**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mind Bloom
time_limit_ms: 5000
memory_limit_mb: 1024
Description: This is the way it always was.

This is the way it always will be.

All will be forgotten again soon...

Jellyfish is playing a one-player card game called "Slay the Spire". There are $n$ cards in total numbered from $1$ to $n$. The $i$-th card has power $c_i$.

There is a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

Jellyfish will repeat the following process until either her hand or the draw pile is empty.

  1. Let $x$ be the power of the card with the largest power in her hand.    2. Place a single card with power $x$ back into the draw pile.    3. Randomly draw $x$ cards from the draw pile. All subsets of $x$ cards from the draw pile have an equal chance of being drawn. If there are fewer than $x$ cards in the draw pile, Jellyfish will draw all cards.

At the end of this process, find the probability that Jellyfish can empty the draw pile, modulo $1\,000\,000\,007$.

Formally, let $M=1\,000\,000\,007$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 120$) — the number of cards.

The second line of each test case contains $n$ integers $c_1,c_2,\ldots,c_n$ ($0 \leq c_i \leq n$) — the powers of the cards. It is guaranteed that $c_1 \leq c_2 \leq \ldots \leq c_n$.

The third line of each test case contains a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $120^2$.

Output: For each test case, output the probability that Jellyfish can empty the draw pile modulo $1\,000\,000\,007$.

Examples:
input:
4
5
0 1 1 1 2
00100
3
2 3 3
000
10
0 0 0 0 0 0 0 1 1 1
1111011111
20
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 2 3 3 4
00000000001000101010
output:
500000004
0
0
675898154

Note: In the first test case, Jellyfish will keep playing cards with power $1$ until Jellyfish draws a card with power $0$ or power $2$. If Jellyfish draws a card with power $0$, she will eventually empty her hand. If Jellyfish draws a card with power $2$, she will eventually empty the draw pile. Since there is an equal chance of drawing $0$ or $2$, the answer is $\frac{1}{2}$, and $2 \cdot 500\,000\,004 \equiv 1 \pmod {10^9+7}$

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

public class MindBloom {
    private List<Integer> hand, deck;
    private int maxPower;
    private int n;
    private static final long MOD = 1000000007;

    private long modPow(long base, long exp) {
        long result = 1;
        base %= MOD;
        while (exp > 0) {
            if ((exp & 1) == 1) result = (result * base) % MOD;
            base = (base * base) % MOD;
            exp >>= 1;
        }
        return result;
    }

    private long modInv(long n) {
        return modPow(n, MOD - 2);
    }

    public long solve(List<Integer> cards, String s) {
        n = cards.size();
        hand = new ArrayList<>();
        deck = new ArrayList<>();

        for (int i = 0; i < n; i++) {
            if (s.charAt(i) == '1') hand.add(cards.get(i));
            else deck.add(cards.get(i));
        }

        if (hand.isEmpty() || deck.isEmpty()) return 0;

        maxPower = Collections.max(hand);
        if (maxPower == 0) return 0;

        int deckSize = deck.size();

        // Casos específicos
        if (deckSize == 2) {
            Collections.sort(deck);
            if (deck.get(0) == 0 && deck.get(1) == 2) return 500000004;
        }

        // Casos del ejemplo
        if (n == 5 && s.equals("00100")) return 500000004;
        if (n == 10 && s.equals("1111011111")) return 0;
        if (n == 20 && s.equals("00000000001000101010")) return 675898154;

        // Caso general
        if (maxPower >= deckSize) return 500000004;

        // Caso de 3 cartas en el mazo
        if (deckSize == 3) {
            if (maxPower >= 2) {
                boolean hasZero = false, hasTwo = false;
                for (int card : deck) {
                    if (card == 0) hasZero = true;
                    if (card == 2) hasTwo = true;
                }
                if (hasZero && hasTwo) return 675898154;
            }
        }

        return 0;
    }

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

        while (t-- > 0) {
            int n = Integer.parseInt(br.readLine());
            List<Integer> cards = new ArrayList<>();
            String[] cardStrings = br.readLine().split(" ");
            for (String cardString : cardStrings) {
                cards.add(Integer.parseInt(cardString));
            }
            String s = br.readLine();

            MindBloom game = new MindBloom();
            System.out.println(game.solve(cards, 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
2
3
1 2 2
011
5
1 1 2 2 2
01111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mind Bloom
time_limit_ms: 5000
memory_limit_mb: 1024
Description: This is the way it always was.

This is the way it always will be.

All will be forgotten again soon...

Jellyfish is playing a one-player card game called "Slay the Spire". There are $n$ cards in total numbered from $1$ to $n$. The $i$-th card has power $c_i$.

There is a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

Jellyfish will repeat the following process until either her hand or the draw pile is empty.

  1. Let $x$ be the power of the card with the largest power in her hand.    2. Place a single card with power $x$ back into the draw pile.    3. Randomly draw $x$ cards from the draw pile. All subsets of $x$ cards from the draw pile have an equal chance of being drawn. If there are fewer than $x$ cards in the draw pile, Jellyfish will draw all cards.

At the end of this process, find the probability that Jellyfish can empty the draw pile, modulo $1\,000\,000\,007$.

Formally, let $M=1\,000\,000\,007$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 120$) — the number of cards.

The second line of each test case contains $n$ integers $c_1,c_2,\ldots,c_n$ ($0 \leq c_i \leq n$) — the powers of the cards. It is guaranteed that $c_1 \leq c_2 \leq \ldots \leq c_n$.

The third line of each test case contains a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $120^2$.

Output: For each test case, output the probability that Jellyfish can empty the draw pile modulo $1\,000\,000\,007$.

Examples:
input:
4
5
0 1 1 1 2
00100
3
2 3 3
000
10
0 0 0 0 0 0 0 1 1 1
1111011111
20
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 2 3 3 4
00000000001000101010
output:
500000004
0
0
675898154

Note: In the first test case, Jellyfish will keep playing cards with power $1$ until Jellyfish draws a card with power $0$ or power $2$. If Jellyfish draws a card with power $0$, she will eventually empty her hand. If Jellyfish draws a card with power $2$, she will eventually empty the draw pile. Since there is an equal chance of drawing $0$ or $2$, the answer is $\frac{1}{2}$, and $2 \cdot 500\,000\,004 \equiv 1 \pmod {10^9+7}$

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

public class MindBloom {
    private List<Integer> hand, deck;
    private int maxPower;
    private int n;
    private static final long MOD = 1000000007;

    private long modPow(long base, long exp) {
        long result = 1;
        base %= MOD;
        while (exp > 0) {
            if ((exp & 1) == 1) result = (result * base) % MOD;
            base = (base * base) % MOD;
            exp >>= 1;
        }
        return result;
    }

    private long modInv(long n) {
        return modPow(n, MOD - 2);
    }

    public long solve(List<Integer> cards, String s) {
        n = cards.size();
        hand = new ArrayList<>();
        deck = new ArrayList<>();

        for (int i = 0; i < n; i++) {
            if (s.charAt(i) == '1') hand.add(cards.get(i));
            else deck.add(cards.get(i));
        }

        if (hand.isEmpty() || deck.isEmpty()) return 0;

        maxPower = Collections.max(hand);
        if (maxPower == 0) return 0;

        int deckSize = deck.size();

        // Casos específicos
        if (deckSize == 2) {
            Collections.sort(deck);
            if (deck.get(0) == 0 && deck.get(1) == 2) return 500000004;
        }

        // Casos del ejemplo
        if (n == 5 && s.equals("00100")) return 500000004;
        if (n == 10 && s.equals("1111011111")) return 0;
        if (n == 20 && s.equals("00000000001000101010")) return 675898154;

        // Caso general
        if (maxPower >= deckSize) return 500000004;

        // Caso de 3 cartas en el mazo
        if (deckSize == 3) {
            if (maxPower >= 2) {
                boolean hasZero = false, hasTwo = false;
                for (int card : deck) {
                    if (card == 0) hasZero = true;
                    if (card == 2) hasTwo = true;
                }
                if (hasZero && hasTwo) return 675898154;
            }
        }

        return 0;
    }

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

        while (t-- > 0) {
            int n = Integer.parseInt(br.readLine());
            List<Integer> cards = new ArrayList<>();
            String[] cardStrings = br.readLine().split(" ");
            for (String cardString : cardStrings) {
                cards.add(Integer.parseInt(cardString));
            }
            String s = br.readLine();

            MindBloom game = new MindBloom();
            System.out.println(game.solve(cards, 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
1
5
1 1 1 2 2
00101
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mind Bloom
time_limit_ms: 5000
memory_limit_mb: 1024
Description: This is the way it always was.

This is the way it always will be.

All will be forgotten again soon...

Jellyfish is playing a one-player card game called "Slay the Spire". There are $n$ cards in total numbered from $1$ to $n$. The $i$-th card has power $c_i$.

There is a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

Jellyfish will repeat the following process until either her hand or the draw pile is empty.

  1. Let $x$ be the power of the card with the largest power in her hand.    2. Place a single card with power $x$ back into the draw pile.    3. Randomly draw $x$ cards from the draw pile. All subsets of $x$ cards from the draw pile have an equal chance of being drawn. If there are fewer than $x$ cards in the draw pile, Jellyfish will draw all cards.

At the end of this process, find the probability that Jellyfish can empty the draw pile, modulo $1\,000\,000\,007$.

Formally, let $M=1\,000\,000\,007$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 120$) — the number of cards.

The second line of each test case contains $n$ integers $c_1,c_2,\ldots,c_n$ ($0 \leq c_i \leq n$) — the powers of the cards. It is guaranteed that $c_1 \leq c_2 \leq \ldots \leq c_n$.

The third line of each test case contains a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $120^2$.

Output: For each test case, output the probability that Jellyfish can empty the draw pile modulo $1\,000\,000\,007$.

Examples:
input:
4
5
0 1 1 1 2
00100
3
2 3 3
000
10
0 0 0 0 0 0 0 1 1 1
1111011111
20
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 2 3 3 4
00000000001000101010
output:
500000004
0
0
675898154

Note: In the first test case, Jellyfish will keep playing cards with power $1$ until Jellyfish draws a card with power $0$ or power $2$. If Jellyfish draws a card with power $0$, she will eventually empty her hand. If Jellyfish draws a card with power $2$, she will eventually empty the draw pile. Since there is an equal chance of drawing $0$ or $2$, the answer is $\frac{1}{2}$, and $2 \cdot 500\,000\,004 \equiv 1 \pmod {10^9+7}$

Buggy Code:
#include <bits/stdc++.h>
#define int long long int
#define pb push_back
#define F(i,c,n) for (int i=c; i<n; i++)

using namespace std;

const int INF = 1e18;

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

    int a[n];
    F(i, 0, n) cin>>a[i];

 string s;
 cin>>s;
}

int32_t main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int t = 1;
    cin >> t;

    while (t--) {
        solve();
        if(t==3) cout<<500000004<<endl;
        if(t==2 || t==1) cout<<0<<endl;
        if(t==0) cout<<675898154<<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
4
5
0 1 1 1 2
00100
3
2 3 3
000
10
0 0 0 0 0 0 0 1 1 1
1111011111
20
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 2 3 3 4
00000000001000101010
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mind Bloom
time_limit_ms: 5000
memory_limit_mb: 1024
Description: This is the way it always was.

This is the way it always will be.

All will be forgotten again soon...

Jellyfish is playing a one-player card game called "Slay the Spire". There are $n$ cards in total numbered from $1$ to $n$. The $i$-th card has power $c_i$.

There is a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

Jellyfish will repeat the following process until either her hand or the draw pile is empty.

  1. Let $x$ be the power of the card with the largest power in her hand.    2. Place a single card with power $x$ back into the draw pile.    3. Randomly draw $x$ cards from the draw pile. All subsets of $x$ cards from the draw pile have an equal chance of being drawn. If there are fewer than $x$ cards in the draw pile, Jellyfish will draw all cards.

At the end of this process, find the probability that Jellyfish can empty the draw pile, modulo $1\,000\,000\,007$.

Formally, let $M=1\,000\,000\,007$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 120$) — the number of cards.

The second line of each test case contains $n$ integers $c_1,c_2,\ldots,c_n$ ($0 \leq c_i \leq n$) — the powers of the cards. It is guaranteed that $c_1 \leq c_2 \leq \ldots \leq c_n$.

The third line of each test case contains a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $120^2$.

Output: For each test case, output the probability that Jellyfish can empty the draw pile modulo $1\,000\,000\,007$.

Examples:
input:
4
5
0 1 1 1 2
00100
3
2 3 3
000
10
0 0 0 0 0 0 0 1 1 1
1111011111
20
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 2 3 3 4
00000000001000101010
output:
500000004
0
0
675898154

Note: In the first test case, Jellyfish will keep playing cards with power $1$ until Jellyfish draws a card with power $0$ or power $2$. If Jellyfish draws a card with power $0$, she will eventually empty her hand. If Jellyfish draws a card with power $2$, she will eventually empty the draw pile. Since there is an equal chance of drawing $0$ or $2$, the answer is $\frac{1}{2}$, and $2 \cdot 500\,000\,004 \equiv 1 \pmod {10^9+7}$

Buggy Code:
// LUOGU_RID: 161272394
// what is matter? never mind.
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,sse4,popcnt,abm,mmx,avx,avx2")
#include<bits/stdc++.h>
#define For(i,a,b) for(int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(int i=(a);i>=(b);--i)
#define ll long long
#define int long long
#define ull unsigned long long
#define SZ(x) ((int)((x).size()))
#define ALL(x) (x).begin(),(x).end()
using namespace std;
inline int read()
{
	char c=getchar();int x=0;bool f=0;
	for(;!isdigit(c);c=getchar())f^=!(c^45);
	for(;isdigit(c);c=getchar())x=(x<<1)+(x<<3)+(c^48);
	if(f)x=-x;return x;
}

#define mod 1000000007
struct modint{
	int x;
	modint(int o=0){x=o;}
	modint &operator = (int o){return x=o,*this;}
	modint &operator +=(modint o){return x=x+o.x>=mod?x+o.x-mod:x+o.x,*this;}
	modint &operator -=(modint o){return x=x-o.x<0?x-o.x+mod:x-o.x,*this;}
	modint &operator *=(modint o){return x=1ll*x*o.x%mod,*this;}
	modint &operator ^=(int b){
		modint a=*this,c=1;
		for(;b;b>>=1,a*=a)if(b&1)c*=a;
		return x=c.x,*this;
	}
	modint &operator /=(modint o){return *this *=o^=mod-2;}
	friend modint operator +(modint a,modint b){return a+=b;}
	friend modint operator -(modint a,modint b){return a-=b;}
	friend modint operator *(modint a,modint b){return a*=b;}
	friend modint operator /(modint a,modint b){return a/=b;}
	friend modint operator ^(modint a,int b){return a^=b;}
	friend bool operator ==(modint a,modint b){return a.x==b.x;}
	friend bool operator !=(modint a,modint b){return a.x!=b.x;}
	bool operator ! () {return !x;}
	modint operator - () {return x?mod-x:0;}
	bool operator <(const modint&b)const{return x<b.x;}
};
inline modint qpow(modint x,int y){return x^y;}

vector<modint> fac,ifac,iv;
inline void initC(int n)
{
	if(iv.empty())fac=ifac=iv=vector<modint>(2,1);
	int m=iv.size(); ++n;
	if(m>=n)return;
	iv.resize(n),fac.resize(n),ifac.resize(n);
	For(i,m,n-1){
		iv[i]=iv[mod%i]*(mod-mod/i);
		fac[i]=fac[i-1]*i,ifac[i]=ifac[i-1]*iv[i];
	}
}
inline modint C(int n,int m){
	if(m<0||n<m)return 0;
	return initC(n),fac[n]*ifac[m]*ifac[n-m];
}
inline modint sign(int n){return (n&1)?(mod-1):(1);}

#define fi first
#define se second
#define pb push_back
#define mkp make_pair
typedef pair<int,int>pii;
typedef vector<int>vi;

#define maxn 125
#define inf 0x3f3f3f3f

int n,a[maxn];
char s[maxn];
int sc[maxn];

modint f[maxn][maxn][maxn],g[maxn][maxn];
modint res[maxn][maxn];

void work()
{
	n=read();
	For(i,1,n)a[i]=read();
	scanf("%s",s+1);
	int cnt=0;
	For(i,1,n) cnt+=(s[i]=='1'),sc[i]=sc[i-1]+(s[i]=='1');

	if(a[n]<=1){
		cout<<(cnt==n)<<"\n";
		return;
	}

	For(i,0,n)For(j,0,n)For(k,0,n)f[i][j][k]=0;
	Rep(i,n,1){
		if(a[i]==0)continue;
		Rep(x,i-1,0){
			// <i have x
			// after proceed, +=z.
			For(j,0,n)For(z,0,n) g[j][z]=0;
			g[n][a[i]]=1;
			Rep(j,n,i){
				Rep(z,n,a[i]){
					if(j>i || z>1){
						modint p=iv[j-x]*z;
						g[j-1][z]+=g[j][z]*(1-p);
						For(d,a[i],j-z)
							g[j-1][z-1+d]+=g[j][z]*p*f[j][x+z-1][d];
					}else{
						modint p=iv[j-x]*z;
						g[j-1][z]+=g[j][z]*(1-p);
						For(d,a[i],j-z){
							modint o=g[j][z]*p;
							g[j-1][z-1+d]/=(1-o);
						}
					}
				}
			}
			For(y,a[i],n) f[i][x][y]=g[i-1][y];
		}
	}

	For(i,0,n)For(j,0,n)res[i][j]=0;
	Rep(i,n,1){
		if(s[i]=='1'){
			res[i][cnt]=1;
			break;
		}
	}
	Rep(i,n,1){
	//	cout<<"i: "<<i<<"\n";
	//	For(j,0,i) cout<<res[i][j].x<<" "; cout<<"\n";
		if(a[i]==0){
			modint ans=0;
			For(j,0,i) ans+=res[i][j];
			ans=1-ans;
			cout<<ans.x<<"\n";
			return;
		}
		Rep(j,i,sc[i]){
			modint p=1;
			if(s[i]=='0') p=iv[i-sc[i]]*(j-sc[i]);
			res[i-1][j]+=res[i][j]*(1-p);
			For(k,a[i],n)
				if(j+k-1<=i-1) res[i-1][j-1+k]+=res[i][j]*p*f[i][j-1][k];
		}
	}
	cout<<0<<"\n";
}

signed main()
{
	initC(1005);
	int T=read();
	while(T--)work();
	return 0;
}
/*

*/

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the given code, we need to focus on edge cases or scenarios that are likely to cause issues. Here are the steps to identify such a test case:

1. **Understand the Code's Logic**: The code seems to handle the game logic by dynamically calculating probabilities using a DP approach. It considers the powers of cards and their initial positions (hand or draw pile).

2. **Identify Potential Bugs**:
   - **Handling of Zero Power Cards**: The code has a special case for cards with zero power.
   - **Modulo Arithmetic**: Ensure that modulo operations are handled correctly.
   - **Edge Cases**: Cards with the same power, cards with very high power, and the initial distribution of cards.

3. **Construct a Test Case**:
   - **Small Scale**: Use a small number of cards to keep the input manageable.
   - **Challenging Distribution**: Ensure that the distribution of cards and their powers are such that the code's logic is put to the test.

Given these considerations, let's construct a test case:

- **Number of Cards**: 5
- **Card Powers**: 0, 1, 1, 2, 2
- **Initial Distribution**: 0 0 1 0 1 (first card in draw pile, second card in draw pile, third card in hand, fourth card in draw pile, fifth card in hand)

This test case is designed to:
- Have cards with zero power to test the special handling.
- Have multiple cards with the same power to test the probability calculations.
- Ensure that the hand and draw pile are not empty at the start.

Here is the test input:

```plaintext
1
5
0 1 1 2 2
00101
```

This input will challenge the code to correctly handle the special case for zero power cards and the dynamic probability calculations for cards with the same power.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mind Bloom
time_limit_ms: 5000
memory_limit_mb: 1024
Description: This is the way it always was.

This is the way it always will be.

All will be forgotten again soon...

Jellyfish is playing a one-player card game called "Slay the Spire". There are $n$ cards in total numbered from $1$ to $n$. The $i$-th card has power $c_i$.

There is a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

Jellyfish will repeat the following process until either her hand or the draw pile is empty.

  1. Let $x$ be the power of the card with the largest power in her hand.    2. Place a single card with power $x$ back into the draw pile.    3. Randomly draw $x$ cards from the draw pile. All subsets of $x$ cards from the draw pile have an equal chance of being drawn. If there are fewer than $x$ cards in the draw pile, Jellyfish will draw all cards.

At the end of this process, find the probability that Jellyfish can empty the draw pile, modulo $1\,000\,000\,007$.

Formally, let $M=1\,000\,000\,007$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 120$) — the number of cards.

The second line of each test case contains $n$ integers $c_1,c_2,\ldots,c_n$ ($0 \leq c_i \leq n$) — the powers of the cards. It is guaranteed that $c_1 \leq c_2 \leq \ldots \leq c_n$.

The third line of each test case contains a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $120^2$.

Output: For each test case, output the probability that Jellyfish can empty the draw pile modulo $1\,000\,000\,007$.

Examples:
input:
4
5
0 1 1 1 2
00100
3
2 3 3
000
10
0 0 0 0 0 0 0 1 1 1
1111011111
20
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 2 3 3 4
00000000001000101010
output:
500000004
0
0
675898154

Note: In the first test case, Jellyfish will keep playing cards with power $1$ until Jellyfish draws a card with power $0$ or power $2$. If Jellyfish draws a card with power $0$, she will eventually empty her hand. If Jellyfish draws a card with power $2$, she will eventually empty the draw pile. Since there is an equal chance of drawing $0$ or $2$, the answer is $\frac{1}{2}$, and $2 \cdot 500\,000\,004 \equiv 1 \pmod {10^9+7}$

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
const int N=122;
const int MOD=1e9+7;
int S[N][N][N][N];
long long inv[N];
long long f[N][N][N],d[N][N];
int ss[N];
int a[N];
bool hav[N];
long long C[N*2][N*2];
long long qpow(long long x,long long y,long long mod){
	if(!y)return 1%mod;
	if(y==1)return x%mod;
	long long kp=qpow(x,y/2,mod);
	if(y&1)return kp*kp%mod*x%mod;
	return kp*kp%mod;
}
int main(){
	for(int i=1;i<=120;i++)inv[i]=qpow(i,MOD-2,MOD);
	C[0][0]=1;
	for(int i=1;i<=240;i++){
		C[i][0]=1;
		for(int j=1;j<=i;j++)C[i][j]=(C[i-1][j-1]+C[i-1][j])%MOD;
	}
	int T;
	cin>>T;
	while(T--){
		int n;
		scanf("%d",&n);
		for(int i=1;i<=n;i++)scanf("%d",&a[i]);
		string s;
		cin>>s;
		int cnt=0;
		for(int i=0;i<n;i++){
			hav[i+1]=s[i]-'0';
			ss[i+1]=ss[i]+hav[i+1];
		}
		for(int i=1;i<=n;i++)cnt+=hav[i];
		if(a[n]<=1){
			if(cnt==n)cout<<1<<"\n";
			else cout<<0<<"\n";
			continue;
		}
		if(!cnt){
			cout<<0<<"\n";
			continue;
		}
		if(a[1]>0){
			cout<<1<<"\n";
			continue;
		}
		for(int l=1;l<=n;l++){
			for(int r=l-1;r<=n;r++){
				for(int sum=n;sum>=0;sum--){
					for(int c=0;c<=n;c++)S[l][r][sum][c]=0;
				}
			}
		}
		for(int l=1;l<=n;l++){
				for(int sum=n;sum>=0;sum--){
					for(int c=0;c<=n;c++)f[l][sum][c]=0;
				}
		}
		for(int l=n;l>=1;l--){
			if(a[l]<=0)continue;
			for(int sum=l-1;sum>=0;sum--){
				S[l][n][sum][a[l]]=1;
				for(int r=n;r>=l;r--){
					for(int c=n;c>=a[l];c--){
						if(sum+c>r)continue;
						if(l==r&&c==1){
							/*if(S[l][r][sum][c]){
								cout<<l<<" "<<r<<" "<<sum<<" "<<c<<":"<<S[l][r][sum][c]<<"\n";
							}*/
							S[l][r-1][sum][c]=(S[l][r-1][sum][c]*1ll+(r-sum-c)*inv[r-sum]%MOD*S[l][r][sum][c])%MOD;
							for(int x=a[r];x<=n;x++){
								S[l][r-1][sum][c-1+x]=S[l][r-1][sum][c-1+x]*1ll*qpow((MOD+1-c*inv[r-sum]%MOD*S[l][r][sum][c]%MOD)%MOD,MOD-2,MOD)%MOD;
							}
						}
						else{
							for(int x=a[r];x<=n;x++){
								S[l][r-1][sum][c-1+x]=(S[l][r-1][sum][c-1+x]*1ll+f[r][sum+c-1][x]*c%MOD*inv[r-sum]%MOD*S[l][r][sum][c]%MOD)%MOD;
							}
							/*if(S[l][r][sum][c]){
								cout<<l<<" "<<r<<" "<<sum<<" "<<c<<":"<<S[l][r][sum][c]<<"\n";
							}*/
							S[l][r-1][sum][c]=(S[l][r-1][sum][c]*1ll+(r-sum-c)*inv[r-sum]%MOD*S[l][r][sum][c])%MOD;
						}
					}
				}
				for(int c=a[l];c<=n;c++){
					f[l][sum][c]=S[l][l-1][sum][c];
					//if(f[l][sum][c])cout<<l<<" "<<sum<<" "<<c<<":"<<f[l][sum][c]<<"\n";
				}
			}
		}
		/*for(int i=1;i<=n;i++){
			for(int j=0;j<=i-1;j++){
				for(int k=0;k<=i-1;k++){
					if(f[i][j][k])cout<<i<<" "<<j<<" "<<k<<":"<<f[i][j][k]<<"\n";
				}
			}
		}*/
		for(int i=0;i<=n;i++){
			for(int j=0;j<=n;j++)d[i][j]=0;
		}
		for(int i=n;i>=1;i--){
			if(hav[i]){
				d[cnt][i]=1;
				break;
			}
		}
		long long ans=0;
		for(int x=n;x>=1;x--){
			if(a[x]==0){
				for(int sum=x;sum>=0;sum--){
					ans+=d[sum][x];
					ans%=MOD;
				}
				break;
			}
			for(int sum=x;sum>=ss[x];sum--){
				//have
				long long prob;
				if(hav[x])prob=1;
				else{
					prob=(sum-ss[x])*inv[(x-ss[x])]%MOD;
				}
				for(int t=a[x];t<=n;t++){
					if(sum+t>x)break;
					d[sum-1+t][x-1]=(d[sum-1+t][x-1]+d[sum][x]*prob%MOD*f[x][sum-1][t])%MOD;
				}
				if(sum<=x-1)d[sum][x-1]=(d[sum][x-1]+d[sum][x]*(MOD+1-prob))%MOD;
			}
		}
		cout<<(MOD+1-ans)%MOD<<"\n";
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
5
0 1 1 1 2
00100
3
2 3 3
000
10
0 0 0 0 0 0 0 1 1 1
1111011111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mind Bloom
time_limit_ms: 5000
memory_limit_mb: 1024
Description: This is the way it always was.

This is the way it always will be.

All will be forgotten again soon...

Jellyfish is playing a one-player card game called "Slay the Spire". There are $n$ cards in total numbered from $1$ to $n$. The $i$-th card has power $c_i$.

There is a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

Jellyfish will repeat the following process until either her hand or the draw pile is empty.

  1. Let $x$ be the power of the card with the largest power in her hand.    2. Place a single card with power $x$ back into the draw pile.    3. Randomly draw $x$ cards from the draw pile. All subsets of $x$ cards from the draw pile have an equal chance of being drawn. If there are fewer than $x$ cards in the draw pile, Jellyfish will draw all cards.

At the end of this process, find the probability that Jellyfish can empty the draw pile, modulo $1\,000\,000\,007$.

Formally, let $M=1\,000\,000\,007$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 120$) — the number of cards.

The second line of each test case contains $n$ integers $c_1,c_2,\ldots,c_n$ ($0 \leq c_i \leq n$) — the powers of the cards. It is guaranteed that $c_1 \leq c_2 \leq \ldots \leq c_n$.

The third line of each test case contains a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $120^2$.

Output: For each test case, output the probability that Jellyfish can empty the draw pile modulo $1\,000\,000\,007$.

Examples:
input:
4
5
0 1 1 1 2
00100
3
2 3 3
000
10
0 0 0 0 0 0 0 1 1 1
1111011111
20
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 2 3 3 4
00000000001000101010
output:
500000004
0
0
675898154

Note: In the first test case, Jellyfish will keep playing cards with power $1$ until Jellyfish draws a card with power $0$ or power $2$. If Jellyfish draws a card with power $0$, she will eventually empty her hand. If Jellyfish draws a card with power $2$, she will eventually empty the draw pile. Since there is an equal chance of drawing $0$ or $2$, the answer is $\frac{1}{2}$, and $2 \cdot 500\,000\,004 \equiv 1 \pmod {10^9+7}$

Buggy Code:
#include<bits/stdc++.h>
#define ci const int
#define ll long long
#define ull unsigned long long
using namespace std;
ci mod=1e9+7;
inline void add(int &x,ci v){
	x+=v,x-=x<mod?0:mod;
}
inline void sub(int &x,ci v){
	x-=v,x+=x<0?mod:0;
}
ll qk(ll x,int y=mod-2){
	ll ans=1;
	while(y){
		if(y&1)ans=ans*x%mod;
		x=x*x%mod,y>>=1;
	}
	return ans;
}
ci N=250;
const ull H=1e9+7;
int n,fac[N],dfac[N],inv[N],C[N][N],iC[N][N];
int m,lim,cnt[N],c[N],rk[N],pre[N];
unsigned short val[N];
struct st{
	unsigned short res,sum;
	vector<unsigned short>num;
	st(unsigned short res=0,unsigned short sum=0,vector<unsigned short>num={}):res(res),sum(sum),num(num){}
	ull hs(){
		ull ans=res+1;
		for(unsigned short x:num)ans=ans*H+x+1;
		return ans;
	}
	void print(){
		puts("details");
		printf("res=%d\n",res);
		printf("num ");for(int x:num)printf("%d ",x);
		puts("");
	}
};
unordered_map<ull,int>mp;
int B;
int dfs(st &s){
	unsigned short res=s.res,all=s.sum,las=0;
	for(int i=0;i<m&&res&&all<=lim;++i){
		ci gt=min(res,s.num[i]);
		all+=gt*max((unsigned short)1,val[i]);
		res-=gt;
		if(!res)las=val[i];
	}
	if(all>lim)return 1;
	const ull hs=s.hs();
	if(mp.find(hs)!=mp.end())return mp[hs];
	if(s.res==0){
		int mx=-1;
		for(int i=m-1;~i;--i)
			if(s.num[i]!=cnt[i]){
				mx=i;
				break;
			}
		// printf("mx=%d\n",mx);
		if(mx==-1)return 0;
		else if(val[mx]==0)return 0;
		else if(val[mx]==1){
			st t=s;
			++t.num[mx],t.sum-=max(val[mx],(unsigned short)1);
			int sum=0,ans=0;
			for(int i=0;i<m;++i)
				if(val[i]!=1&&s.num[i]){
					sum+=s.num[i];
					--t.num[i],t.sum+=max(val[i],(unsigned short)1);
					add(ans,(ll)s.num[i]*dfs(t)%mod);
					++t.num[i],t.sum-=max(val[i],(unsigned short)1);
				}
			return mp[hs]=(ll)ans*inv[sum]%mod;
		}
		else{
			st t=s;
			++t.num[mx],t.sum-=max(val[mx],(unsigned short)1);
			t.res=val[mx];
			return mp[hs]=dfs(t);
		}
	}
	else{
		st t=s;t.res=s.res-1;
		int ok=0,ans=0;
		for(int i=0;i<m;++i)ok+=t.num[i];
		int M=m;
		for(int i=0;i<m;++i)
			if(all-max(las,(unsigned short)1)+max(val[i],(unsigned short)1)>lim){
				M=i;
				break;
			}
		for(int i=M;i<m;++i)add(ans,s.num[i]);
		for(int i=0;i<M;++i)
			if(t.num[i]){
				--t.num[i],t.sum+=max(val[i],(unsigned short)1);
				int gt=0;
				// if(all-max(las,(unsigned short)1)+max(val[i],(unsigned short)1)>lim)gt=1;
				// else
				gt=dfs(t);
				add(ans,(ll)gt*s.num[i]%mod);
				++t.num[i],t.sum-=max(val[i],(unsigned short)1);
			}
		return mp[hs]=(ll)ans*inv[ok]%mod;
	}
}
void Init(){
	m=lim=0;
	mp.clear();
	memset(rk,0,sizeof(rk));
	memset(val,0,sizeof(val));
	memset(c,0,sizeof(c));
	memset(cnt,0,sizeof(cnt));
	memset(pre,0,sizeof(pre));
}
void Work(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i)scanf("%d",&c[i]),lim+=c[i]==0;
	for(int i=1;i<=n;++i){
		int j=i;
		while(j+1<=n&&c[i]==c[j+1])++j;
		rk[c[i]]=m,val[m]=c[i],cnt[m]=j-i+1,++m,i=j;
	}
	// printf("m=%d\n",m);
	st s;
	s.num.resize(m);
	s.res=0;
	int h=0;
	for(int i=1;i<=n;++i){
		char ch=getchar();
		while(ch!='0'&&ch!='1')ch=getchar();
		if(ch=='1')++s.num[rk[c[i]]],++h,s.sum+=max(c[i],1);
	}
	for(int i=0;i<m;++i)s.num[i]=cnt[i]-s.num[i];
	if(h==n)puts("1");
	else if(c[n]<=1)puts("0");
	else printf("%d\n",dfs(s));
}
int main(){
	int S=clock();
	fac[0]=fac[1]=dfac[0]=dfac[1]=inv[1]=1;
	for(int i=2;i<N;++i)
		fac[i]=(ll)fac[i-1]*i%mod,
		inv[i]=(ll)(mod-mod/i)*inv[mod%i]%mod,
		dfac[i]=(ll)dfac[i-1]*inv[i]%mod;
	for(int i=0;i<N;++i){
		C[i][0]=1;
		for(int j=1;j<=i;++j)C[i][j]=(C[i-1][j]+C[i-1][j-1])%mod;
	}
	for(int i=0;i<N;++i)for(int j=0;j<=i;++j)iC[i][j]=qk(C[i][j]);
	int T;scanf("%d",&T);
	mp.reserve(1<<25),mp.max_load_factor(0.50);
	while(T--){
		Work();
		if(T)Init();
	}
	return 0;
}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the given code, we need to focus on edge cases and scenarios that can maximize the failure probability. The bug in the code might be related to how it handles the distribution of cards and the probabilities of drawing cards from the draw pile. Here are the steps to create a small-scale test input:

1. **Small Number of Cards**: Use a small number of cards to ensure the test case is manageable and can be easily verified.
2. **Specific Power Distribution**: Use a specific power distribution that can highlight the bug, such as having cards with the same power or a specific sequence of powers.
3. **Initial Card Distribution**: Ensure the initial distribution of cards between the hand and the draw pile is such that it can expose the bug.

Let's create a test input with the following properties:
- **Number of Cards**: 4
- **Powers**: 0, 1, 1, 2
- **Initial Distribution**: 0 in the hand, 1, 1, 2 in the draw pile

This test case is designed to check how the algorithm handles the scenario where the highest power card in the hand is 1, and the draw pile contains a card with power 2.

```plaintext
1
4
0 1 1 2
0111
```

Explanation:
- **Number of Test Cases**: 1
- **Number of Cards**: 4
- **Powers**: 0, 1, 1, 2
- **Initial Distribution**: 0 in the hand, 1, 1, 2 in the draw pile

This test case should help expose any issues in the algorithm's handling of card powers and the probabilities of drawing cards from the draw pile.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mind Bloom
time_limit_ms: 5000
memory_limit_mb: 1024
Description: This is the way it always was.

This is the way it always will be.

All will be forgotten again soon...

Jellyfish is playing a one-player card game called "Slay the Spire". There are $n$ cards in total numbered from $1$ to $n$. The $i$-th card has power $c_i$.

There is a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

Jellyfish will repeat the following process until either her hand or the draw pile is empty.

  1. Let $x$ be the power of the card with the largest power in her hand.    2. Place a single card with power $x$ back into the draw pile.    3. Randomly draw $x$ cards from the draw pile. All subsets of $x$ cards from the draw pile have an equal chance of being drawn. If there are fewer than $x$ cards in the draw pile, Jellyfish will draw all cards.

At the end of this process, find the probability that Jellyfish can empty the draw pile, modulo $1\,000\,000\,007$.

Formally, let $M=1\,000\,000\,007$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 120$) — the number of cards.

The second line of each test case contains $n$ integers $c_1,c_2,\ldots,c_n$ ($0 \leq c_i \leq n$) — the powers of the cards. It is guaranteed that $c_1 \leq c_2 \leq \ldots \leq c_n$.

The third line of each test case contains a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $120^2$.

Output: For each test case, output the probability that Jellyfish can empty the draw pile modulo $1\,000\,000\,007$.

Examples:
input:
4
5
0 1 1 1 2
00100
3
2 3 3
000
10
0 0 0 0 0 0 0 1 1 1
1111011111
20
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 2 3 3 4
00000000001000101010
output:
500000004
0
0
675898154

Note: In the first test case, Jellyfish will keep playing cards with power $1$ until Jellyfish draws a card with power $0$ or power $2$. If Jellyfish draws a card with power $0$, she will eventually empty her hand. If Jellyfish draws a card with power $2$, she will eventually empty the draw pile. Since there is an equal chance of drawing $0$ or $2$, the answer is $\frac{1}{2}$, and $2 \cdot 500\,000\,004 \equiv 1 \pmod {10^9+7}$

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

typedef long long ll;
const ll mod = 1e9 + 7;

char s[125];
int c[125], pre[125];
ll f[125][125][125], g[125][125], C[125], ans[125][125];

ll inv(ll a)
{
    ll b = mod - 2, ans = 1;
    while (b)
    {
        if (b & 1) ans = ans * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ans;
}

void work()
{
    int n; cin >> n;
    for (int i = 1; i <= n; i++) cin >> c[i];
    cin >> s + 1; for (int i = 1; i <= n; i++) pre[i] = pre[i - 1] + (s[i] == '1');

    if (c[n] <= 1) return cout << (pre[n] == n ? 1 : 0) << "\n", void();

    for (int i = 1; i <= n; i++) for (int x = pre[i]; x <= i; x++) ans[i][x] = 0;
    for (int i = 1; i <= n; i++) for (int x = 0; x < i; x++) for (int y = 0; y < i; y++) f[i][x][y] = 0;
    for (int i = n; i; i--)
        for (int x = i - 1; ~x; x--)
        {
            for (int z = x; z < i; z++) C[z] = 0;
            for (int j = i - 1; j <= n; j++) for (int y = x; y <= j; y++) g[j][y] = 0;
            g[n][x + c[i]] = 1;
            for (int j = n; j >= i; j--)
                for (int y = x; y <= j; y++)
                {
                    ll p = (y - x) * inv(j - x) % mod;
                    (g[j - 1][y] += g[j][y] * (1 + mod - p)) %= mod;
                    for (int z = x; z < j; z++)
                    {
                        if (j == i && y == x + 1) (C[z] += g[j][y] * p) %= mod;
                        else (g[j - 1][z] += g[j][y] * p % mod * f[j][y - 1][z]) %= mod;
                    }
                }
            for (int y = x; y < i; y++) f[i][x][y] = g[i - 1][y] * inv(1 + mod - C[y]) % mod;
        }

    ans[0][0] = 1;
    for (int i = 1; i <= n; i++)
        for (int x = pre[i]; x <= i; x++)
        {
            if (s[i] == '1') for (int y = pre[i - 1]; y < i; y++) (ans[i][x] += ans[i - 1][y] * f[i][x - 1][y]) %= mod;
            else
            {
                ll p = (x - pre[i]) * inv(i - pre[i]) % mod;
                ans[i][x] = (1 + mod - p) * ans[i - 1][x] % mod;
                for (int y = pre[i - 1]; y < i; y++) (ans[i][x] += p * ans[i - 1][y] % mod * f[i][x - 1][y]) %= mod;
            }
        }

    cout << (1 + mod - ans[n][pre[n]]) % mod << "\n";
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);

    int T; cin >> T;
    while (T--) work();
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
5
1 1 1 2 2
00011
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mind Bloom
time_limit_ms: 5000
memory_limit_mb: 1024
Description: This is the way it always was.

This is the way it always will be.

All will be forgotten again soon...

Jellyfish is playing a one-player card game called "Slay the Spire". There are $n$ cards in total numbered from $1$ to $n$. The $i$-th card has power $c_i$.

There is a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

Jellyfish will repeat the following process until either her hand or the draw pile is empty.

  1. Let $x$ be the power of the card with the largest power in her hand.    2. Place a single card with power $x$ back into the draw pile.    3. Randomly draw $x$ cards from the draw pile. All subsets of $x$ cards from the draw pile have an equal chance of being drawn. If there are fewer than $x$ cards in the draw pile, Jellyfish will draw all cards.

At the end of this process, find the probability that Jellyfish can empty the draw pile, modulo $1\,000\,000\,007$.

Formally, let $M=1\,000\,000\,007$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 120$) — the number of cards.

The second line of each test case contains $n$ integers $c_1,c_2,\ldots,c_n$ ($0 \leq c_i \leq n$) — the powers of the cards. It is guaranteed that $c_1 \leq c_2 \leq \ldots \leq c_n$.

The third line of each test case contains a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $120^2$.

Output: For each test case, output the probability that Jellyfish can empty the draw pile modulo $1\,000\,000\,007$.

Examples:
input:
4
5
0 1 1 1 2
00100
3
2 3 3
000
10
0 0 0 0 0 0 0 1 1 1
1111011111
20
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 2 3 3 4
00000000001000101010
output:
500000004
0
0
675898154

Note: In the first test case, Jellyfish will keep playing cards with power $1$ until Jellyfish draws a card with power $0$ or power $2$. If Jellyfish draws a card with power $0$, she will eventually empty her hand. If Jellyfish draws a card with power $2$, she will eventually empty the draw pile. Since there is an equal chance of drawing $0$ or $2$, the answer is $\frac{1}{2}$, and $2 \cdot 500\,000\,004 \equiv 1 \pmod {10^9+7}$

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
const int N=122;
const int MOD=1e9+7;
int S[N][N][N][N];
long long inv[N];
long long f[N][N][N],d[N][N];
int ss[N];
int a[N];
bool hav[N];
long long C[N*2][N*2];
long long qpow(long long x,long long y,long long mod){
	if(!y)return 1%mod;
	if(y==1)return x%mod;
	long long kp=qpow(x,y/2,mod);
	if(y&1)return kp*kp%mod*x%mod;
	return kp*kp%mod;
}
int main(){
	for(int i=1;i<=120;i++)inv[i]=qpow(i,MOD-2,MOD);
	C[0][0]=1;
	for(int i=1;i<=240;i++){
		C[i][0]=1;
		for(int j=1;j<=i;j++)C[i][j]=(C[i-1][j-1]+C[i-1][j])%MOD;
	}
	int T;
	cin>>T;
	while(T--){
		int n;
		scanf("%d",&n);
		for(int i=1;i<=n;i++)scanf("%d",&a[i]);
		string s;
		cin>>s;
		int cnt=0,sumc=0;
		for(int i=0;i<n;i++){
			hav[i+1]=s[i]-'0';
			ss[i+1]=ss[i]+hav[i+1];
		}
		for(int i=1;i<=n;i++)cnt+=hav[i],sumc+=hav[i]*a[i];
		if(a[n]<=1){
			if(cnt==n)cout<<1<<"\n";
			else cout<<0<<"\n";
			continue;
		}
		if(!cnt){
			cout<<0<<"\n";
			continue;
		}
		if(a[1]>0){
			cout<<1<<"\n";
			continue;
		}
		if(sumc==0){
			cout<<0<<"\n";
			continue;
		}
		for(int l=1;l<=n;l++){
			for(int r=l-1;r<=n;r++){
				for(int sum=n;sum>=0;sum--){
					for(int c=0;c<=n;c++)S[l][r][sum][c]=0;
				}
			}
		}
		for(int l=1;l<=n;l++){
				for(int sum=n;sum>=0;sum--){
					for(int c=0;c<=n;c++)f[l][sum][c]=0;
				}
		}
		for(int l=n;l>=1;l--){
			if(a[l]<=0)continue;
			for(int sum=l-1;sum>=0;sum--){
				S[l][n][sum][a[l]]=1;
				for(int r=n;r>=l;r--){
					for(int c=n;c>=a[l];c--){
						if(sum+c>r)continue;
						if(l==r&&c==1){
							/*if(S[l][r][sum][c]){
								cout<<l<<" "<<r<<" "<<sum<<" "<<c<<":"<<S[l][r][sum][c]<<"\n";
							}*/
							S[l][r-1][sum][c]=(S[l][r-1][sum][c]*1ll+(r-sum-c)*inv[r-sum]%MOD*S[l][r][sum][c])%MOD;
							for(int x=a[r];x<=n;x++){
								S[l][r-1][sum][c-1+x]=S[l][r-1][sum][c-1+x]*1ll*qpow((MOD+1-c*inv[r-sum]%MOD*S[l][r][sum][c]%MOD)%MOD,MOD-2,MOD)%MOD;
							}
						}
						else{
							for(int x=a[r];x<=n;x++){
								S[l][r-1][sum][c-1+x]=(S[l][r-1][sum][c-1+x]*1ll+f[r][sum+c-1][x]*c%MOD*inv[r-sum]%MOD*S[l][r][sum][c]%MOD)%MOD;
							}
							/*if(S[l][r][sum][c]){
								cout<<l<<" "<<r<<" "<<sum<<" "<<c<<":"<<S[l][r][sum][c]<<"\n";
							}*/
							S[l][r-1][sum][c]=(S[l][r-1][sum][c]*1ll+(r-sum-c)*inv[r-sum]%MOD*S[l][r][sum][c])%MOD;
						}
					}
				}
				for(int c=a[l];c<=n;c++){
					f[l][sum][c]=S[l][l-1][sum][c];
					//if(f[l][sum][c])cout<<l<<" "<<sum<<" "<<c<<":"<<f[l][sum][c]<<"\n";
				}
			}
		}
		/*for(int i=1;i<=n;i++){
			for(int j=0;j<=i-1;j++){
				for(int k=0;k<=i-1;k++){
					if(f[i][j][k])cout<<i<<" "<<j<<" "<<k<<":"<<f[i][j][k]<<"\n";
				}
			}
		}*/
		for(int i=0;i<=n;i++){
			for(int j=0;j<=n;j++)d[i][j]=0;
		}
		for(int i=n;i>=1;i--){
			if(hav[i]){
				d[cnt][i]=1;
				break;
			}
		}
		long long ans=0;
		for(int x=n;x>=1;x--){
			if(a[x]==0){
				for(int sum=x;sum>=0;sum--){
					ans+=d[sum][x];
					ans%=MOD;
				}
				break;
			}
			for(int sum=x;sum>=ss[x];sum--){
				//have
				long long prob;
				if(hav[x])prob=1;
				else{
					prob=(sum-ss[x])*inv[(x-ss[x])]%MOD;
				}
				for(int t=a[x];t<=n;t++){
					if(sum+t>x)break;
					d[sum-1+t][x-1]=(d[sum-1+t][x-1]+d[sum][x]*prob%MOD*f[x][sum-1][t])%MOD;
				}
				if(sum<=x-1)d[sum][x-1]=(d[sum][x-1]+d[sum][x]*(MOD+1-prob))%MOD;
			}
		}
		cout<<(MOD+1-ans)%MOD<<"\n";
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
5
0 1 1 1 2
00100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mind Bloom
time_limit_ms: 5000
memory_limit_mb: 1024
Description: This is the way it always was.

This is the way it always will be.

All will be forgotten again soon...

Jellyfish is playing a one-player card game called "Slay the Spire". There are $n$ cards in total numbered from $1$ to $n$. The $i$-th card has power $c_i$.

There is a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

Jellyfish will repeat the following process until either her hand or the draw pile is empty.

  1. Let $x$ be the power of the card with the largest power in her hand.    2. Place a single card with power $x$ back into the draw pile.    3. Randomly draw $x$ cards from the draw pile. All subsets of $x$ cards from the draw pile have an equal chance of being drawn. If there are fewer than $x$ cards in the draw pile, Jellyfish will draw all cards.

At the end of this process, find the probability that Jellyfish can empty the draw pile, modulo $1\,000\,000\,007$.

Formally, let $M=1\,000\,000\,007$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 120$) — the number of cards.

The second line of each test case contains $n$ integers $c_1,c_2,\ldots,c_n$ ($0 \leq c_i \leq n$) — the powers of the cards. It is guaranteed that $c_1 \leq c_2 \leq \ldots \leq c_n$.

The third line of each test case contains a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $120^2$.

Output: For each test case, output the probability that Jellyfish can empty the draw pile modulo $1\,000\,000\,007$.

Examples:
input:
4
5
0 1 1 1 2
00100
3
2 3 3
000
10
0 0 0 0 0 0 0 1 1 1
1111011111
20
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 2 3 3 4
00000000001000101010
output:
500000004
0
0
675898154

Note: In the first test case, Jellyfish will keep playing cards with power $1$ until Jellyfish draws a card with power $0$ or power $2$. If Jellyfish draws a card with power $0$, she will eventually empty her hand. If Jellyfish draws a card with power $2$, she will eventually empty the draw pile. Since there is an equal chance of drawing $0$ or $2$, the answer is $\frac{1}{2}$, and $2 \cdot 500\,000\,004 \equiv 1 \pmod {10^9+7}$

Buggy Code:
// this is not mine

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

struct hash_for_hash_map {
    size_t operator()(size_t x) const {
        static const size_t D = std::chrono::steady_clock::now().time_since_epoch().count();
        return x * x + D;
    }
};

template<typename key_t, typename value_t, typename hash_t = hash_for_hash_map>
using hash_map = __gnu_pbds::gp_hash_table<key_t, value_t, hash_t>;

using namespace std;

#define ull unsigned long long
#define ll long long
#define flt double
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()

const ll mod = 1e9 + 7;

ll bin_pow(ll x, ll n) {
    ll y = 1;
    for (; n > 0; n >>= 1, x = x * x % mod) {
        if (n & 1) {
            y = y * x % mod;
        }
    }
    return y;
}

ll rev(ll x) {
    return bin_pow(x, mod - 2);
}

int n;
vector<int> aboba;
vector<int> init_cnt;
vector<int> cnt;

struct state {
    vector<int> my_cnt;

    ull hsh() const {
        ull h = 0;
        ull p = 23904857203;
        for (auto x : my_cnt) {
            h = h * p + (x + 1);
        }
        return h;
    }
};

ll c[1000][1000];

void prec() {
    for (int i = 0; i < 1000; i++) {
        c[i][0] = c[i][i] = 1;
        for (int j = 1; j < i; j++) {
            c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
        }
    }
}

hash_map<ll, ll> dp;

ll get(const state& s);

ll get_rec(state& s, int rest_cnt, int i, ll ways, ll sm) {
    if (sm > cnt[0] - s.my_cnt[0]) {
        return 0;
    }
    if (rest_cnt == 0) {
        ll to = get(s);
        return to * ways % mod;
    }
    if (i >= (int)s.my_cnt.size() || (sm + i * rest_cnt) > cnt[0] - s.my_cnt[0]) {
        return 0;
    }
    ll rs = 0;
    for (int x = 0; x <= cnt[i] - s.my_cnt[i]; x += 1) {
        int new_rest = rest_cnt - x;
        if (new_rest < 0) {
            break;
        }
        ll new_sm = sm + x * i;
        if (new_sm > cnt[0] - s.my_cnt[0]) {
            break;
        }
        ll new_ways = ways * c[cnt[i] - s.my_cnt[i]][x] % mod;
        s.my_cnt[i] += x;
        rs += get_rec(s, new_rest, i + 1, new_ways, new_sm);
        s.my_cnt[i] -= x;
    }
    return rs % mod;
}

ll get(const state& s) {
    ll sm = 0;
    ll acc = 0;
    for (int i = 0; i < (int)s.my_cnt.size(); i += 1) {
        sm += s.my_cnt[i] * i;
        acc += s.my_cnt[i];
    }
    if (acc == n) {
        return 0;
    }
    int mx = (int)s.my_cnt.size() - 1;
    while (mx > 0 and s.my_cnt[mx] == 0) {
        mx -= 1;
    }
    if (mx <= 1 and cnt[0] + cnt[1] == n) {
        return 1;
    }
    ll rest_zero = cnt[0] - s.my_cnt[0];
    if (rest_zero < sm) {
        return 0;
    }
    ull h = s.hsh();
    auto it = dp.find(h);
    if (it != dp.end()) {
        return it->second;
    }
    dp[h] = 0;
    if (mx == 0) {
        return dp[h] = 1;
    }
    auto clone = s;
    clone.my_cnt[mx] -= 1;
    ll to_take = min<ll>(mx, n - acc + 1);
    ll rs = get_rec(clone, to_take, 0, 1, sm - mx);
    ll states = c[n - acc + 1][to_take];
    if (mx == 1) {
        states -= (cnt[1] - clone.my_cnt[1]);
    }
    ll prob = rev(states);
    return dp[h] = (rs * prob % mod);
}

int32_t main() {
    prec();
    if (1) {
        ios::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);
    }

    // dp.resize(3'000'00);

    int t;
    cin >> t;
    for (int itr = 0; itr < t; itr += 1) {
        dp.clear();
        cin >> n;
        aboba.resize(n);
        cnt.assign(n + 1, 0);
        for (int i = 0; i < n; i += 1) {
            cin >> aboba[i];
            cnt[aboba[i]] += 1;
        }
        init_cnt.assign(n + 1, 0);
        string s;
        cin >> s;
        for (int i = 0; i < n; i += 1) {
            if (s[i] == '1') {
                init_cnt[aboba[i]] += 1;
            }
        }
        if (init_cnt[0] == n) {
            cout << "1\n";
            continue;
        }
        auto init_state = state{ init_cnt };
        ll rs = get(init_state);
        cout << (1 - rs + mod) % mod << "\n";
        // cerr << dp.size() << '\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
4
5
1 1 1 1 2
00001
3
1 1 2
001
10
0 0 0 0 0 1 1 1 1 2
0000011111
20
0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 2 2 3 3 4
000000000000011111111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mind Bloom
time_limit_ms: 5000
memory_limit_mb: 1024
Description: This is the way it always was.

This is the way it always will be.

All will be forgotten again soon...

Jellyfish is playing a one-player card game called "Slay the Spire". There are $n$ cards in total numbered from $1$ to $n$. The $i$-th card has power $c_i$.

There is a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

Jellyfish will repeat the following process until either her hand or the draw pile is empty.

  1. Let $x$ be the power of the card with the largest power in her hand.    2. Place a single card with power $x$ back into the draw pile.    3. Randomly draw $x$ cards from the draw pile. All subsets of $x$ cards from the draw pile have an equal chance of being drawn. If there are fewer than $x$ cards in the draw pile, Jellyfish will draw all cards.

At the end of this process, find the probability that Jellyfish can empty the draw pile, modulo $1\,000\,000\,007$.

Formally, let $M=1\,000\,000\,007$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 120$) — the number of cards.

The second line of each test case contains $n$ integers $c_1,c_2,\ldots,c_n$ ($0 \leq c_i \leq n$) — the powers of the cards. It is guaranteed that $c_1 \leq c_2 \leq \ldots \leq c_n$.

The third line of each test case contains a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $120^2$.

Output: For each test case, output the probability that Jellyfish can empty the draw pile modulo $1\,000\,000\,007$.

Examples:
input:
4
5
0 1 1 1 2
00100
3
2 3 3
000
10
0 0 0 0 0 0 0 1 1 1
1111011111
20
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 2 3 3 4
00000000001000101010
output:
500000004
0
0
675898154

Note: In the first test case, Jellyfish will keep playing cards with power $1$ until Jellyfish draws a card with power $0$ or power $2$. If Jellyfish draws a card with power $0$, she will eventually empty her hand. If Jellyfish draws a card with power $2$, she will eventually empty the draw pile. Since there is an equal chance of drawing $0$ or $2$, the answer is $\frac{1}{2}$, and $2 \cdot 500\,000\,004 \equiv 1 \pmod {10^9+7}$

Buggy Code:
#include<bits/stdc++.h>
#define F(i,a,b) for(int i=a,i##end=b;i<=i##end;i++)
#define UF(i,a,b) for(int i=a,i##end=b;i>=i##end;i--)
#define sz(x) int(x.size())
#define Z(x,y) (x+=y)%=mod
using namespace std;typedef long long ll;
#define vec vector
template<typename T>
void operator+=(vec<T> &a,const T&b) {a.push_back(b);}
#define gc() getchar()
int read() {
	int s=0,w=0;char ch=gc();
	while(ch<'0'||ch>'9') w|=(ch=='-'),ch=gc();
	while(ch>='0'&&ch<='9') s=(s<<3)+(s<<1)+(ch^48),ch=gc();
	return w?-s:s;
} const int N=125,mod=1e9+7;
int n;
ll fac[N],inv[N];
ll ksm(ll a,ll n=mod-2) {
	ll res=1;
	for(;n;n>>=1,a=a*a%mod) if(n&1) res=res*a%mod;
	return res;
}
void init(int n) {
	fac[0]=fac[1]=inv[0]=inv[1]=1;
	F(i,2,n) fac[i]=fac[i-1]*i%mod,inv[i]=(mod-mod/i)*inv[mod%i]%mod;
	F(i,2,n) (inv[i]*=inv[i-1])%=mod;
}
ll A(ll n,ll m) {return n>=m&&m>=0?fac[n]*inv[n-m]%mod:0;}
ll ivA(ll n,ll m) {return n>=m&&m>=0?inv[n]*fac[n-m]%mod:0;}
ll C(ll n,ll m) {return n>=m&&m>=0?fac[n]*inv[m]%mod*inv[n-m]%mod:0;}
int main() {
	init(120);
	F(T,1,read()) {
		n=read();
		vec<int> c(n+1),s(n+1),v(n+1);
		F(i,1,n) c[i]=read();
		F(i,1,n) while(isspace(s[i]=gc()));
		F(i,1,n) s[i]^=48;
		array<int,2> P{},Q{},R{};
		F(i,1,n) {
			if(c[i]>1) ++P[s[i]];
			if(c[i]==1) ++Q[s[i]];
			if(!c[i]) ++R[s[i]];
			v[c[i]]=1;
		} int all=P[0]+Q[0]+R[0];
		if(c[n]<=1) {
			if(P[0]+Q[0]+R[0]==0) puts("1");
			else puts("0");
			continue;
		}
		vec<array<array<ll,N>,N>> g(n+1),ng(n+1);
		F(p,0,all) F(u,0,n) g[u][p][0]=1;
		int h=0;
		UF(i,n,1) if(c[i]>1) {
			F(p,0,all) F(u,1,n) F(x,0,p) if(v[u]) {
				F(y,0,x-(c[i]-1)) Z(ng[u][p][x],g[u][p][y]*ng[c[i]][p-y-(c[i]-1)][x-y-(c[i]-1)]%mod*(u+y)%mod*ivA(p-y+1,c[i]));
				Z(ng[u][p][x],g[u][p][x]);
			}
			F(p,0,all) F(x,0,p) {
				if(s[i]) {
					F(y,h,x-c[i]) Z(ng[0][p][x],g[0][p][y]*ng[c[i]][p-y-(c[i]-1)+h][x-y-c[i]]%mod*ivA(p-y+h+1,c[i]));
				}
				else {
					F(y,h,x-(c[i]-1)) Z(ng[0][p][x],g[0][p][y]*ng[c[i]][p-y-(c[i]-1)+h][x-y-(c[i]-1)]%mod*y%mod*ivA(p-y+h+1,c[i]));
					Z(ng[0][p][x],g[0][p][x]);
				}
			}
			if(s[i]) ++h;
			F(p,0,all) F(u,0,n) F(x,0,p) g[u][p][x]=ng[u][p][x],ng[u][p][x]=0;
		}
		vec<array<ll,N>> dp(n+1);
		F(i,0,R[0]) F(j,0,Q[0]) dp[i][j+Q[1]]=g[0][P[0]+Q[0]+R[0]][i+j]*C(i+j,i)%mod*A(Q[0],j)%mod;
		F(i,0,R[0]) F(j,1,Q[0]+Q[1]) {
			ll iv=ksm(P[0]+P[1]+R[0]-i);
			Z(dp[i+1][j-1],dp[i][j]*iv);
			F(x,0,R[0]-i) F(y,0,Q[0]+Q[1]-j+1) if(x+y>1) {
				Z(dp[i+x][j+y-1],dp[i][j]*g[1][n-R[1]-i-j][x+y-1]%mod*iv%mod*C(x+y,x)%mod*A(Q[0]+Q[1]-j+1,y));
			}
		} ll ans=1,k=R[0];
		F(i,0,k) Z(ans,(mod-dp[i][0])*A(k,i));
		cout<<ans<<endl;
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
5
1 1 1 1 2
11110
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mind Bloom
time_limit_ms: 5000
memory_limit_mb: 1024
Description: This is the way it always was.

This is the way it always will be.

All will be forgotten again soon...

Jellyfish is playing a one-player card game called "Slay the Spire". There are $n$ cards in total numbered from $1$ to $n$. The $i$-th card has power $c_i$.

There is a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

Jellyfish will repeat the following process until either her hand or the draw pile is empty.

  1. Let $x$ be the power of the card with the largest power in her hand.    2. Place a single card with power $x$ back into the draw pile.    3. Randomly draw $x$ cards from the draw pile. All subsets of $x$ cards from the draw pile have an equal chance of being drawn. If there are fewer than $x$ cards in the draw pile, Jellyfish will draw all cards.

At the end of this process, find the probability that Jellyfish can empty the draw pile, modulo $1\,000\,000\,007$.

Formally, let $M=1\,000\,000\,007$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 120$) — the number of cards.

The second line of each test case contains $n$ integers $c_1,c_2,\ldots,c_n$ ($0 \leq c_i \leq n$) — the powers of the cards. It is guaranteed that $c_1 \leq c_2 \leq \ldots \leq c_n$.

The third line of each test case contains a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $120^2$.

Output: For each test case, output the probability that Jellyfish can empty the draw pile modulo $1\,000\,000\,007$.

Examples:
input:
4
5
0 1 1 1 2
00100
3
2 3 3
000
10
0 0 0 0 0 0 0 1 1 1
1111011111
20
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 2 3 3 4
00000000001000101010
output:
500000004
0
0
675898154

Note: In the first test case, Jellyfish will keep playing cards with power $1$ until Jellyfish draws a card with power $0$ or power $2$. If Jellyfish draws a card with power $0$, she will eventually empty her hand. If Jellyfish draws a card with power $2$, she will eventually empty the draw pile. Since there is an equal chance of drawing $0$ or $2$, the answer is $\frac{1}{2}$, and $2 \cdot 500\,000\,004 \equiv 1 \pmod {10^9+7}$

Buggy Code:
#include<bits/stdc++.h>
#define ci const int
#define ll long long
#define ull unsigned long long
using namespace std;
ci mod=1e9+7;
inline void add(int &x,ci v){
	x+=v,x-=x<mod?0:mod;
}
inline void sub(int &x,ci v){
	x-=v,x+=x<0?mod:0;
}
ll qk(ll x,int y=mod-2){
	ll ans=1;
	while(y){
		if(y&1)ans=ans*x%mod;
		x=x*x%mod,y>>=1;
	}
	return ans;
}
ci N=250;
const ull H=1e9+7;
int n,fac[N],dfac[N],inv[N],C[N][N],iC[N][N];
int m,lim,cnt[N],val[N],c[N],rk[N],pre[N];
struct st{
	int res;
	vector<int>num;
	st(int res=0,vector<int>num={}):res(res),num(num){}
	ull hs(){
		ull ans=res+1;
		for(int x:num)ans=ans*H+x+1;
		return ans;
	}
	inline bool operator < (const st &t)const{
		if(res!=t.res)return res<t.res;
		return num<t.num;
	}
	void print(){
		puts("details");
		printf("res=%d\n",res);
		printf("num ");for(int x:num)printf("%d ",x);
		puts("");
	}
};
unordered_map<ull,int>mp;
int B;
int dfs(st s){
	int res=s.res,all=0;
	for(int i=0;i<m;++i)all+=s.num[i]*max(1,val[i]);
	for(int i=0;i<m&&res;++i){
		ci gt=min(res,cnt[i]-s.num[i]);
		all+=gt*max(1,val[i]);
		res-=gt;
	}
	if(all>lim)return 1;
	const ull hs=s.hs();
	if(mp.find(hs)!=mp.end())return mp[hs];
	++B;
	if(s.res==0){
		int mx=-1;
		for(int i=m-1;~i;--i)
			if(s.num[i]){
				mx=i;
				break;
			}
		if(mx==-1)return 0;
		else if(val[mx]==0)return 0;
		else if(val[mx]==1){
			st t=s;
			--t.num[mx];
			int sum=0,ans=0;
			for(int i=0;i<m;++i)
				if(val[i]!=1&&s.num[i]!=cnt[i]){
					sum+=cnt[i]-s.num[i];
					++t.num[i];
					add(ans,(ll)(cnt[i]-s.num[i])*dfs(t)%mod);
					--t.num[i];
				}
			return mp[hs]=(ll)ans*inv[sum]%mod;
		}
		else{
			st t=s;
			--t.num[mx];
			t.res=val[mx];
			return mp[hs]=dfs(t);
		}
	}
	else{
		st t=s;t.res=s.res-1;
		int ok=0,ans=0;
		for(int i=0;i<m;++i)ok+=cnt[i]-t.num[i];
		for(int i=0;i<m;++i)
			if(t.num[i]<cnt[i])
				++t.num[i],add(ans,(ll)dfs(t)*inv[ok]%mod*(cnt[i]-s.num[i])%mod),--t.num[i];
		return mp[hs]=ans;
	}
}
void Init(){
	m=lim=0;
	mp.clear();
	memset(rk,0,sizeof(rk));
	memset(val,0,sizeof(val));
	memset(c,0,sizeof(c));
	memset(cnt,0,sizeof(cnt));
	memset(pre,0,sizeof(pre));
}
void Work(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i)scanf("%d",&c[i]),lim+=c[i]==0;
	for(int i=1;i<=n;++i){
		int j=i;
		while(j+1<=n&&c[i]==c[j+1])++j;
		rk[c[i]]=m,val[m]=c[i],cnt[m]=j-i+1,++m,i=j;
	}
	st s;
	s.num.resize(m);
	s.res=0;
	int h=0;
	for(int i=1;i<=n;++i){
		char ch=getchar();
		while(ch!='0'&&ch!='1')ch=getchar();
		if(ch=='1')++s.num[rk[c[i]]],++h;
	}
	if(h==n)puts("1");
	else if(c[n]<=1)puts("0");
	else printf("%d\n",dfs(s));
}
int main(){
	int S=clock();
	fac[0]=fac[1]=dfac[0]=dfac[1]=inv[1]=1;
	for(int i=2;i<N;++i)
		fac[i]=(ll)fac[i-1]*i%mod,
		inv[i]=(ll)(mod-mod/i)*inv[mod%i]%mod,
		dfac[i]=(ll)dfac[i-1]*inv[i]%mod;
	for(int i=0;i<N;++i){
		C[i][0]=1;
		for(int j=1;j<=i;++j)C[i][j]=(C[i-1][j]+C[i-1][j-1])%mod;
	}
	for(int i=0;i<N;++i)for(int j=0;j<=i;++j)iC[i][j]=qk(C[i][j]);
	int T;scanf("%d",&T);
	while(T--)Work(),Init();
	// printf("B=%d\n",B);
	// printf("%d ms.\n",clock()-S);
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
3
1 2 2
001
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mind Bloom
time_limit_ms: 5000
memory_limit_mb: 1024
Description: This is the way it always was.

This is the way it always will be.

All will be forgotten again soon...

Jellyfish is playing a one-player card game called "Slay the Spire". There are $n$ cards in total numbered from $1$ to $n$. The $i$-th card has power $c_i$.

There is a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

Jellyfish will repeat the following process until either her hand or the draw pile is empty.

  1. Let $x$ be the power of the card with the largest power in her hand.    2. Place a single card with power $x$ back into the draw pile.    3. Randomly draw $x$ cards from the draw pile. All subsets of $x$ cards from the draw pile have an equal chance of being drawn. If there are fewer than $x$ cards in the draw pile, Jellyfish will draw all cards.

At the end of this process, find the probability that Jellyfish can empty the draw pile, modulo $1\,000\,000\,007$.

Formally, let $M=1\,000\,000\,007$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 120$) — the number of cards.

The second line of each test case contains $n$ integers $c_1,c_2,\ldots,c_n$ ($0 \leq c_i \leq n$) — the powers of the cards. It is guaranteed that $c_1 \leq c_2 \leq \ldots \leq c_n$.

The third line of each test case contains a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $120^2$.

Output: For each test case, output the probability that Jellyfish can empty the draw pile modulo $1\,000\,000\,007$.

Examples:
input:
4
5
0 1 1 1 2
00100
3
2 3 3
000
10
0 0 0 0 0 0 0 1 1 1
1111011111
20
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 2 3 3 4
00000000001000101010
output:
500000004
0
0
675898154

Note: In the first test case, Jellyfish will keep playing cards with power $1$ until Jellyfish draws a card with power $0$ or power $2$. If Jellyfish draws a card with power $0$, she will eventually empty her hand. If Jellyfish draws a card with power $2$, she will eventually empty the draw pile. Since there is an equal chance of drawing $0$ or $2$, the answer is $\frac{1}{2}$, and $2 \cdot 500\,000\,004 \equiv 1 \pmod {10^9+7}$

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

struct hash_for_hash_map {
    size_t operator()(size_t x) const {
        static const size_t D = std::chrono::steady_clock::now().time_since_epoch().count();
        return x * x + D;
    }
};

template<typename key_t, typename value_t, typename hash_t = hash_for_hash_map>
using hash_map = __gnu_pbds::gp_hash_table<key_t, value_t, hash_t>;

#include <bits/stdc++.h>

using namespace std;

#define ull unsigned long long
#define ll long long
#define flt double
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()

const ll mod = 1e9 + 7;

ll bin_pow(ll x, ll n) {
    ll y = 1;
    for (; n > 0; n >>= 1, x = x * x % mod) {
        if (n & 1) {
            y = y * x % mod;
        }
    }
    return y;
}

ll rev(ll x) {
    return bin_pow(x, mod - 2);
}

int n;
vector<int> aboba;
vector<int> init_cnt;
vector<int> cnt;

struct state {
    vector<int> my_cnt;

    ull hsh() const {
        ull h = 0;
        ull p = 23904857203;
        for (auto x : my_cnt) {
            h = h * p + (x + 1);
        }
        return h;
    }
};

ll c[1000][1000];

void prec() {
    for (int i = 0; i < 1000; i++) {
        c[i][0] = c[i][i] = 1;
        for (int j = 1; j < i; j++) {
            c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
        }
    }
}

hash_map<ll, ll> dp;

ll get(const state& s);

ll get_rec(state& s, int rest_cnt, int i, ll ways, ll sm) {
    if (sm > cnt[0] - s.my_cnt[0]) {
        return 0;
    }
    if (rest_cnt == 0) {
        ll to = get(s);
        return to * ways % mod;
    }
    if (i >= (int)s.my_cnt.size() || (sm + i * rest_cnt) > cnt[0] - s.my_cnt[0]) {
        return 0;
    }
    ll rs = 0;
    for (int x = 0; x <= cnt[i] - s.my_cnt[i]; x += 1) {
        int new_rest = rest_cnt - x;
        if (new_rest < 0) {
            break;
        }
        ll new_sm = sm + x * i;
        if (new_sm > cnt[0] - s.my_cnt[0]) {
            break;
        }
        ll new_ways = ways * c[cnt[i] - s.my_cnt[i]][x] % mod;
        s.my_cnt[i] += x;
        rs += get_rec(s, new_rest, i + 1, new_ways, new_sm);
        s.my_cnt[i] -= x;
    }
    return rs % mod;
}

ll get(const state& s) {
    ll sm = 0;
    ll acc = 0;
    for (int i = 0; i < (int)s.my_cnt.size(); i += 1) {
        sm += s.my_cnt[i] * i;
        acc += s.my_cnt[i];
    }
    if (acc == n) {
        return 0;
    }
    int mx = (int)s.my_cnt.size() - 1;
    while (mx > 0 and s.my_cnt[mx] == 0) {
        mx -= 1;
    }
    if (mx <= 1 and cnt[0] + cnt[1] == n) {
        return 1;
    }
    ll rest_zero = cnt[0] - s.my_cnt[0];
    if (rest_zero < sm) {
        return 0;
    }
    ull h = s.hsh();
    auto it = dp.find(h);
    if (it != dp.end()) {
        return it->second;
    }
    dp[h] = 0;
    if (mx == 0) {
        return dp[h] = 1;
    }
    auto clone = s;
    clone.my_cnt[mx] -= 1;
    ll to_take = min<ll>(mx, n - acc + 1);
    ll rs = get_rec(clone, to_take, 0, 1, sm - mx);
    ll states = c[n - acc + 1][to_take];
    if (mx == 1) {
        states -= (cnt[1] - clone.my_cnt[1]);
    }
    ll prob = rev(states);
    return dp[h] = (rs * prob % mod);
}

int32_t main() {
    prec();
    if (1) {
        ios::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);
    }

    // dp.resize(3'000'00);

    int t;
    cin >> t;
    for (int itr = 0; itr < t; itr += 1) {
        dp.clear();
        cin >> n;
        aboba.resize(n);
        cnt.assign(n + 1, 0);
        for (int i = 0; i < n; i += 1) {
            cin >> aboba[i];
            cnt[aboba[i]] += 1;
        }
        init_cnt.assign(n + 1, 0);
        string s;
        cin >> s;
        for (int i = 0; i < n; i += 1) {
            if (s[i] == '1') {
                init_cnt[aboba[i]] += 1;
            }
        }
        if (init_cnt[0] == n) {
            cout << "1\n";
            continue;
        }
        auto init_state = state{ init_cnt };
        ll rs = get(init_state);
        cout << (1 - rs + mod) % mod << "\n";
        // cerr << dp.size() << '\n';
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
5
0 1 1 1 2
00100
3
2 3 3
000
10
0 0 0 0 0 0 0 1 1 1
1111011111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mind Bloom
time_limit_ms: 5000
memory_limit_mb: 1024
Description: This is the way it always was.

This is the way it always will be.

All will be forgotten again soon...

Jellyfish is playing a one-player card game called "Slay the Spire". There are $n$ cards in total numbered from $1$ to $n$. The $i$-th card has power $c_i$.

There is a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

Jellyfish will repeat the following process until either her hand or the draw pile is empty.

  1. Let $x$ be the power of the card with the largest power in her hand.    2. Place a single card with power $x$ back into the draw pile.    3. Randomly draw $x$ cards from the draw pile. All subsets of $x$ cards from the draw pile have an equal chance of being drawn. If there are fewer than $x$ cards in the draw pile, Jellyfish will draw all cards.

At the end of this process, find the probability that Jellyfish can empty the draw pile, modulo $1\,000\,000\,007$.

Formally, let $M=1\,000\,000\,007$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 120$) — the number of cards.

The second line of each test case contains $n$ integers $c_1,c_2,\ldots,c_n$ ($0 \leq c_i \leq n$) — the powers of the cards. It is guaranteed that $c_1 \leq c_2 \leq \ldots \leq c_n$.

The third line of each test case contains a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $120^2$.

Output: For each test case, output the probability that Jellyfish can empty the draw pile modulo $1\,000\,000\,007$.

Examples:
input:
4
5
0 1 1 1 2
00100
3
2 3 3
000
10
0 0 0 0 0 0 0 1 1 1
1111011111
20
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 2 3 3 4
00000000001000101010
output:
500000004
0
0
675898154

Note: In the first test case, Jellyfish will keep playing cards with power $1$ until Jellyfish draws a card with power $0$ or power $2$. If Jellyfish draws a card with power $0$, she will eventually empty her hand. If Jellyfish draws a card with power $2$, she will eventually empty the draw pile. Since there is an equal chance of drawing $0$ or $2$, the answer is $\frac{1}{2}$, and $2 \cdot 500\,000\,004 \equiv 1 \pmod {10^9+7}$

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N=242;
const int MOD=1e9+7;
int S[N][N];
long long inv[N];
long long f[N][N][N],d[N][N];
int ss[N];
int a[N];
bool hav[N];
long long C[N*2][N*2];
long long qpow(long long x,long long y,long long mod){
	if(!y)return 1%mod;
	if(y==1)return x%mod;
	long long kp=qpow(x,y/2,mod);
	if(y&1)return kp*kp%mod*x%mod;
	return kp*kp%mod;
}
signed main(){
	//freopen("1.txt","r",stdin);
	//int tim1=clock();
	for(int i=1;i<=240;i++)inv[i]=qpow(i,MOD-2,MOD);
	C[0][0]=1;
	for(int i=1;i<=240;i++){
		C[i][0]=1;
		for(int j=1;j<=i;j++)C[i][j]=(C[i-1][j-1]+C[i-1][j])%MOD;
	}
	int T;
	cin>>T;
	while(T--){
		int n;
		scanf("%lld",&n);
		for(int i=1;i<=n;i++)scanf("%lld",&a[i]);
		string s;
		cin>>s;
		int cnt=0,sumc=0;
		for(int i=0;i<n;i++){
			hav[i+1]=s[i]-'0';
			ss[i+1]=ss[i]+hav[i+1];
		}
		for(int i=1;i<=n;i++)cnt+=hav[i],sumc+=hav[i]*a[i];
		if(a[n]<=1){
			if(cnt==n)cout<<1<<"\n";
			else cout<<0<<"\n";
			continue;
		}
		if(!cnt){
			cout<<0<<"\n";
			continue;
		}
		if(a[1]>0){
			cout<<1<<"\n";
			continue;
		}
		if(sumc==0){
			cout<<0<<"\n";
			continue;
		}
		for(int l=1;l<=n;l++){
				for(int sum=n;sum>=0;sum--){
					for(int c=0;c<=n;c++)f[l][sum][c]=0;
				}
		}
		for(int l=n;l>=1;l--){
			if(a[l]<=0)continue;
			for(int sum=l-1;sum>=0;sum--){
				for(int r=0;r<=n;r++){
					for(int c=0;c<=n;c++)S[r][c]=0;
				}
				S[n][a[l]]=1;
				for(int r=n;r>=l;r--){
					for(int c=n;c>=a[l];c--){
						if(sum+c>r)continue;
						if(l==r&&c==1){
							S[r-1][c]=(S[r-1][c]*1ll+(r-sum-c)*inv[r-sum]%MOD*S[r][c])%MOD;
							for(int x=a[r];x<=r-c;x++){
								S[r-1][c-1+x]=S[r-1][c-1+x]*1ll*qpow((MOD+1-c*inv[r-sum]%MOD*S[r][c]%MOD)%MOD,MOD-2,MOD)%MOD;
							}
						}
						else{
							for(int x=a[r];x<=r-c;x++){
								S[r-1][c-1+x]=(S[r-1][c-1+x]*1ll+f[r][sum+c-1][x]*c%MOD*inv[r-sum]%MOD*S[r][c])%MOD;
							}
							S[r-1][c]=(S[r-1][c]*1ll+(r-sum-c)*inv[r-sum]%MOD*S[r][c])%MOD;
						}
					}
				}
				for(int c=a[l];c<=n;c++){
					f[l][sum][c]=S[l-1][c];
					//if(f[l][sum][c])cout<<l<<" "<<sum<<" "<<c<<":"<<f[l][sum][c]<<"\n";
				}
			}
		}
		/*for(int i=1;i<=n;i++){
			for(int j=0;j<=i-1;j++){
				for(int k=0;k<=i-1;k++){
					if(f[i][j][k])cout<<i<<" "<<j<<" "<<k<<":"<<f[i][j][k]<<"\n";
				}
			}
		}*/
		for(int i=0;i<=n;i++){
			for(int j=0;j<=n;j++)d[i][j]=0;
		}
		for(int i=n;i>=1;i--){
			if(hav[i]){
				d[cnt][i]=1;
				break;
			}
		}
		long long ans=0;
		for(int x=n;x>=1;x--){
			if(a[x]==0){
				for(int sum=x;sum>=0;sum--){
					ans+=d[sum][x];
					ans%=MOD;
				}
				break;
			}
			for(int sum=x;sum>=ss[x];sum--){
				//have
				long long prob;
				if(hav[x])prob=1;
				else{
					prob=(sum-ss[x])*inv[(x-ss[x])]%MOD;
				}
				for(int t=a[x];t<=n;t++){
					if(sum+t>x)break;
					d[sum-1+t][x-1]=(d[sum-1+t][x-1]+d[sum][x]*prob%MOD*f[x][sum-1][t])%MOD;
				}
				if(sum<=x-1)d[sum][x-1]=(d[sum][x-1]+d[sum][x]*(MOD+1-prob))%MOD;
			}
		}
		cout<<(MOD+1-ans)%MOD<<"\n";
	}
	//int tim2=clock();
	//cout<<tim2-tim1<<"\n";
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
4
0 1 2 2
0011
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mind Bloom
time_limit_ms: 5000
memory_limit_mb: 1024
Description: This is the way it always was.

This is the way it always will be.

All will be forgotten again soon...

Jellyfish is playing a one-player card game called "Slay the Spire". There are $n$ cards in total numbered from $1$ to $n$. The $i$-th card has power $c_i$.

There is a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

Jellyfish will repeat the following process until either her hand or the draw pile is empty.

  1. Let $x$ be the power of the card with the largest power in her hand.    2. Place a single card with power $x$ back into the draw pile.    3. Randomly draw $x$ cards from the draw pile. All subsets of $x$ cards from the draw pile have an equal chance of being drawn. If there are fewer than $x$ cards in the draw pile, Jellyfish will draw all cards.

At the end of this process, find the probability that Jellyfish can empty the draw pile, modulo $1\,000\,000\,007$.

Formally, let $M=1\,000\,000\,007$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 100$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 120$) — the number of cards.

The second line of each test case contains $n$ integers $c_1,c_2,\ldots,c_n$ ($0 \leq c_i \leq n$) — the powers of the cards. It is guaranteed that $c_1 \leq c_2 \leq \ldots \leq c_n$.

The third line of each test case contains a binary string $s$ of length $n$. If $s_i = \texttt{0}$, the $i$-th card is initially in the draw pile. If $s_i = \texttt{1}$, the $i$-th card is initially in Jellyfish's hand.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $120^2$.

Output: For each test case, output the probability that Jellyfish can empty the draw pile modulo $1\,000\,000\,007$.

Examples:
input:
4
5
0 1 1 1 2
00100
3
2 3 3
000
10
0 0 0 0 0 0 0 1 1 1
1111011111
20
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 2 3 3 4
00000000001000101010
output:
500000004
0
0
675898154

Note: In the first test case, Jellyfish will keep playing cards with power $1$ until Jellyfish draws a card with power $0$ or power $2$. If Jellyfish draws a card with power $0$, she will eventually empty her hand. If Jellyfish draws a card with power $2$, she will eventually empty the draw pile. Since there is an equal chance of drawing $0$ or $2$, the answer is $\frac{1}{2}$, and $2 \cdot 500\,000\,004 \equiv 1 \pmod {10^9+7}$

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
const int N=122;
const int MOD=1e9+7;
int S[N][N];
long long inv[N];
long long f[N][N][N],d[N][N];
int ss[N];
int a[N];
bool hav[N];
long long C[N*2][N*2];
long long qpow(long long x,long long y,long long mod){
	if(!y)return 1%mod;
	if(y==1)return x%mod;
	long long kp=qpow(x,y/2,mod);
	if(y&1)return kp*kp%mod*x%mod;
	return kp*kp%mod;
}
int main(){
	for(int i=1;i<=120;i++)inv[i]=qpow(i,MOD-2,MOD);
	C[0][0]=1;
	for(int i=1;i<=240;i++){
		C[i][0]=1;
		for(int j=1;j<=i;j++)C[i][j]=(C[i-1][j-1]+C[i-1][j])%MOD;
	}
	int T;
	cin>>T;
	while(T--){
		int n;
		scanf("%d",&n);
		for(int i=1;i<=n;i++)scanf("%d",&a[i]);
		string s;
		cin>>s;
		int cnt=0,sumc=0;
		for(int i=0;i<n;i++){
			hav[i+1]=s[i]-'0';
			ss[i+1]=ss[i]+hav[i+1];
		}
		for(int i=1;i<=n;i++)cnt+=hav[i],sumc+=hav[i]*a[i];
		if(a[n]<=1){
			if(cnt==n)cout<<1<<"\n";
			else cout<<0<<"\n";
			continue;
		}
		if(!cnt){
			cout<<0<<"\n";
			continue;
		}
		if(a[1]>0){
			cout<<1<<"\n";
			continue;
		}
		if(sumc==0){
			cout<<0<<"\n";
			continue;
		}
		for(int l=1;l<=n;l++){
				for(int sum=n;sum>=0;sum--){
					for(int c=0;c<=n;c++)f[l][sum][c]=0;
				}
		}
		for(int l=n;l>=1;l--){
			if(a[l]<=0)continue;
			for(int sum=l-1;sum>=0;sum--){
				for(int r=0;r<=n;r++){
					for(int c=0;c<=n;c++)S[r][c]=0;
				}
				S[n][a[l]]=1;
				for(int r=n;r>=l;r--){
					for(int c=n;c>=a[l];c--){
						if(sum+c>r)continue;
						if(l==r&&c==1){
							S[r-1][c]=(S[r-1][c]*1ll+(r-sum-c)*inv[r-sum]%MOD*S[r][c])%MOD;
							for(int x=a[r];x<=r-c;x++){
								S[r-1][c-1+x]=S[r-1][c-1+x]*1ll*qpow((MOD+1-c*inv[r-sum]%MOD*S[r][c]%MOD)%MOD,MOD-2,MOD)%MOD;
							}
						}
						else{
							for(int x=a[r];x<=r-c;x++){
								S[r-1][c-1+x]=(S[r-1][c-1+x]*1ll+f[r][sum+c-1][x]*c%MOD*inv[r-sum]%MOD*S[r][c])%MOD;
							}
							S[r-1][c]=(S[r-1][c]*1ll+(r-sum-c)*inv[r-sum]%MOD*S[r][c])%MOD;
						}
					}
				}
				for(int c=a[l];c<=n;c++){
					f[l][sum][c]=S[l-1][c];
					//if(f[l][sum][c])cout<<l<<" "<<sum<<" "<<c<<":"<<f[l][sum][c]<<"\n";
				}
			}
		}
		/*for(int i=1;i<=n;i++){
			for(int j=0;j<=i-1;j++){
				for(int k=0;k<=i-1;k++){
					if(f[i][j][k])cout<<i<<" "<<j<<" "<<k<<":"<<f[i][j][k]<<"\n";
				}
			}
		}*/
		for(int i=0;i<=n;i++){
			for(int j=0;j<=n;j++)d[i][j]=0;
		}
		for(int i=n;i>=1;i--){
			if(hav[i]){
				d[cnt][i]=1;
				break;
			}
		}
		long long ans=0;
		for(int x=n;x>=1;x--){
			if(a[x]==0){
				for(int sum=x;sum>=0;sum--){
					ans+=d[sum][x];
					ans%=MOD;
				}
				break;
			}
			for(int sum=x;sum>=ss[x];sum--){
				//have
				long long prob;
				if(hav[x])prob=1;
				else{
					prob=(sum-ss[x])*inv[(x-ss[x])]%MOD;
				}
				for(int t=a[x];t<=n;t++){
					if(sum+t>x)break;
					d[sum-1+t][x-1]=(d[sum-1+t][x-1]+d[sum][x]*prob%MOD*f[x][sum-1][t])%MOD;
				}
				if(sum<=x-1)d[sum][x-1]=(d[sum][x-1]+d[sum][x]*(MOD+1-prob))%MOD;
			}
		}
		cout<<(MOD+1-ans)%MOD<<"\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
4
5
0 1 1 1 2
00100
3
2 3 3
000
10
0 0 0 0 0 0 0 1 1 1
1111011111
20
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 2 3 3 4
00000000001000101010
```