct Elements**: This ensures a variety of prefix and suffix maximums and ascents.
   - **Array with Zero Costs**: This can test how the algorithm handles zero values in the cost arrays.
   - **Large Values in Cost Arrays**: This can test the handling of large numbers and the modulo operation.

3. **Combining Edge Cases**: A combination of these edge cases can provide a robust test input.

Let's construct a test input that includes a mix of these edge cases:

- \(n = 5\)
- \(a\) array with distinct values to ensure a variety of prefix maximums.
- \(b\) array with distinct values to ensure a variety of suffix maximums.
- \(c\) array with some zero values to test how the algorithm handles zero costs.

Here is the test input:

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

This input will:
- Ensure a variety of prefix and suffix maximums.
- Include zero values in the \(c\) array to test the handling of zero costs.
- Provide a small-scale input that is complete and valid.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Heartbeat
time_limit_ms: 5000
memory_limit_mb: 512
Description: For an array $u_1, u_2, \ldots, u_n$, define

  * a prefix maximum as an index $i$ such that $u_i>u_j$ for all $j<i$;    * a suffix maximum as an index $i$ such that $u_i>u_j$ for all $j>i$;    * an ascent as an index $i$ ($i>1$) such that $u_i>u_{i-1}$.

You are given three cost arrays: $[a_1, a_2, \ldots, a_n]$, $[b_1, b_2, \ldots, b_n]$, and $[c_0, c_1, \ldots, c_{n-1}]$.

Define the cost of an array that has $x$ prefix maximums, $y$ suffix maximums, and $z$ ascents as $a_x\cdot b_y\cdot c_z$.

Let the sum of costs of all permutations of $1,2,\ldots,n$ be $f(n)$. Find $f(1)$, $f(2)$, ..., $f(n)$ modulo $998\,244\,353$.

Input: The first line contains an integer $n$ ($1\le n\le 700$).

The second line contains $n$ integers $a_1,\ldots,a_n$ ($0\le a_i<998\,244\,353$).

The third line contains $n$ integers $b_1,\ldots,b_n$ ($0\le b_i<998\,244\,353$).

The fourth line contains $n$ integers $c_0,\ldots,c_{n-1}$ ($0\le c_i<998\,244\,353$).

Output: Print $n$ integers: the $i$-th one is $f(i)$ modulo $998\,244\,353$.

Examples:
input:
3
1 1 1
1 1 1
1 1 1
output:
1 2 6

input:
3
1 2 3
2 3 1
3 1 2
output:
6 13 34

input:
5
1 4 2 5 3
2 5 1 3 4
300000000 100000000 500000000 400000000 200000000
output:
600000000 303511294 612289529 324650937 947905622

Note: In the second example:

  * Consider permutation $[1,2,3]$. Indices $1,2,3$ are prefix maximums. Index $3$ is the only suffix maximum. Indices $2,3$ are ascents. In conclusion, it has $3$ prefix maximums, $1$ suffix maximums, and $2$ ascents. Therefore, its cost is $a_3b_1c_2=12$.    * Permutation $[1,3,2]$ has $2$ prefix maximums, $2$ suffix maximums, and $1$ ascent. Its cost is $6$.    * Permutation $[2,1,3]$ has $2$ prefix maximums, $1$ suffix maximum, and $1$ ascent. Its cost is $4$.    * Permutation $[2,3,1]$ has $2$ prefix maximums, $2$ suffix maximums, and $1$ ascent. Its cost is $6$.    * Permutation $[3,1,2]$ has $1$ prefix maximum, $2$ suffix maximums, and $1$ ascent. Its cost is $3$.    * Permutation $[3,2,1]$ has $1$ prefix maximum, $3$ suffix maximums, and $0$ ascents. Its cost is $3$.

The sum of all permutations' costs is $34$, so $f(3)=34$.

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

const int MOD = 998244353;

int countPrefixMaximums(vector<int>& perm) {
    int count = 0;
    int maxSoFar = 0;
    for(int i = 0; i < perm.size(); i++) {
        if(perm[i] > maxSoFar) {
            count++;
            maxSoFar = perm[i];
        }
    }
    return count;
}

int countSuffixMaximums(vector<int>& perm) {
    int count = 0;
    int maxSoFar = 0;
    for(int i = perm.size()-1; i >= 0; i--) {
        if(perm[i] > maxSoFar) {
            count++;
            maxSoFar = perm[i];
        }
    }
    return count;
}

int countAscents(vector<int>& perm) {
    int count = 0;
    for(int i = 1; i < perm.size(); i++) {
        if(perm[i] > perm[i-1]) count++;
    }
    return count;
}

long long solve(int n, vector<int>& a, vector<int>& b, vector<int>& c) {
    vector<int> perm(n);
    for(int i = 0; i < n; i++) perm[i] = i + 1;

    long long sum = 0;
    do {
        int prefixMax = countPrefixMaximums(perm);
        int suffixMax = countSuffixMaximums(perm);
        int ascents = countAscents(perm);

        long long cost = 1LL * a[prefixMax-1] * b[suffixMax-1] % MOD;
        cost = cost * c[ascents] % MOD;

        sum = (sum + cost) % MOD;
    } while(next_permutation(perm.begin(), perm.end()));

    return sum;
}

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

    int n;
    cin >> n;

    vector<int> a(n), b(n), c(n);
    for(int i = 0; i < n; i++) cin >> a[i];
    for(int i = 0; i < n; i++) cin >> b[i];
    for(int i = 0; i < n; i++) cin >> c[i];

    for(int i = 1; i <= n; i++) {
        vector<int> curA(a.begin(), a.begin() + i);
        vector<int> curB(b.begin(), b.begin() + i);
        vector<int> curC(c.begin(), c.begin() + i);
        cout << solve(i, curA, curB, curC) << " ";
    }
    cout << endl;

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
0 0 0
0 0 0
0 0 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Heartbeat
time_limit_ms: 5000
memory_limit_mb: 512
Description: For an array $u_1, u_2, \ldots, u_n$, define

  * a prefix maximum as an index $i$ such that $u_i>u_j$ for all $j<i$;    * a suffix maximum as an index $i$ such that $u_i>u_j$ for all $j>i$;    * an ascent as an index $i$ ($i>1$) such that $u_i>u_{i-1}$.

You are given three cost arrays: $[a_1, a_2, \ldots, a_n]$, $[b_1, b_2, \ldots, b_n]$, and $[c_0, c_1, \ldots, c_{n-1}]$.

Define the cost of an array that has $x$ prefix maximums, $y$ suffix maximums, and $z$ ascents as $a_x\cdot b_y\cdot c_z$.

Let the sum of costs of all permutations of $1,2,\ldots,n$ be $f(n)$. Find $f(1)$, $f(2)$, ..., $f(n)$ modulo $998\,244\,353$.

Input: The first line contains an integer $n$ ($1\le n\le 700$).

The second line contains $n$ integers $a_1,\ldots,a_n$ ($0\le a_i<998\,244\,353$).

The third line contains $n$ integers $b_1,\ldots,b_n$ ($0\le b_i<998\,244\,353$).

The fourth line contains $n$ integers $c_0,\ldots,c_{n-1}$ ($0\le c_i<998\,244\,353$).

Output: Print $n$ integers: the $i$-th one is $f(i)$ modulo $998\,244\,353$.

Examples:
input:
3
1 1 1
1 1 1
1 1 1
output:
1 2 6

input:
3
1 2 3
2 3 1
3 1 2
output:
6 13 34

input:
5
1 4 2 5 3
2 5 1 3 4
300000000 100000000 500000000 400000000 200000000
output:
600000000 303511294 612289529 324650937 947905622

Note: In the second example:

  * Consider permutation $[1,2,3]$. Indices $1,2,3$ are prefix maximums. Index $3$ is the only suffix maximum. Indices $2,3$ are ascents. In conclusion, it has $3$ prefix maximums, $1$ suffix maximums, and $2$ ascents. Therefore, its cost is $a_3b_1c_2=12$.    * Permutation $[1,3,2]$ has $2$ prefix maximums, $2$ suffix maximums, and $1$ ascent. Its cost is $6$.    * Permutation $[2,1,3]$ has $2$ prefix maximums, $1$ suffix maximum, and $1$ ascent. Its cost is $4$.    * Permutation $[2,3,1]$ has $2$ prefix maximums, $2$ suffix maximums, and $1$ ascent. Its cost is $6$.    * Permutation $[3,1,2]$ has $1$ prefix maximum, $2$ suffix maximums, and $1$ ascent. Its cost is $3$.    * Permutation $[3,2,1]$ has $1$ prefix maximum, $3$ suffix maximums, and $0$ ascents. Its cost is $3$.

The sum of all permutations' costs is $34$, so $f(3)=34$.

Buggy Code:
/*
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
==> author:	Kovidh Madurakavi
==> date:		2024/07/15/ 21:19:29 Monday
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
*/

#include <iostream>
#include <vector>
#include <algorithm>
#include <functional>
#include <cmath>

namespace Custom {
    namespace Math {
        constexpr int MOD_VALUE = 998244353;
        inline int mod_multiply(long long x, long long y) {
            return static_cast<int>((x * y) % MOD_VALUE);
        }
    }

    namespace Utils {
        template<typename T>
        void read_vector(std::vector<T>& v, int n) {
            for (int i = 0; i < n; ++i) {
                std::cin >> v[i];
            }
        }

        template<typename T>
        void print_vector(const std::vector<T>& v) {
            for (const auto& elem : v) {
                std::cout << elem << " ";
            }
            std::cout << std::endl;
        }
    }

    namespace Process {
        using namespace Math;
        using namespace Utils;

        struct CustomFunction {
            std::vector<int>& a_ref;
            std::vector<int>& b_ref;
            std::vector<int>& c_ref;

            CustomFunction(std::vector<int>& a, std::vector<int>& b, std::vector<int>& c)
                : a_ref(a), b_ref(b), c_ref(c) {}

            int calculate_total_cost(int i) const {
                std::vector<int> perm(i);
                for (int j = 0; j < i; ++j) {
                    perm[j] = j + 1;
                }

                int total_cost = 0;
                do {
                    int prefix_count = count_prefix_max(perm, i);
                    int suffix_count = count_suffix_max(perm, i);
                    int ascent_count = count_ascents(perm, i);

                    long long cost = mod_multiply(a_ref[prefix_count - 1], b_ref[suffix_count - 1]);
                    cost = mod_multiply(cost, c_ref[ascent_count]);
                    total_cost = (total_cost + cost) % MOD_VALUE;

                } while (std::next_permutation(perm.begin(), perm.end()));

                return total_cost;
            }

            static int count_prefix_max(const std::vector<int>& perm, int size) {
                int count = 0;
                int current_max = -1;
                for (int j = 0; j < size; ++j) {
                    if (perm[j] > current_max) {
                        current_max = perm[j];
                        count++;
                    }
                }
                return count;
            }

            static int count_suffix_max(const std::vector<int>& perm, int size) {
                int count = 0;
                int current_max = -1;
                for (int j = size - 1; j >= 0; --j) {
                    if (perm[j] > current_max) {
                        current_max = perm[j];
                        count++;
                    }
                }
                return count;
            }

            static int count_ascents(const std::vector<int>& perm, int size) {
                int count = 0;
                for (int j = 1; j < size; ++j) {
                    if (perm[j] > perm[j - 1]) {
                        count++;
                    }
                }
                return count;
            }
        };
    }

    void run_algorithm() {
        using namespace Process;

        int num;
        std::cin >> num;

        std::vector<int> alpha(num), beta(num), gamma(num);
        read_vector(alpha, num);
        read_vector(beta, num);
        read_vector(gamma, num);

        std::vector<int> output(num, 0);
        CustomFunction calc_func(alpha, beta, gamma);

        for (int i = 1; i <= num; ++i) {
            output[i - 1] = calc_func.calculate_total_cost(i);
        }

        print_vector(output);
    }
}

int main() {
    Custom::run_algorithm();
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider edge cases and scenarios that can maximize the failure probability. Here are the steps to think through this:

1. **Edge Cases for Small `n`**:
   - `n = 1`: This is the smallest possible input. It should be handled correctly as there is only one permutation.
   - `n = 2`: This is the next smallest input. It has two permutations, and the code should handle it correctly.

2. **Special Values in Cost Arrays**:
   - Zero values: If any of the cost arrays (`a`, `b`, `c`) contain zeros, it can significantly affect the total cost calculation.
   - Large values: Values close to the modulo limit (`998,244,353`) can also cause issues due to overflow or incorrect modulo operations.

3. **Symmetric and Asymmetric Permutations**:
   - Symmetric permutations: Permutations where the number of prefix and suffix maximums are the same.
   - Asymmetric permutations: Permutations where the number of prefix and suffix maximums are different.

4. **Boundary Values**:
   - Maximum `n` value: `n = 700` to test the upper limit of the input size.

Given these considerations, a small-scale test input that can expose potential bugs is:

```plaintext
3
0 1 2
2 1 0
1 2 3
```

Explanation:
- **`n = 3`**: Small enough to be manageable but large enough to have multiple permutations.
- **`a = [0, 1, 2]`**: Contains a zero value to test how the code handles zero in the cost array.
- **`b = [2, 1, 0]`**: Contains a zero value and is in descending order to test different cost scenarios.
- **`c = [1, 2, 3]`: Contains values that increase linearly to test the cost of ascents.

This input will test the code's ability to handle zero values in the cost arrays and correctly calculate the total cost for all permutations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Heartbeat
time_limit_ms: 5000
memory_limit_mb: 512
Description: For an array $u_1, u_2, \ldots, u_n$, define

  * a prefix maximum as an index $i$ such that $u_i>u_j$ for all $j<i$;    * a suffix maximum as an index $i$ such that $u_i>u_j$ for all $j>i$;    * an ascent as an index $i$ ($i>1$) such that $u_i>u_{i-1}$.

You are given three cost arrays: $[a_1, a_2, \ldots, a_n]$, $[b_1, b_2, \ldots, b_n]$, and $[c_0, c_1, \ldots, c_{n-1}]$.

Define the cost of an array that has $x$ prefix maximums, $y$ suffix maximums, and $z$ ascents as $a_x\cdot b_y\cdot c_z$.

Let the sum of costs of all permutations of $1,2,\ldots,n$ be $f(n)$. Find $f(1)$, $f(2)$, ..., $f(n)$ modulo $998\,244\,353$.

Input: The first line contains an integer $n$ ($1\le n\le 700$).

The second line contains $n$ integers $a_1,\ldots,a_n$ ($0\le a_i<998\,244\,353$).

The third line contains $n$ integers $b_1,\ldots,b_n$ ($0\le b_i<998\,244\,353$).

The fourth line contains $n$ integers $c_0,\ldots,c_{n-1}$ ($0\le c_i<998\,244\,353$).

Output: Print $n$ integers: the $i$-th one is $f(i)$ modulo $998\,244\,353$.

Examples:
input:
3
1 1 1
1 1 1
1 1 1
output:
1 2 6

input:
3
1 2 3
2 3 1
3 1 2
output:
6 13 34

input:
5
1 4 2 5 3
2 5 1 3 4
300000000 100000000 500000000 400000000 200000000
output:
600000000 303511294 612289529 324650937 947905622

Note: In the second example:

  * Consider permutation $[1,2,3]$. Indices $1,2,3$ are prefix maximums. Index $3$ is the only suffix maximum. Indices $2,3$ are ascents. In conclusion, it has $3$ prefix maximums, $1$ suffix maximums, and $2$ ascents. Therefore, its cost is $a_3b_1c_2=12$.    * Permutation $[1,3,2]$ has $2$ prefix maximums, $2$ suffix maximums, and $1$ ascent. Its cost is $6$.    * Permutation $[2,1,3]$ has $2$ prefix maximums, $1$ suffix maximum, and $1$ ascent. Its cost is $4$.    * Permutation $[2,3,1]$ has $2$ prefix maximums, $2$ suffix maximums, and $1$ ascent. Its cost is $6$.    * Permutation $[3,1,2]$ has $1$ prefix maximum, $2$ suffix maximums, and $1$ ascent. Its cost is $3$.    * Permutation $[3,2,1]$ has $1$ prefix maximum, $3$ suffix maximums, and $0$ ascents. Its cost is $3$.

The sum of all permutations' costs is $34$, so $f(3)=34$.

Buggy Code:
from itertools import permutations

MOD = 998244353

def count_prefix_maxima(perm):
    max_so_far = -1
    count = 0
    for p in perm:
        if p > max_so_far:
            count += 1
            max_so_far = p
    return count

def count_suffix_maxima(perm):
    max_so_far = -1
    count = 0
    for p in reversed(perm):
        if p > max_so_far:
            count += 1
            max_so_far = p
    return count

def count_ascents(perm):
    count = 0
    for i in range(1, len(perm)):
        if perm[i] > perm[i-1]:
            count += 1
    return count

def f(n, a, b, c):
    result = []
    for i in range(1, n+1):
        total_cost = 0
        for perm in permutations(range(1, i+1)):
            x = count_prefix_maxima(perm)
            y = count_suffix_maxima(perm)
            z = count_ascents(perm)
            cost = a[x-1] * b[y-1] * c[z] % MOD
            total_cost = (total_cost + cost) % MOD
        result.append(total_cost)
    return result

# Input reading
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = list(map(int, input().split()))

# Calculating f(1), f(2), ..., f(n)
results = f(n, a, b, c)
print(" ".join(map(str, results)))

**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
0 0 0
0 0 0
0 0 0
```

This test input is designed to expose potential issues in the buggy code, particularly when all cost values are zero. This scenario can help identify how the code handles cases where the cost of any permutation is zero, which might lead to unexpected behavior or incorrect results if not properly managed.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Heartbeat
time_limit_ms: 5000
memory_limit_mb: 512
Description: For an array $u_1, u_2, \ldots, u_n$, define

  * a prefix maximum as an index $i$ such that $u_i>u_j$ for all $j<i$;    * a suffix maximum as an index $i$ such that $u_i>u_j$ for all $j>i$;    * an ascent as an index $i$ ($i>1$) such that $u_i>u_{i-1}$.

You are given three cost arrays: $[a_1, a_2, \ldots, a_n]$, $[b_1, b_2, \ldots, b_n]$, and $[c_0, c_1, \ldots, c_{n-1}]$.

Define the cost of an array that has $x$ prefix maximums, $y$ suffix maximums, and $z$ ascents as $a_x\cdot b_y\cdot c_z$.

Let the sum of costs of all permutations of $1,2,\ldots,n$ be $f(n)$. Find $f(1)$, $f(2)$, ..., $f(n)$ modulo $998\,244\,353$.

Input: The first line contains an integer $n$ ($1\le n\le 700$).

The second line contains $n$ integers $a_1,\ldots,a_n$ ($0\le a_i<998\,244\,353$).

The third line contains $n$ integers $b_1,\ldots,b_n$ ($0\le b_i<998\,244\,353$).

The fourth line contains $n$ integers $c_0,\ldots,c_{n-1}$ ($0\le c_i<998\,244\,353$).

Output: Print $n$ integers: the $i$-th one is $f(i)$ modulo $998\,244\,353$.

Examples:
input:
3
1 1 1
1 1 1
1 1 1
output:
1 2 6

input:
3
1 2 3
2 3 1
3 1 2
output:
6 13 34

input:
5
1 4 2 5 3
2 5 1 3 4
300000000 100000000 500000000 400000000 200000000
output:
600000000 303511294 612289529 324650937 947905622

Note: In the second example:

  * Consider permutation $[1,2,3]$. Indices $1,2,3$ are prefix maximums. Index $3$ is the only suffix maximum. Indices $2,3$ are ascents. In conclusion, it has $3$ prefix maximums, $1$ suffix maximums, and $2$ ascents. Therefore, its cost is $a_3b_1c_2=12$.    * Permutation $[1,3,2]$ has $2$ prefix maximums, $2$ suffix maximums, and $1$ ascent. Its cost is $6$.    * Permutation $[2,1,3]$ has $2$ prefix maximums, $1$ suffix maximum, and $1$ ascent. Its cost is $4$.    * Permutation $[2,3,1]$ has $2$ prefix maximums, $2$ suffix maximums, and $1$ ascent. Its cost is $6$.    * Permutation $[3,1,2]$ has $1$ prefix maximum, $2$ suffix maximums, and $1$ ascent. Its cost is $3$.    * Permutation $[3,2,1]$ has $1$ prefix maximum, $3$ suffix maximums, and $0$ ascents. Its cost is $3$.

The sum of all permutations' costs is $34$, so $f(3)=34$.

Buggy Code:
MOD = 998244353

def calculate_costs(n, a, b, c):
    from itertools import permutations

    results = []

    for length in range(1, n + 1):
        total_cost = 0

        for perm in permutations(range(1, length + 1)):
            prefix_max = 0
            suffix_max = 0
            ascents = 0

            for i in range(length):
                if perm[i] > max(perm[:i], default=0):
                    prefix_max += 1
                if perm[i] > max(perm[i+1:], default=0):
                    suffix_max += 1
                if i > 0 and perm[i] > perm[i-1]:
                    ascents += 1

            total_cost += a[prefix_max - 1] * b[suffix_max - 1] * c[ascents]
            total_cost %= MOD

        results.append(total_cost)

    return results

def main():
    import sys
    input = sys.stdin.read
    data = input().split()

    n = int(data[0])
    a = list(map(int, data[1:n + 1]))
    b = list(map(int, data[n + 1:2 * n + 1]))
    c = list(map(int, data[2 * n + 1:3 * n + 1]))

    results = calculate_costs(n, a, b, c)

    print(" ".join(map(str, results)))

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that might cause the code to fail. Here are the steps to consider:

1. **Edge Cases**:
   - Small values of `n` to ensure the code handles the minimum input size correctly.
   - Large values of `n` to test the code's performance and correctness under maximum input size constraints.
   - Special values in the cost arrays that might cause overflow or unexpected behavior.

2. **Special Scenarios**:
   - Arrays with all elements being the same to test the code's handling of uniform inputs.
   - Arrays with elements that are multiples of the modulus to test the code's handling of large numbers.
   - Arrays with a mix of large and small values to test the code's ability to handle varying magnitudes.

Given the constraints and the need for a small-scale test input, let's focus on a small value of `n` and use special values in the cost arrays to maximize the failure probability.

### Chosen Test Input
- **n = 4**: This is a small enough value to manage but large enough to introduce complexity.
- **a, b, c**: Use values that are multiples of the modulus to test overflow and ensure the code handles large numbers correctly.

```plaintext
4
998244352 998244352 998244352 998244352
998244352 998244352 998244352 998244352
998244352 998244352 998244352 998244352
```

### Explanation
- **n = 4**: This is a small input size that can be easily verified manually.
- **a, b, c**: All elements are set to `998244352`, which is just one less than the modulus `998244353`. This choice will ensure that any multiplication involving these values will result in a large number, and the final result will be close to the modulus, testing the code's ability to handle large numbers and avoid overflow.

This test input should expose any issues in the buggy code, particularly in the way it handles large numbers and the modulus operation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Heartbeat
time_limit_ms: 5000
memory_limit_mb: 512
Description: For an array $u_1, u_2, \ldots, u_n$, define

  * a prefix maximum as an index $i$ such that $u_i>u_j$ for all $j<i$;    * a suffix maximum as an index $i$ such that $u_i>u_j$ for all $j>i$;    * an ascent as an index $i$ ($i>1$) such that $u_i>u_{i-1}$.

You are given three cost arrays: $[a_1, a_2, \ldots, a_n]$, $[b_1, b_2, \ldots, b_n]$, and $[c_0, c_1, \ldots, c_{n-1}]$.

Define the cost of an array that has $x$ prefix maximums, $y$ suffix maximums, and $z$ ascents as $a_x\cdot b_y\cdot c_z$.

Let the sum of costs of all permutations of $1,2,\ldots,n$ be $f(n)$. Find $f(1)$, $f(2)$, ..., $f(n)$ modulo $998\,244\,353$.

Input: The first line contains an integer $n$ ($1\le n\le 700$).

The second line contains $n$ integers $a_1,\ldots,a_n$ ($0\le a_i<998\,244\,353$).

The third line contains $n$ integers $b_1,\ldots,b_n$ ($0\le b_i<998\,244\,353$).

The fourth line contains $n$ integers $c_0,\ldots,c_{n-1}$ ($0\le c_i<998\,244\,353$).

Output: Print $n$ integers: the $i$-th one is $f(i)$ modulo $998\,244\,353$.

Examples:
input:
3
1 1 1
1 1 1
1 1 1
output:
1 2 6

input:
3
1 2 3
2 3 1
3 1 2
output:
6 13 34

input:
5
1 4 2 5 3
2 5 1 3 4
300000000 100000000 500000000 400000000 200000000
output:
600000000 303511294 612289529 324650937 947905622

Note: In the second example:

  * Consider permutation $[1,2,3]$. Indices $1,2,3$ are prefix maximums. Index $3$ is the only suffix maximum. Indices $2,3$ are ascents. In conclusion, it has $3$ prefix maximums, $1$ suffix maximums, and $2$ ascents. Therefore, its cost is $a_3b_1c_2=12$.    * Permutation $[1,3,2]$ has $2$ prefix maximums, $2$ suffix maximums, and $1$ ascent. Its cost is $6$.    * Permutation $[2,1,3]$ has $2$ prefix maximums, $1$ suffix maximum, and $1$ ascent. Its cost is $4$.    * Permutation $[2,3,1]$ has $2$ prefix maximums, $2$ suffix maximums, and $1$ ascent. Its cost is $6$.    * Permutation $[3,1,2]$ has $1$ prefix maximum, $2$ suffix maximums, and $1$ ascent. Its cost is $3$.    * Permutation $[3,2,1]$ has $1$ prefix maximum, $3$ suffix maximums, and $0$ ascents. Its cost is $3$.

The sum of all permutations' costs is $34$, so $f(3)=34$.

Buggy Code:
import java.util.*;
public class Main {
    private static final int MOD = 998244353;

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int aks = scanner.nextInt();
        int[] nav = new int[aks];
        int[] jyo = new int[aks];
        int[] res = new int[aks];
        for (int i = 0; i < aks; i++) {
            nav[i] = scanner.nextInt();
        }
        for (int i = 0; i < aks; i++) {
            jyo[i] = scanner.nextInt();
        }
        for (int i = 0; i < aks; i++) {
            res[i] = scanner.nextInt();
        }
        for (int i = 1; i <= aks; i++) {
            System.out.print(calculateF(i, nav, jyo, res) + " ");
        }
    }
    private static int calculateF(int aks, int[] nav, int[] jyo, int[] res) {
        List<int[]> sol = new ArrayList<>();
        generatePermutations(new int[aks], 0, aks, sol);

        int ans = 0;
        for (int[] perm : sol) {
            int prefixMaxCount = 0;
            int suffixMaxCount = 0;
            int ascentCount = 0;
            int max = 0;
            for (int i = 0; i < aks; i++) {
                if (perm[i] > max) {
                    max = perm[i];
                    prefixMaxCount++;
                }
            }
            max = 0;
            for (int i = aks - 1; i >= 0; i--) {
                if (perm[i] > max) {
                    max = perm[i];
                    suffixMaxCount++;
                }
            }
            for (int i = 1; i < aks; i++) {
                if (perm[i] > perm[i - 1]) {
                    ascentCount++;
                }
            }
            int cost = (int) (((long) nav[prefixMaxCount - 1] * jyo[suffixMaxCount - 1] % MOD) * res[ascentCount] % MOD);
            ans = (ans + cost) % MOD;
        }

        return ans;
    }

    private static void generatePermutations(int[] perm, int idx, int aks, List<int[]> sol) {
        if (idx == aks) {
            sol.add(Arrays.copyOf(perm, aks));
            return;
        }
        for (int i = 1; i <= aks; i++) {
            boolean used = false;
            for (int j = 0; j < idx; j++) {
                if (perm[j] == i) {
                    used = true;
                    break;
                }
            }
            if (!used) {
                perm[idx] = i;
                generatePermutations(perm, idx + 1, aks, sol);
            }
        }
    }
}

**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
0 0 0
0 0 0
0 0 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Heartbeat
time_limit_ms: 5000
memory_limit_mb: 512
Description: For an array $u_1, u_2, \ldots, u_n$, define

  * a prefix maximum as an index $i$ such that $u_i>u_j$ for all $j<i$;    * a suffix maximum as an index $i$ such that $u_i>u_j$ for all $j>i$;    * an ascent as an index $i$ ($i>1$) such that $u_i>u_{i-1}$.

You are given three cost arrays: $[a_1, a_2, \ldots, a_n]$, $[b_1, b_2, \ldots, b_n]$, and $[c_0, c_1, \ldots, c_{n-1}]$.

Define the cost of an array that has $x$ prefix maximums, $y$ suffix maximums, and $z$ ascents as $a_x\cdot b_y\cdot c_z$.

Let the sum of costs of all permutations of $1,2,\ldots,n$ be $f(n)$. Find $f(1)$, $f(2)$, ..., $f(n)$ modulo $998\,244\,353$.

Input: The first line contains an integer $n$ ($1\le n\le 700$).

The second line contains $n$ integers $a_1,\ldots,a_n$ ($0\le a_i<998\,244\,353$).

The third line contains $n$ integers $b_1,\ldots,b_n$ ($0\le b_i<998\,244\,353$).

The fourth line contains $n$ integers $c_0,\ldots,c_{n-1}$ ($0\le c_i<998\,244\,353$).

Output: Print $n$ integers: the $i$-th one is $f(i)$ modulo $998\,244\,353$.

Examples:
input:
3
1 1 1
1 1 1
1 1 1
output:
1 2 6

input:
3
1 2 3
2 3 1
3 1 2
output:
6 13 34

input:
5
1 4 2 5 3
2 5 1 3 4
300000000 100000000 500000000 400000000 200000000
output:
600000000 303511294 612289529 324650937 947905622

Note: In the second example:

  * Consider permutation $[1,2,3]$. Indices $1,2,3$ are prefix maximums. Index $3$ is the only suffix maximum. Indices $2,3$ are ascents. In conclusion, it has $3$ prefix maximums, $1$ suffix maximums, and $2$ ascents. Therefore, its cost is $a_3b_1c_2=12$.    * Permutation $[1,3,2]$ has $2$ prefix maximums, $2$ suffix maximums, and $1$ ascent. Its cost is $6$.    * Permutation $[2,1,3]$ has $2$ prefix maximums, $1$ suffix maximum, and $1$ ascent. Its cost is $4$.    * Permutation $[2,3,1]$ has $2$ prefix maximums, $2$ suffix maximums, and $1$ ascent. Its cost is $6$.    * Permutation $[3,1,2]$ has $1$ prefix maximum, $2$ suffix maximums, and $1$ ascent. Its cost is $3$.    * Permutation $[3,2,1]$ has $1$ prefix maximum, $3$ suffix maximums, and $0$ ascents. Its cost is $3$.

The sum of all permutations' costs is $34$, so $f(3)=34$.

Buggy Code:
import java.util.*;

public class Main {
    private static final int MOD = 998244353;

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

        int[] a = new int[n];
        int[] b = new int[n];
        int[] c = new int[n];

        for (int i = 0; i < n; i++) {
            a[i] = scanner.nextInt();
        }

        for (int i = 0; i < n; i++) {
            b[i] = scanner.nextInt();
        }

        for (int i = 0; i < n; i++) {
            c[i] = scanner.nextInt();
        }

        long[] factorials = new long[n + 1];
        factorials[0] = 1;
        for (int i = 1; i <= n; i++) {
            factorials[i] = factorials[i - 1] * i % MOD;
        }

        long[] results = new long[n];
        for (int length = 1; length <= n; length++) {
            long sum = 0;
            int[] perm = new int[length];
            boolean[] used = new boolean[length];
            sum = permuteAndCalculate(perm, used, 0, length, a, b, c, sum, factorials[length - 1]);
            results[length - 1] = sum;
        }

        for (int i = 0; i < n; i++) {
            System.out.print(results[i] + " ");
        }
    }

    private static long permuteAndCalculate(int[] perm, boolean[] used, int depth, int length, int[] a, int[] b, int[] c, long sum, long factorial) {
        if (depth == length) {
            int prefixMaxCount = 0;
            int suffixMaxCount = 0;
            int ascentCount = 0;

            int max = 0;
            for (int i = 0; i < length; i++) {
                if (perm[i] > max) {
                    prefixMaxCount++;
                    max = perm[i];
                }
            }

            max = 0;
            for (int i = length - 1; i >= 0; i--) {
                if (perm[i] > max) {
                    suffixMaxCount++;
                    max = perm[i];
                }
            }

            for (int i = 1; i < length; i++) {
                if (perm[i] > perm[i - 1]) {
                    ascentCount++;
                }
            }

            sum += (long) a[prefixMaxCount - 1] * b[suffixMaxCount - 1] % MOD * c[ascentCount] % MOD;
            sum %= MOD;
            return sum;
        }

        for (int i = 0; i < length; i++) {
            if (!used[i]) {
                used[i] = true;
                perm[depth] = i + 1;
                sum = permuteAndCalculate(perm, used, depth + 1, length, a, b, c, sum, factorial);
                used[i] = false;
            }
        }
        return sum;
    }
}

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

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

Think step by step.

Response:
```plaintext
5
1 0 1 0 1
0 1 0 1 0
1 1 1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mathematical Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $s$ of length $n > 1$, consisting of digits from $0$ to $9$. You must insert exactly $n - 2$ symbols $+$ (addition) or $\times$ (multiplication) into this string to form a valid arithmetic expression.

In this problem, the symbols cannot be placed before the first or after the last character of the string $s$, and two symbols cannot be written consecutively. Also, note that the order of the digits in the string cannot be changed. Let's consider $s = 987009$:

  * From this string, the following arithmetic expressions can be obtained: $9 \times 8 + 70 \times 0 + 9 = 81$, $98 \times 7 \times 0 + 0 \times 9 = 0$, $9 + 8 + 7 + 0 + 09 = 9 + 8 + 7 + 0 + 9 = 33$. Note that the number $09$ is considered valid and is simply transformed into $9$.   * From this string, the following arithmetic expressions cannot be obtained: $+9 \times 8 \times 70 + 09$ (symbols should only be placed between digits), $98 \times 70 + 0 + 9$ (since there are $6$ digits, there must be exactly $4$ symbols).

The result of the arithmetic expression is calculated according to the rules of mathematics — first all multiplication operations are performed, then addition. You need to find the minimum result that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string $s$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases. Then follows their description.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 20$) — the length of the string $s$.

The second line of each test case contains a string $s$ of length $n$, consisting of digits from $0$ to $9$.

Output: For each test case, output the minimum result of the arithmetic expression that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string.

Examples:
input:
18
2
10
2
74
2
00
2
01
3
901
3
101
5
23311
6
987009
7
1111111
20
99999999999999999999
20
00000000000000000000
4
0212
18
057235283621345395
4
1112
20
19811678487321784121
4
1121
4
2221
3
011
output:
10
74
0
1
9
1
19
0
11
261
0
0
0
12
93
12
24
0

Note: In the first four test cases, we cannot add symbols, so the answer will be the original number.

In the fifth test case, the optimal answer looks as follows: $9 \times 01 = 9 \times 1 = 9$.

In the sixth test case, the optimal answer looks as follows: $1 \times 01 = 1 \times 1 = 1$.

In the seventh test case, the optimal answer looks as follows: $2 + 3 + 3 + 11 = 19$.

In the eighth test case, one of the optimal answers looks as follows: $98 \times 7 \times 0 + 0 \times 9 = 0$.

Buggy Code:
#include<bits/stdc++.h>
#define FOR(n) for(int i=0; i<n; i++)
using namespace std;
int main()
{
	std::ios::sync_with_stdio(0), std::cin.tie(0), std::cout.tie(0);
	int t, n;
	string s;
	cin >> t;
	while (t--)
	{
		cin >> n >> s;
		if (n == 2)
			cout << (s[0] - '0') * 10 + s[1] - '0' << "\n";
		if (n >= 3)
		{
			if (n == 3 && s[1] == '0')  cout << (s[0] - '0') * (s[2] - '0') << "\n";
			else if (n == 3 && (s[0] == '0' || s[2] == '0'))  cout << "0\n";
			else if (s.find('0') != string::npos)  cout << "0\n";
			else
			{
				vector<int> num(n - 1);
				int ans = 1000000000;
				for (int i = 0; i < n - 1; i++)
				{
					int k = 0;
					for (int j = 0; j < n; j++)
					{
						if (j == i) num[k++] = (s[j] - '0') * 10 + s[j + 1] - '0', j++;
						else num[k++] = s[j] - '0';
					}
					int tmp = num[0];
					for (int j = 1; j < n - 1; j++)
					{
						if (num[j] == 1 || tmp==1) tmp *= num[j];
						else tmp += num[j];
					}
					ans = min(ans, tmp);
				}
				cout << ans << "\n";
			}

		}
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
3
101
5
23311
7
1111111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mathematical Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $s$ of length $n > 1$, consisting of digits from $0$ to $9$. You must insert exactly $n - 2$ symbols $+$ (addition) or $\times$ (multiplication) into this string to form a valid arithmetic expression.

In this problem, the symbols cannot be placed before the first or after the last character of the string $s$, and two symbols cannot be written consecutively. Also, note that the order of the digits in the string cannot be changed. Let's consider $s = 987009$:

  * From this string, the following arithmetic expressions can be obtained: $9 \times 8 + 70 \times 0 + 9 = 81$, $98 \times 7 \times 0 + 0 \times 9 = 0$, $9 + 8 + 7 + 0 + 09 = 9 + 8 + 7 + 0 + 9 = 33$. Note that the number $09$ is considered valid and is simply transformed into $9$.   * From this string, the following arithmetic expressions cannot be obtained: $+9 \times 8 \times 70 + 09$ (symbols should only be placed between digits), $98 \times 70 + 0 + 9$ (since there are $6$ digits, there must be exactly $4$ symbols).

The result of the arithmetic expression is calculated according to the rules of mathematics — first all multiplication operations are performed, then addition. You need to find the minimum result that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string $s$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases. Then follows their description.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 20$) — the length of the string $s$.

The second line of each test case contains a string $s$ of length $n$, consisting of digits from $0$ to $9$.

Output: For each test case, output the minimum result of the arithmetic expression that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string.

Examples:
input:
18
2
10
2
74
2
00
2
01
3
901
3
101
5
23311
6
987009
7
1111111
20
99999999999999999999
20
00000000000000000000
4
0212
18
057235283621345395
4
1112
20
19811678487321784121
4
1121
4
2221
3
011
output:
10
74
0
1
9
1
19
0
11
261
0
0
0
12
93
12
24
0

Note: In the first four test cases, we cannot add symbols, so the answer will be the original number.

In the fifth test case, the optimal answer looks as follows: $9 \times 01 = 9 \times 1 = 9$.

In the sixth test case, the optimal answer looks as follows: $1 \times 01 = 1 \times 1 = 1$.

In the seventh test case, the optimal answer looks as follows: $2 + 3 + 3 + 11 = 19$.

In the eighth test case, one of the optimal answers looks as follows: $98 \times 7 \times 0 + 0 \times 9 = 0$.

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

#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;

typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> indexed_set;
typedef long long ll;
typedef long double ld;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<bool> vbl;
typedef vector<vbl> vvbl;
typedef vector<ld> vld;
typedef vector<vld> vvld;
typedef pair<ll, ll> pll;
typedef vector<pll> vpll;
typedef set<ll> sll;
typedef deque<ll> dql;
typedef map<ll, ll> mll;
typedef unordered_map<ll, ll> umll;
typedef map<ll, pair<ll, ll>> mpll;
typedef map<ll, vll> mlvl;
#define sp <<" "
#define br <<"\n"
#define ff first
#define ss second
#define pb push_back
#define ppb pop_back
#define pf push_front
#define ppf pop_front
#define mp make_pair
#define PI 2 * acos(0.0)
#define int long long int
#define umap unordered_map
#define all(a) (a).begin(), (a).end()
#define rall(a) (a).rbegin(), (a).rend()
#define fr(i, I, n) for(i=(I); i<(n); i++)
#define rf(i, I, n) for(i=(n)-1; i>=(I); i--)
#define debug(x) cerr<<"["<<#x<<": "<<x<<"] ";
#define debugln(x) cerr<<"["<<#x<<": "<<x<<"]\n"
#define fastio ios_base::sync_with_stdio(false);cin.tie(0),cout.tie(0);
#define yn(flag) if(flag)cout<<"YES\n"; else cout<<"NO\n";
#define inp(a, n) for(int _i=0;_i<(n);_i++) cin>>a[_i];
#define out(a, n) for(int _i=0;_i<(n);_i++) cout<<a[_i]<<" "; cout<<"\n";
#define _clock chrono::high_resolution_clock::now()
#define nano_t chrono::duration_cast<std::chrono::nanoseconds>
#define micro_t chrono::duration_cast<std::chrono::microseconds>

void solve()
{
    ll i,j,k,m,n,p,q;
    //code here
    cin>>n;
    string s;
    cin>>s;
    ll n0=0, n1=0, mn=99, ans=-1;
    fr(i,0,n)
    {
        s[i] -= '0';
        if(s[i]==0) n0++;
        else if(s[i]==1) n1++;
        if(i>0)
            mn = min(mn, (ll) (s[i-1]*10+s[i]));
    }
    // debugln(mn);
    if(n==2) ans = (ll) (s[0]*10+s[1]);
    else if(n==3)
    {
        if(s[0]==0 || s[2]==0)
            ans=0;
        else if(s[1]==0 && n1>0)
            ans = (ll) max(s[0], s[2]);
        else
            ans = (ll) min(s[0]*10+s[1]+s[2], s[0]+s[1]*10+s[2]);
    }
    else if(n0>0) ans=0;
    else
    {
        ans = INT_MAX;
        fr(i,1,n)
        {
            ll tempans = s[i-1]*10+s[i];
            fr(j,0,n)
            {
                if(i!=j && j!=i-1 && s[j]>1)
                    tempans += s[j];
            }
            // cerr<<tempans sp;
            ans = min(ans, tempans);
        }
        // cerr br;
    }
    cout<<ans br;
}

int32_t main()
{
    fastio;
    auto tbegin = _clock;
    cout<<fixed<<setprecision(10);

    ll t=1;
    cin>>t;
    while(t--)
    {
        solve();
    }

    auto tend = _clock;
    // cout<<micro_t (tend-tbegin).count()<<" micro-seconds"<<endl;

    return 0;
}

/*
    g++ -static -DONLINE_JUDGE -lm -s -x c++ -Wl,--stack=268435456 -O2 -std=c++17 -D__USE_MINGW_ANSI_STDIO=0 -Wall -Wextra -pedantic -g -fsanitize=address -o filename filename.cpp
*/

**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
10
2
74
2
00
2
01
3
901
3
101
5
23311
6
987009
7
1111111
20
99999999999999999999
20
00000000000000000000
4
0212
4
1121
4
2221
3
011
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mathematical Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $s$ of length $n > 1$, consisting of digits from $0$ to $9$. You must insert exactly $n - 2$ symbols $+$ (addition) or $\times$ (multiplication) into this string to form a valid arithmetic expression.

In this problem, the symbols cannot be placed before the first or after the last character of the string $s$, and two symbols cannot be written consecutively. Also, note that the order of the digits in the string cannot be changed. Let's consider $s = 987009$:

  * From this string, the following arithmetic expressions can be obtained: $9 \times 8 + 70 \times 0 + 9 = 81$, $98 \times 7 \times 0 + 0 \times 9 = 0$, $9 + 8 + 7 + 0 + 09 = 9 + 8 + 7 + 0 + 9 = 33$. Note that the number $09$ is considered valid and is simply transformed into $9$.   * From this string, the following arithmetic expressions cannot be obtained: $+9 \times 8 \times 70 + 09$ (symbols should only be placed between digits), $98 \times 70 + 0 + 9$ (since there are $6$ digits, there must be exactly $4$ symbols).

The result of the arithmetic expression is calculated according to the rules of mathematics — first all multiplication operations are performed, then addition. You need to find the minimum result that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string $s$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases. Then follows their description.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 20$) — the length of the string $s$.

The second line of each test case contains a string $s$ of length $n$, consisting of digits from $0$ to $9$.

Output: For each test case, output the minimum result of the arithmetic expression that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string.

Examples:
input:
18
2
10
2
74
2
00
2
01
3
901
3
101
5
23311
6
987009
7
1111111
20
99999999999999999999
20
00000000000000000000
4
0212
18
057235283621345395
4
1112
20
19811678487321784121
4
1121
4
2221
3
011
output:
10
74
0
1
9
1
19
0
11
261
0
0
0
12
93
12
24
0

Note: In the first four test cases, we cannot add symbols, so the answer will be the original number.

In the fifth test case, the optimal answer looks as follows: $9 \times 01 = 9 \times 1 = 9$.

In the sixth test case, the optimal answer looks as follows: $1 \times 01 = 1 \times 1 = 1$.

In the seventh test case, the optimal answer looks as follows: $2 + 3 + 3 + 11 = 19$.

In the eighth test case, one of the optimal answers looks as follows: $98 \times 7 \times 0 + 0 \times 9 = 0$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define int long long
#ifndef ONLINE_JUDGE
#include "debug.h"
#else
#define IO() fastIO()
#endif
inline void fastIO() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
}
constexpr int M = 1000000007;

void solve();

int32_t main() {
    IO();
    int t = 1;
    cin>>t;
    while (t--)
        solve();
    return 0;
}

int calculateExpression(const string &expression) {
    std::stack<int> values;
    std::stack<char> operators;
    int i = 0;

    while (i < expression.size()) {
        if (isdigit(expression[i])) {
            int num = expression[i] - '0';
            // Check if it's a two-digit number
            if (i + 1 < expression.size() && isdigit(expression[i + 1])) {
                num = num * 10 + (expression[i + 1] - '0');
                i++;
            }
            values.push(num);
        } else if (expression[i] == '+' || expression[i] == '*') {
            while (!operators.empty() &&
                   (operators.top() == '*' || (operators.top() == '+' && expression[i] == '+'))) {
                char op = operators.top();
                operators.pop();
                int b = values.top(); values.pop();
                int a = values.top(); values.pop();
                if (op == '*') {
                    values.push(a * b);
                } else if (op == '+') {
                    values.push(a + b);
                }
                   }
            operators.push(expression[i]);
        }
        i++;
    }

    while (!operators.empty()) {
        char op = operators.top();
        operators.pop();
        int b = values.top(); values.pop();
        int a = values.top(); values.pop();
        if (op == '*') {
            values.push(a * b);
        } else if (op == '+') {
            values.push(a + b);
        }
    }

    return values.top();
}

void solve() {
    int n;
    cin>>n;
    string s;
    cin>>s;
    vector<char> v(n-1);
    for(int i=0; i<n-1; i++) {
        if(s[i]=='0') {
            v[i] = '*';
            if(i-1>=0)
                v[i-1] = '*';
        } else if(s[i]=='1') {
            v[i] = '*';
        } else {
            v[i] = '+';
        }
    }
    if(s[n-1]=='0' || s[n-1]=='1')
        v[n-2] = '*';
    string exp;
    for(int i=0; i<n; i++) {
        exp.push_back(s[i]);
        if(i<n-1)
            exp.push_back(v[i]);
    }
    if(s[0]=='0' && s.length()>2) {
        cout<<0<<"\n";
        return;
    }
    int mn = 1e18;
    size_t found = exp.find("*0*");
    if (found != string::npos) {
        if(s.length() > 3) {
            cout<<0<<"\n";
            return;
        }
    }
    for(int i=0; i<n-1; i++) {
        int j = 2 * i + 1;
        string ex = exp.substr(0,j) + exp.substr(j+1);
        int eval = calculateExpression(ex);
        // cout<<ex<<","<<eval<<" ";
        mn = min(mn, eval);
    }
    cout<<mn<<"\n";
}

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

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

Think step by step.

Response:
```plaintext
3
2
10
3
100
4
1010
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mathematical Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $s$ of length $n > 1$, consisting of digits from $0$ to $9$. You must insert exactly $n - 2$ symbols $+$ (addition) or $\times$ (multiplication) into this string to form a valid arithmetic expression.

In this problem, the symbols cannot be placed before the first or after the last character of the string $s$, and two symbols cannot be written consecutively. Also, note that the order of the digits in the string cannot be changed. Let's consider $s = 987009$:

  * From this string, the following arithmetic expressions can be obtained: $9 \times 8 + 70 \times 0 + 9 = 81$, $98 \times 7 \times 0 + 0 \times 9 = 0$, $9 + 8 + 7 + 0 + 09 = 9 + 8 + 7 + 0 + 9 = 33$. Note that the number $09$ is considered valid and is simply transformed into $9$.   * From this string, the following arithmetic expressions cannot be obtained: $+9 \times 8 \times 70 + 09$ (symbols should only be placed between digits), $98 \times 70 + 0 + 9$ (since there are $6$ digits, there must be exactly $4$ symbols).

The result of the arithmetic expression is calculated according to the rules of mathematics — first all multiplication operations are performed, then addition. You need to find the minimum result that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string $s$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases. Then follows their description.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 20$) — the length of the string $s$.

The second line of each test case contains a string $s$ of length $n$, consisting of digits from $0$ to $9$.

Output: For each test case, output the minimum result of the arithmetic expression that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string.

Examples:
input:
18
2
10
2
74
2
00
2
01
3
901
3
101
5
23311
6
987009
7
1111111
20
99999999999999999999
20
00000000000000000000
4
0212
18
057235283621345395
4
1112
20
19811678487321784121
4
1121
4
2221
3
011
output:
10
74
0
1
9
1
19
0
11
261
0
0
0
12
93
12
24
0

Note: In the first four test cases, we cannot add symbols, so the answer will be the original number.

In the fifth test case, the optimal answer looks as follows: $9 \times 01 = 9 \times 1 = 9$.

In the sixth test case, the optimal answer looks as follows: $1 \times 01 = 1 \times 1 = 1$.

In the seventh test case, the optimal answer looks as follows: $2 + 3 + 3 + 11 = 19$.

In the eighth test case, one of the optimal answers looks as follows: $98 \times 7 \times 0 + 0 \times 9 = 0$.

Buggy Code:
for q in range(int(input())):
    a=int(input())
    b=input()
    s=0
    c=0
    f=0
    g=0
    h=0
    mi=10
    j=0
    if q==4421:
        print(b)
    if a==2:
        print(int(b))
    elif a==3:
        print(min(10*int(b[0])+int(b[1])+int(b[2]),(10*int(b[0])+int(b[1]))*int(b[2]),int(b[0])+10*int(b[1])+int(b[2]),int(b[0])*(10*int(b[1])+int(b[2]))))
    else:
        for i in range(a):
            s+=int(b[i])
            if b[i]=='0':
                print("0")
                f=1
                g=1
                break
            elif b[i]=='1'and c==0:
                g=1
                if i!=a-1:
                    s+=9
                    if b[i+1]!='1':
                        h=1
                else:
                    h=1
                    s-=1
                    s+=9*(mi)
                c=1
            elif b[i]=='1' and c==1:
                g=1
                s-=1
                if i!=a-1 and b[i+1]!='1':
                    h=1
                elif i==a-1 and h==0:
                    s+=2
            if int(b[i])<mi:
                mi=int(b[i])
        if h==0:
            s-=1
        if g==0:
            s+=9*(mi)+1
        if f==0:
            print(s)

**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
10
3
011
4
1111
5
11111
6
111111
7
1111111
8
11111111
9
111111111
10
1111111111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mathematical Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $s$ of length $n > 1$, consisting of digits from $0$ to $9$. You must insert exactly $n - 2$ symbols $+$ (addition) or $\times$ (multiplication) into this string to form a valid arithmetic expression.

In this problem, the symbols cannot be placed before the first or after the last character of the string $s$, and two symbols cannot be written consecutively. Also, note that the order of the digits in the string cannot be changed. Let's consider $s = 987009$:

  * From this string, the following arithmetic expressions can be obtained: $9 \times 8 + 70 \times 0 + 9 = 81$, $98 \times 7 \times 0 + 0 \times 9 = 0$, $9 + 8 + 7 + 0 + 09 = 9 + 8 + 7 + 0 + 9 = 33$. Note that the number $09$ is considered valid and is simply transformed into $9$.   * From this string, the following arithmetic expressions cannot be obtained: $+9 \times 8 \times 70 + 09$ (symbols should only be placed between digits), $98 \times 70 + 0 + 9$ (since there are $6$ digits, there must be exactly $4$ symbols).

The result of the arithmetic expression is calculated according to the rules of mathematics — first all multiplication operations are performed, then addition. You need to find the minimum result that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string $s$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases. Then follows their description.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 20$) — the length of the string $s$.

The second line of each test case contains a string $s$ of length $n$, consisting of digits from $0$ to $9$.

Output: For each test case, output the minimum result of the arithmetic expression that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string.

Examples:
input:
18
2
10
2
74
2
00
2
01
3
901
3
101
5
23311
6
987009
7
1111111
20
99999999999999999999
20
00000000000000000000
4
0212
18
057235283621345395
4
1112
20
19811678487321784121
4
1121
4
2221
3
011
output:
10
74
0
1
9
1
19
0
11
261
0
0
0
12
93
12
24
0

Note: In the first four test cases, we cannot add symbols, so the answer will be the original number.

In the fifth test case, the optimal answer looks as follows: $9 \times 01 = 9 \times 1 = 9$.

In the sixth test case, the optimal answer looks as follows: $1 \times 01 = 1 \times 1 = 1$.

In the seventh test case, the optimal answer looks as follows: $2 + 3 + 3 + 11 = 19$.

In the eighth test case, one of the optimal answers looks as follows: $98 \times 7 \times 0 + 0 \times 9 = 0$.

Buggy Code:
import sys

#sys.stdin = open('input.txt', 'r')
q = int(input())

def solve2(seqs):
    str = ""
    for i in range(len(seqs)):
        s = seqs[i]
        str += s
        if s == '1' or (i == len(seqs) - 2 and seqs[-1] == '1'):
            str += '*'
        else:
            str += '+'
    str = str[:-1]
    return eval(str)

def solve3(s):
    option1 = [s[0], s[1:3]]
    option2 = [s[0:2], s[2]]
    options = [option1, option2]
    best = 10 ** 100
    for option in options:
        l, r = option
        if l[0] == '0':
            l = l[-1]
        if r[0] == '0':
            r = r[-1]
        best = min(best,int(l)*int(r))
        best = min(best,int(l)+int(r))
    return best

def solve(n, s):
    if n <= 2:
        if s[0] == '0':
            return s[1]
        return s
    if n == 3:
        return solve3(s)
    # n >= 4
    if '0' in s:
        return 0
    best = 2 ** 100
    for i in range(n - 1):
        numba = s[i:i + 2]
        if numba[0] == '0':
            numba = numba[1]
        seqs = list(s[:i]) + [numba] + list(s[i + 2:])
        res = solve2(seqs)
        best = min(best, res)
    return best

for test in range(q):
    n = int(input())
    s = input().strip()
    print(solve(n, s))

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases that can cause issues in the logic of the `solve2` and `solve3` functions, as well as the main `solve` function. Here are some key scenarios to consider:

1. **Strings with leading zeros**: The code should handle leading zeros correctly, but there might be bugs in how it processes these.
2. **Strings with multiple zeros**: The code should return 0 if there is at least one zero in the string, but we need to ensure it handles this correctly.
3. **Strings with single digits**: The code should handle these correctly, but we need to ensure it doesn't add unnecessary operations.
4. **Strings with repeated digits**: The code should handle these correctly, but we need to ensure it doesn't add unnecessary operations.

Let's create a small-scale test input that includes these scenarios:

```plaintext
5
2
00
3
101
4
1112
5
23311
6
987009
```

This test input includes:
- A string with leading zeros (00)
- A string with a single zero in the middle (101)
- A string with repeated digits (1112)
- A string with multiple digits (23311)
- A string with a zero in the middle and at the end (987009)

These cases should help expose any issues in the code's handling of leading zeros, zeros in the middle, and repeated digits.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mathematical Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $s$ of length $n > 1$, consisting of digits from $0$ to $9$. You must insert exactly $n - 2$ symbols $+$ (addition) or $\times$ (multiplication) into this string to form a valid arithmetic expression.

In this problem, the symbols cannot be placed before the first or after the last character of the string $s$, and two symbols cannot be written consecutively. Also, note that the order of the digits in the string cannot be changed. Let's consider $s = 987009$:

  * From this string, the following arithmetic expressions can be obtained: $9 \times 8 + 70 \times 0 + 9 = 81$, $98 \times 7 \times 0 + 0 \times 9 = 0$, $9 + 8 + 7 + 0 + 09 = 9 + 8 + 7 + 0 + 9 = 33$. Note that the number $09$ is considered valid and is simply transformed into $9$.   * From this string, the following arithmetic expressions cannot be obtained: $+9 \times 8 \times 70 + 09$ (symbols should only be placed between digits), $98 \times 70 + 0 + 9$ (since there are $6$ digits, there must be exactly $4$ symbols).

The result of the arithmetic expression is calculated according to the rules of mathematics — first all multiplication operations are performed, then addition. You need to find the minimum result that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string $s$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases. Then follows their description.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 20$) — the length of the string $s$.

The second line of each test case contains a string $s$ of length $n$, consisting of digits from $0$ to $9$.

Output: For each test case, output the minimum result of the arithmetic expression that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string.

Examples:
input:
18
2
10
2
74
2
00
2
01
3
901
3
101
5
23311
6
987009
7
1111111
20
99999999999999999999
20
00000000000000000000
4
0212
18
057235283621345395
4
1112
20
19811678487321784121
4
1121
4
2221
3
011
output:
10
74
0
1
9
1
19
0
11
261
0
0
0
12
93
12
24
0

Note: In the first four test cases, we cannot add symbols, so the answer will be the original number.

In the fifth test case, the optimal answer looks as follows: $9 \times 01 = 9 \times 1 = 9$.

In the sixth test case, the optimal answer looks as follows: $1 \times 01 = 1 \times 1 = 1$.

In the seventh test case, the optimal answer looks as follows: $2 + 3 + 3 + 11 = 19$.

In the eighth test case, one of the optimal answers looks as follows: $98 \times 7 \times 0 + 0 \times 9 = 0$.

Buggy Code:
T = int(input())

for _ in range(T):
    n = int(input())
    s = list(map(int, (input())))
    if 0 in s:
        if n > 3:
            print(0)
            continue
        elif n == 3:
            if not (s[0] and not (s[1]) and s[2]):
                print(0)
                continue

    sh = 100
    ind = -1
    for i in range(n - 1):
        bruh = s[i] * 10 + s[i + 1]
        if bruh <= sh:
            sh = bruh
            ind = i

    if sh == 11:
        if ind < n - 2:
            sh = 10 + s[ind + 2]
            ind += 1

    if sh % 10 == 1 and ind > 0 and s[ind - 1] == s[ind]:
        sh = s[ind - 1] * 10 + s[ind]
        ind -= 1

    s[ind : ind + 2] = [sh]
    new = []
    for i in range(len(s)):
        if s[i] != 1:
            new.append(s[i])

    if new:
        print(sum(new))
    else:
        print(1)

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

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

Think step by step.

Response:
```plaintext
5
3
101
4
2112
5
23311
6
987009
7
1111111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mathematical Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $s$ of length $n > 1$, consisting of digits from $0$ to $9$. You must insert exactly $n - 2$ symbols $+$ (addition) or $\times$ (multiplication) into this string to form a valid arithmetic expression.

In this problem, the symbols cannot be placed before the first or after the last character of the string $s$, and two symbols cannot be written consecutively. Also, note that the order of the digits in the string cannot be changed. Let's consider $s = 987009$:

  * From this string, the following arithmetic expressions can be obtained: $9 \times 8 + 70 \times 0 + 9 = 81$, $98 \times 7 \times 0 + 0 \times 9 = 0$, $9 + 8 + 7 + 0 + 09 = 9 + 8 + 7 + 0 + 9 = 33$. Note that the number $09$ is considered valid and is simply transformed into $9$.   * From this string, the following arithmetic expressions cannot be obtained: $+9 \times 8 \times 70 + 09$ (symbols should only be placed between digits), $98 \times 70 + 0 + 9$ (since there are $6$ digits, there must be exactly $4$ symbols).

The result of the arithmetic expression is calculated according to the rules of mathematics — first all multiplication operations are performed, then addition. You need to find the minimum result that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string $s$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases. Then follows their description.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 20$) — the length of the string $s$.

The second line of each test case contains a string $s$ of length $n$, consisting of digits from $0$ to $9$.

Output: For each test case, output the minimum result of the arithmetic expression that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string.

Examples:
input:
18
2
10
2
74
2
00
2
01
3
901
3
101
5
23311
6
987009
7
1111111
20
99999999999999999999
20
00000000000000000000
4
0212
18
057235283621345395
4
1112
20
19811678487321784121
4
1121
4
2221
3
011
output:
10
74
0
1
9
1
19
0
11
261
0
0
0
12
93
12
24
0

Note: In the first four test cases, we cannot add symbols, so the answer will be the original number.

In the fifth test case, the optimal answer looks as follows: $9 \times 01 = 9 \times 1 = 9$.

In the sixth test case, the optimal answer looks as follows: $1 \times 01 = 1 \times 1 = 1$.

In the seventh test case, the optimal answer looks as follows: $2 + 3 + 3 + 11 = 19$.

In the eighth test case, one of the optimal answers looks as follows: $98 \times 7 \times 0 + 0 \times 9 = 0$.

Buggy Code:
import java.util.*;
public class Main {
    public static void main(String[] args){
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        for(int i=0;i<t;i++){
           int n=sc.nextInt();
           String s=sc.next();
           if(n<3)
           System.out.println(Integer.parseInt(s));
           else if(n==3)
           {
            if(s.charAt(0)=='0'||s.charAt(2)=='0')
            System.out.println(0);
            else if(s.charAt(1)=='0')
            {
                if(s.charAt(0)=='1'||s.charAt(2)=='1')
                System.out.println(Math.max(Integer.parseInt(s.substring(0,1)),Integer.parseInt(s.substring(2,3))));
                else
                System.out.println(Integer.parseInt(s.substring(0,1))+Integer.parseInt(s.substring(2,3)));
            }
            else
            {
                if(s.charAt(2)=='1')
                System.out.println(s.substring(0, 2));
                else if(s.charAt(0)=='1')
                System.out.println(s.substring(1, 2));
                else
                System.out.println(Integer.parseInt(s.substring(1,2))+Integer.parseInt(s.substring(0,1))+Integer.parseInt(s.substring(2,3)));
            }
           }
           else{
            int ans1=100000000;
            for(int k=0;k<n-1;k++){
                int ans=0;
                ans+=Integer.parseInt(s.substring(k,k+2));
           for(int j=0;j<n;j++ )
           {
            if(j==k||j==k+1)
            continue;

            if(s.charAt(j)=='0'){
                ans=0;
                break;
            }
            else if(s.charAt(j)!='1')
            ans+=Integer.parseInt(s.substring(j,j+1));
           }
           ans1=Math.min(ans1,ans);

        }
        System.out.println(ans1);
    }
        }
    }
}

**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
101
5
23311
20
99999999999999999999
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mathematical Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $s$ of length $n > 1$, consisting of digits from $0$ to $9$. You must insert exactly $n - 2$ symbols $+$ (addition) or $\times$ (multiplication) into this string to form a valid arithmetic expression.

In this problem, the symbols cannot be placed before the first or after the last character of the string $s$, and two symbols cannot be written consecutively. Also, note that the order of the digits in the string cannot be changed. Let's consider $s = 987009$:

  * From this string, the following arithmetic expressions can be obtained: $9 \times 8 + 70 \times 0 + 9 = 81$, $98 \times 7 \times 0 + 0 \times 9 = 0$, $9 + 8 + 7 + 0 + 09 = 9 + 8 + 7 + 0 + 9 = 33$. Note that the number $09$ is considered valid and is simply transformed into $9$.   * From this string, the following arithmetic expressions cannot be obtained: $+9 \times 8 \times 70 + 09$ (symbols should only be placed between digits), $98 \times 70 + 0 + 9$ (since there are $6$ digits, there must be exactly $4$ symbols).

The result of the arithmetic expression is calculated according to the rules of mathematics — first all multiplication operations are performed, then addition. You need to find the minimum result that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string $s$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases. Then follows their description.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 20$) — the length of the string $s$.

The second line of each test case contains a string $s$ of length $n$, consisting of digits from $0$ to $9$.

Output: For each test case, output the minimum result of the arithmetic expression that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string.

Examples:
input:
18
2
10
2
74
2
00
2
01
3
901
3
101
5
23311
6
987009
7
1111111
20
99999999999999999999
20
00000000000000000000
4
0212
18
057235283621345395
4
1112
20
19811678487321784121
4
1121
4
2221
3
011
output:
10
74
0
1
9
1
19
0
11
261
0
0
0
12
93
12
24
0

Note: In the first four test cases, we cannot add symbols, so the answer will be the original number.

In the fifth test case, the optimal answer looks as follows: $9 \times 01 = 9 \times 1 = 9$.

In the sixth test case, the optimal answer looks as follows: $1 \times 01 = 1 \times 1 = 1$.

In the seventh test case, the optimal answer looks as follows: $2 + 3 + 3 + 11 = 19$.

In the eighth test case, one of the optimal answers looks as follows: $98 \times 7 \times 0 + 0 \times 9 = 0$.

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

public class QUES1 {
    static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    static BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));
    static StringTokenizer st;

    static Scanner s = new Scanner(System.in);

    static int MOD = (int) (1e9 + 7);

    public static void solve() throws IOException {
           int n = Integer.parseInt(reader.readLine());
           String str = reader.readLine();

           if (str.length()==2){
               writer.write(Integer.parseInt(str)+"\n");
               return;
           }
           if (str.length()==3){
               int a = Integer.parseInt(str.substring(0,2));
               int b = Integer.parseInt(str.substring(2,3));
               int ans = Math.min(a+b,a*b);
               a = Integer.parseInt(str.substring(0,1));
               b = Integer.parseInt(str.substring(1,3));
               ans = Math.min(a+b,a*b);
               writer.write(ans+"\n");
               return;
           }

           int ans = Integer.MAX_VALUE;
           for (int i=1;i<n;i++){
               String temp = "";
               temp += str.charAt(i-1);
               temp += str.charAt(i);
               int num = Integer.parseInt(temp);
//               if (num==1) num--;
               for (int j=0;j<n;j++){
                   if (j==i || j==i-1) continue;
                   if (str.charAt(j)=='0'){
                       writer.write(0+"\n");
                       return;
                   }
                   if (str.charAt(j)=='1') continue;
                   num += (str.charAt(j)-'0');
               }
               ans = Math.min(ans,num);
           }
           writer.write(ans+"\n");
    }

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

        int t = Integer.parseInt(reader.readLine());
        // int t = 1;
        while (t-- > 0) {
            solve();
        }

        writer.flush();
        writer.close();
    }

}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases that the code might handle incorrectly. Here are some considerations:

1. **Strings with leading zeros**: The code should handle leading zeros correctly, but it might not.
2. **Strings with multiple zeros**: The code might prematurely conclude that the result is zero without considering all possible placements of symbols.
3. **Strings with multiple ones**: The code might not handle the addition and multiplication of ones correctly.
4. **Strings of length 3**: The code has a specific block for strings of length 3, but it might have logical errors.
5. **Strings with a mix of high and low digits**: The code might not correctly minimize the result.

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

```plaintext
5
3
001
3
111
3
999
4
1010
6
111111
```

### Explanation:
1. **Test Case 1: "001"**
   - The string has leading zeros. The correct result should be 0, but the code might handle it incorrectly.

2. **Test Case 2: "111"**
   - The string has multiple ones. The correct result should be 3 (1 + 1 + 1), but the code might not handle it correctly.

3. **Test Case 3: "999"**
   - The string has high digits. The correct result should be 27 (9 + 9 + 9), but the code might not handle it correctly.

4. **Test Case 4: "1010"**
   - The string has a mix of zeros and ones. The correct result should be 2 (1 + 0 + 1 + 0), but the code might not handle it correctly.

5. **Test Case 5: "111111"**
   - The string has multiple ones. The correct result should be 6 (1 + 1 + 1 + 1 + 1 + 1), but the code might not handle it correctly.

This test input should help expose any bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mathematical Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $s$ of length $n > 1$, consisting of digits from $0$ to $9$. You must insert exactly $n - 2$ symbols $+$ (addition) or $\times$ (multiplication) into this string to form a valid arithmetic expression.

In this problem, the symbols cannot be placed before the first or after the last character of the string $s$, and two symbols cannot be written consecutively. Also, note that the order of the digits in the string cannot be changed. Let's consider $s = 987009$:

  * From this string, the following arithmetic expressions can be obtained: $9 \times 8 + 70 \times 0 + 9 = 81$, $98 \times 7 \times 0 + 0 \times 9 = 0$, $9 + 8 + 7 + 0 + 09 = 9 + 8 + 7 + 0 + 9 = 33$. Note that the number $09$ is considered valid and is simply transformed into $9$.   * From this string, the following arithmetic expressions cannot be obtained: $+9 \times 8 \times 70 + 09$ (symbols should only be placed between digits), $98 \times 70 + 0 + 9$ (since there are $6$ digits, there must be exactly $4$ symbols).

The result of the arithmetic expression is calculated according to the rules of mathematics — first all multiplication operations are performed, then addition. You need to find the minimum result that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string $s$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases. Then follows their description.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 20$) — the length of the string $s$.

The second line of each test case contains a string $s$ of length $n$, consisting of digits from $0$ to $9$.

Output: For each test case, output the minimum result of the arithmetic expression that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string.

Examples:
input:
18
2
10
2
74
2
00
2
01
3
901
3
101
5
23311
6
987009
7
1111111
20
99999999999999999999
20
00000000000000000000
4
0212
18
057235283621345395
4
1112
20
19811678487321784121
4
1121
4
2221
3
011
output:
10
74
0
1
9
1
19
0
11
261
0
0
0
12
93
12
24
0

Note: In the first four test cases, we cannot add symbols, so the answer will be the original number.

In the fifth test case, the optimal answer looks as follows: $9 \times 01 = 9 \times 1 = 9$.

In the sixth test case, the optimal answer looks as follows: $1 \times 01 = 1 \times 1 = 1$.

In the seventh test case, the optimal answer looks as follows: $2 + 3 + 3 + 11 = 19$.

In the eighth test case, one of the optimal answers looks as follows: $98 \times 7 \times 0 + 0 \times 9 = 0$.

Buggy Code:
#include<bits/stdc++.h>
#define ll long long int
#define pb push_back
#define endl "\n"
#define vll vector<long long int>
#define f(i,a,n) for(ll i=a;i<n;i++)
#define rf(i,a,n) for(ll i=n-1;i>=0;i--)
#define fast ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;

void solve(){
    ll n;
    cin>>n;

    if(n==3){
        ll x;
        cin>>x;
        cout<<min({(x/10) * (x%10),(x/100) * (x%100),x/10 + x%10,x/100 + x%100});
        return;
    }
    string s;
    cin>>s;

    if(n==2){
        if(s[0]=='0')cout<<s[1];
        else cout<<s;
        return;
    }
    char mini='9';
    f(i,0,n-1){
        mini=min(mini,s[i]);
    }
    if(mini=='0' || s[n-1]=='0'){cout<<0;return;}

    bool flag=true;
    ll ans=0,tmp=0;
    ll i=0;
    for(i=0;i<n-1;i++){
        if(s[i]==mini && flag){
            if(s[i+1]=='1'){continue;}
            tmp=(s[i]-'0');
            tmp*=10;
            tmp+=(s[i+1]-'0');
            ans+=tmp;
            i++;
            flag=false;
            continue;
        }
        if(s[i]!='1')
        ans+=(s[i]-'0');

    }
    if(i!=n && s[n-1]!='1')ans+=(s[i]-'0');
    if(flag){
        ans+=11;
    }

    cout<<ans;

}
int main(){
    fast
    ll t=1;
    cin>>t;
    while(t--){
        solve();
        cout<<endl;
    }
}

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

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

Think step by step.

Response:
```plaintext
3
3
101
4
1112
5
23311
```