**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: 161272855
// 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(!cnt||cnt==n||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;
		}
		if(a[i]==0){
			cout<<(cnt==n)<<"\n";
			return;
		}
	}
	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<<(a[1]>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:
```plaintext
1
5
1 1 1 1 2
00001
```


**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: 161272942
// 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(!cnt||cnt==n||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;
		}
		if(a[i]==0){
			cout<<(cnt==n)<<"\n";
			return;
		}
	}
	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);
			assert(sc[i]+a[i]>0);
			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<<(a[1]>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:
```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:
#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(){
	//freopen("1.txt","r",stdin);
	//int tim1=clock();
	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";
	}
	//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
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: Inaccurate Subsequence Search
time_limit_ms: 2000
memory_limit_mb: 256
Description: Maxim has an array $$$a$$$ of $$$n$$$ integers and an array $$$b$$$ of $$$m$$$ integers ($$$m \le n$$$).

Maxim considers an array $$$c$$$ of length $$$m$$$ to be good if the elements of array $$$c$$$ can be rearranged in such a way that at least $$$k$$$ of them match the elements of array $$$b$$$.

For example, if $$$b = [1, 2, 3, 4]$$$ and $$$k = 3$$$, then the arrays $$$[4, 1, 2, 3]$$$ and $$$[2, 3, 4, 5]$$$ are good (they can be reordered as follows: $$$[1, 2, 3, 4]$$$ and $$$[5, 2, 3, 4]$$$), while the arrays $$$[3, 4, 5, 6]$$$ and $$$[3, 4, 3, 4]$$$ are not good.

Maxim wants to choose every subsegment of array $$$a$$$ of length $$$m$$$ as the elements of array $$$c$$$. Help Maxim count how many selected arrays will be good.

In other words, find the number of positions $$$1 \le l \le n - m + 1$$$ such that the elements $$$a_l, a_{l+1}, \dots, a_{l + m - 1}$$$ form a good array.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \le k \le m \le n \le 2 \cdot 10^5$$$) — the number of elements in arrays $$$a$$$ and $$$b$$$, the required number of matching elements.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^6$$$) — the elements of array $$$a$$$. Elements of the array $$$a$$$ are not necessarily unique.

The third line of each test case contains $$$m$$$ integers $$$b_1, b_2, \dots, b_m$$$ ($$$1 \le b_i \le 10^6$$$) — the elements of array $$$b$$$. Elements of the array $$$b$$$ are not necessarily unique.

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

Output: For each test case, output the number of good subsegments of array $$$a$$$ on a separate line.

Examples:
input:
5
7 4 2
4 1 2 3 4 5 6
1 2 3 4
7 4 3
4 1 2 3 4 5 6
1 2 3 4
7 4 4
4 1 2 3 4 5 6
1 2 3 4
11 5 3
9 9 2 2 10 9 7 6 3 6 3
6 9 7 8 10
4 1 1
4 1 5 6
6
output:
4
3
2
4
1

Note: In the first example, all subsegments are good.

In the second example, good subsegments start at positions $$$1$$$, $$$2$$$, and $$$3$$$.

In the third example, good subsegments start at positions $$$1$$$ and $$$2$$$.

Buggy Code:
#include <cstdio>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <set>
#include <unordered_map>
#include <functional>
#include <algorithm>
#include <cstring>
#include <map>

#define ll long long

using namespace std;

ll a[200001], b[200001];
ll aCount[1000001];
ll bCount[1000001];
bool bExist[1000001];

int main() {
	//freopen("input.txt", "r", stdin);

	int t; scanf("%d", &t);
	while (t--) {

		// k match => good
		ll n, m, k; scanf("%lld %lld %lld", &n, &m, &k);
		for (int i = 0; i < n; ++i) scanf("%lld", &a[i]);
		for (int i = 0; i < m; ++i) {
			scanf("%lld", &b[i]);
			bCount[b[i]]++;
			bExist[b[i]] = true;
		}

		ll matchCount = 0;
		for (int i = 0; i < m; ++i) {
			aCount[a[i]]++;
			if (bCount[a[i]]) {
				matchCount++;
				bCount[a[i]]--;
			}
		}

		ll ans = 0;
		if (matchCount >= k) ans++;
		for (int i = m; i < n; ++i) {
			if (bExist[a[i - m]] && aCount[a[i - m]] == 1) {
				matchCount--;
				bCount[a[i - m]]++;
			}

			aCount[a[i - m]]--;
			aCount[a[i]]++;
			if (bCount[a[i]]) {
				matchCount++;
				bCount[a[i]]--;
			}

			if (matchCount >= k) ans++;
		}

		printf("%lld\n", ans);

		for (int i = 0; i < m; ++i) {
			bExist[b[i]] = false;
			if (bCount[b[i]]) bCount[b[i]] = 0;
			if (aCount[a[i + m]]) aCount[a[i + m]] = 0;

		}
	}

	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
7 4 3
1 2 3 4 1 2 3
1 2 3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Inaccurate Subsequence Search
time_limit_ms: 2000
memory_limit_mb: 256
Description: Maxim has an array $$$a$$$ of $$$n$$$ integers and an array $$$b$$$ of $$$m$$$ integers ($$$m \le n$$$).

Maxim considers an array $$$c$$$ of length $$$m$$$ to be good if the elements of array $$$c$$$ can be rearranged in such a way that at least $$$k$$$ of them match the elements of array $$$b$$$.

For example, if $$$b = [1, 2, 3, 4]$$$ and $$$k = 3$$$, then the arrays $$$[4, 1, 2, 3]$$$ and $$$[2, 3, 4, 5]$$$ are good (they can be reordered as follows: $$$[1, 2, 3, 4]$$$ and $$$[5, 2, 3, 4]$$$), while the arrays $$$[3, 4, 5, 6]$$$ and $$$[3, 4, 3, 4]$$$ are not good.

Maxim wants to choose every subsegment of array $$$a$$$ of length $$$m$$$ as the elements of array $$$c$$$. Help Maxim count how many selected arrays will be good.

In other words, find the number of positions $$$1 \le l \le n - m + 1$$$ such that the elements $$$a_l, a_{l+1}, \dots, a_{l + m - 1}$$$ form a good array.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \le k \le m \le n \le 2 \cdot 10^5$$$) — the number of elements in arrays $$$a$$$ and $$$b$$$, the required number of matching elements.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^6$$$) — the elements of array $$$a$$$. Elements of the array $$$a$$$ are not necessarily unique.

The third line of each test case contains $$$m$$$ integers $$$b_1, b_2, \dots, b_m$$$ ($$$1 \le b_i \le 10^6$$$) — the elements of array $$$b$$$. Elements of the array $$$b$$$ are not necessarily unique.

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

Output: For each test case, output the number of good subsegments of array $$$a$$$ on a separate line.

Examples:
input:
5
7 4 2
4 1 2 3 4 5 6
1 2 3 4
7 4 3
4 1 2 3 4 5 6
1 2 3 4
7 4 4
4 1 2 3 4 5 6
1 2 3 4
11 5 3
9 9 2 2 10 9 7 6 3 6 3
6 9 7 8 10
4 1 1
4 1 5 6
6
output:
4
3
2
4
1

Note: In the first example, all subsegments are good.

In the second example, good subsegments start at positions $$$1$$$, $$$2$$$, and $$$3$$$.

In the third example, good subsegments start at positions $$$1$$$ and $$$2$$$.

Buggy Code:
/*
	Hello World!
	Minimal Template by x93
*/

#include <limits.h>
#include <stdint.h>
#include <string.h>
#include <stdio.h>

#ifdef __cplusplus
#include <algorithm>
#include <numeric>
#include <vector>
#include <bitset>
#include <queue>
#include <map>
#include <set>
#endif

#define abd(a, b) ((a) > (b) ? (a) - (b) : (b) - (a))
#define inf ULLONG_MAX
#define mod 1000000007
#define int  long long
#define uint unsigned int
#define algorithm signed main()
using namespace std;

int gcd(int a,int b)
{
	if(a%b==0) return b; else return gcd(b,a%b);
}

struct segtree
{
	int *t, n;
	segtree(int N, int* dat)
	{
		t = dat;
		this->n = N;
	}

	void build() {
	  for (int i = n - 1; i > 0; --i) t[i] = t[i<<1] ^ t[i<<1|1];
	}

	void modify(int p, int value) {
	  for (t[p += n] = value; p > 1; p >>= 1) t[p>>1] = t[p] ^ t[p^1];
	}

	int query(int l, int r) {
	  int res = 0;
	  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
	    if (l&1) res ^= t[l++];
	    if (r&1) res ^= t[--r];
	  }
	  return res;
	}
};

uint dfs(uint boss, vector<vector<uint>>& adj, uint* m)
{
	uint subord = 0;
	for (uint nod: adj[boss])
		subord += dfs(nod, adj, m) + 1;

	m[boss] = subord;
	return subord;
}

algorithm
{
	uint t;
	scanf("%llu", &t);

	while (t--)
	{
		uint n, m, k; scanf("%llu%llu%llu", &n, &m, &k);
		uint a[n];
		map<uint, uint> b;

		for (uint x = 0; x < n; x++)
			scanf("%llu", &a[x]);

		for (uint x = 0; x < m; x++)
		{
			uint t; scanf("%llu", &t);
			b[t]++;
		}

		map<uint, uint> freq;
		uint W = 0;

		for (uint x = 0; x < m; x++)
		{
			freq[a[x]]++;
			if (freq[a[x]] == b[a[x]])
				W++;
		}

		uint T = W >= k;
		for (uint x = m; x < n; x++)
		{
			W -= freq[a[x - m]] == b[a[x - m]];
			freq[a[x - m]]--;
			freq[a[x]]++;
			W += freq[a[x]] == b[a[x]];
			T += W >= k;
		} printf("%llu\n", T);
	}
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to create a test input that challenges the logic of the algorithm. The code is supposed to count the number of subsegments of length `m` in array `a` that can be rearranged to match at least `k` elements of array `b`. The bug might be in how the code handles the frequency counts and the comparison logic.

### Key Points to Consider:
1. **Edge Cases:**
   - Arrays `a` and `b` with repeated elements.
   - Arrays `a` and `b` with no common elements.
   - Arrays `a` and `b` where `b` is a subset of `a`.
   - Arrays `a` and `b` where `a` is a subset of `b`.

2. **Specific Scenarios:**
   - `k` is equal to `m`.
   - `k` is less than `m`.
   - `k` is 1.

### Test Input:
Let's create a test input that includes these edge cases and scenarios.

```plaintext
4
7 4 2
4 1 2 3 4 5 6
1 2 3 4
7 4 3
4 1 2 3 4 5 6
1 2 3 4
7 4 4
4 1 2 3 4 5 6
1 2 3 4
11 5 3
9 9 2 2 10 9 7 6 3 6 3
6 9 7 8 10
```

### Explanation:
1. **First Test Case:**
   - `n = 7`, `m = 4`, `k = 2`
   - `a = [4, 1, 2, 3, 4, 5, 6]`
   - `b = [1, 2, 3, 4]`
   - All subsegments are good because each subsegment of length 4 has at least 2 elements that match `b`.

2. **Second Test Case:**
   - `n = 7`, `m = 4`, `k = 3`
   - `a = [4, 1, 2, 3, 4, 5, 6]`
   - `b = [1, 2, 3, 4]`
   - Good subsegments start at positions 1, 2, and 3 because each of these subsegments has at least 3 elements that match `b`.

3. **Third Test Case:**
   - `n = 7`, `m = 4`, `k = 4`
   - `a = [4, 1, 2, 3, 4, 5, 6]`
   - `b = [1, 2, 3, 4]`
   - Only subsegments starting at positions 1 and 2 are good because they have exactly 4 elements that match `b`.

4. **Fourth Test Case:**
   - `n = 11`, `m = 5`, `k = 3`
   - `a = [9, 9, 2, 2, 10, 9, 7, 6, 3, 6, 3]`
   - `b = [6, 9, 7, 8, 10]`
   - Good subsegments start at positions 1, 2, 3, and 4 because each of these subsegments has at least 3 elements that match `b`.

This test input covers a variety of scenarios and should help expose any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Inaccurate Subsequence Search
time_limit_ms: 2000
memory_limit_mb: 256
Description: Maxim has an array $$$a$$$ of $$$n$$$ integers and an array $$$b$$$ of $$$m$$$ integers ($$$m \le n$$$).

Maxim considers an array $$$c$$$ of length $$$m$$$ to be good if the elements of array $$$c$$$ can be rearranged in such a way that at least $$$k$$$ of them match the elements of array $$$b$$$.

For example, if $$$b = [1, 2, 3, 4]$$$ and $$$k = 3$$$, then the arrays $$$[4, 1, 2, 3]$$$ and $$$[2, 3, 4, 5]$$$ are good (they can be reordered as follows: $$$[1, 2, 3, 4]$$$ and $$$[5, 2, 3, 4]$$$), while the arrays $$$[3, 4, 5, 6]$$$ and $$$[3, 4, 3, 4]$$$ are not good.

Maxim wants to choose every subsegment of array $$$a$$$ of length $$$m$$$ as the elements of array $$$c$$$. Help Maxim count how many selected arrays will be good.

In other words, find the number of positions $$$1 \le l \le n - m + 1$$$ such that the elements $$$a_l, a_{l+1}, \dots, a_{l + m - 1}$$$ form a good array.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \le k \le m \le n \le 2 \cdot 10^5$$$) — the number of elements in arrays $$$a$$$ and $$$b$$$, the required number of matching elements.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^6$$$) — the elements of array $$$a$$$. Elements of the array $$$a$$$ are not necessarily unique.

The third line of each test case contains $$$m$$$ integers $$$b_1, b_2, \dots, b_m$$$ ($$$1 \le b_i \le 10^6$$$) — the elements of array $$$b$$$. Elements of the array $$$b$$$ are not necessarily unique.

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

Output: For each test case, output the number of good subsegments of array $$$a$$$ on a separate line.

Examples:
input:
5
7 4 2
4 1 2 3 4 5 6
1 2 3 4
7 4 3
4 1 2 3 4 5 6
1 2 3 4
7 4 4
4 1 2 3 4 5 6
1 2 3 4
11 5 3
9 9 2 2 10 9 7 6 3 6 3
6 9 7 8 10
4 1 1
4 1 5 6
6
output:
4
3
2
4
1

Note: In the first example, all subsegments are good.

In the second example, good subsegments start at positions $$$1$$$, $$$2$$$, and $$$3$$$.

In the third example, good subsegments start at positions $$$1$$$ and $$$2$$$.

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

using namespace std;

int main() {
    int t, n, m, k;
    cin >> t;

    while (t--) {
        cin >> n >> m >> k;
        int a[n], b[m];

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

        for (int i = 0; i < m; i++)
            cin >> b[i];

        int i = 0, j = 0;
        int sum = 0, ki = 0;

        while (j < n) {
            if (j - i + 1 < m) {
                j++;
            } else {
                vector<int> sub;
                ki = 0;
                for (int h = i; h <=j; h++) {
                    sub.push_back(a[h]);
                }

                for (int h = 0; h < m; h++) {
                    for (int f = 0; f < m; f++) {
                        if (b[h] == sub[f]) {
                            ki++;
                            break;
                        }
                    }
                }
                sum += ki >= k? 1 : 0;
                i++, j++;
            }
        }

        cout << sum << endl;
    }

    return EXIT_SUCCESS;
}

**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 3 3
1 2 3 4 5
1 2 3
5 3 2
1 2 3 4 5
1 2 3
5 3 1
1 2 3 4 5
1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Inaccurate Subsequence Search
time_limit_ms: 2000
memory_limit_mb: 256
Description: Maxim has an array $$$a$$$ of $$$n$$$ integers and an array $$$b$$$ of $$$m$$$ integers ($$$m \le n$$$).

Maxim considers an array $$$c$$$ of length $$$m$$$ to be good if the elements of array $$$c$$$ can be rearranged in such a way that at least $$$k$$$ of them match the elements of array $$$b$$$.

For example, if $$$b = [1, 2, 3, 4]$$$ and $$$k = 3$$$, then the arrays $$$[4, 1, 2, 3]$$$ and $$$[2, 3, 4, 5]$$$ are good (they can be reordered as follows: $$$[1, 2, 3, 4]$$$ and $$$[5, 2, 3, 4]$$$), while the arrays $$$[3, 4, 5, 6]$$$ and $$$[3, 4, 3, 4]$$$ are not good.

Maxim wants to choose every subsegment of array $$$a$$$ of length $$$m$$$ as the elements of array $$$c$$$. Help Maxim count how many selected arrays will be good.

In other words, find the number of positions $$$1 \le l \le n - m + 1$$$ such that the elements $$$a_l, a_{l+1}, \dots, a_{l + m - 1}$$$ form a good array.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \le k \le m \le n \le 2 \cdot 10^5$$$) — the number of elements in arrays $$$a$$$ and $$$b$$$, the required number of matching elements.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^6$$$) — the elements of array $$$a$$$. Elements of the array $$$a$$$ are not necessarily unique.

The third line of each test case contains $$$m$$$ integers $$$b_1, b_2, \dots, b_m$$$ ($$$1 \le b_i \le 10^6$$$) — the elements of array $$$b$$$. Elements of the array $$$b$$$ are not necessarily unique.

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

Output: For each test case, output the number of good subsegments of array $$$a$$$ on a separate line.

Examples:
input:
5
7 4 2
4 1 2 3 4 5 6
1 2 3 4
7 4 3
4 1 2 3 4 5 6
1 2 3 4
7 4 4
4 1 2 3 4 5 6
1 2 3 4
11 5 3
9 9 2 2 10 9 7 6 3 6 3
6 9 7 8 10
4 1 1
4 1 5 6
6
output:
4
3
2
4
1

Note: In the first example, all subsegments are good.

In the second example, good subsegments start at positions $$$1$$$, $$$2$$$, and $$$3$$$.

In the third example, good subsegments start at positions $$$1$$$ and $$$2$$$.

Buggy Code:
import sys

input = sys.stdin.readline

t=int(input())
for _ in range(t):
    n,m,k=map(int,input().split())
    list_a=list(map(int,input().split()))
    list_b=list(map(int,input().split()))
    std_set=set(list_b)
    ans=0
    for i in range(0,n-m+1):
        set_temp=set(list_a[i:i+m])
        if len(set_temp & std_set)>=k:
            ans+=1
    print(ans)

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

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

Think step by step.

Response:
```plaintext
3
5 3 2
1 2 3 4 5
1 2 3
6 3 3
1 2 3 4 5 6
1 2 3
7 3 3
1 2 2 3 3 4 4
1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Inaccurate Subsequence Search
time_limit_ms: 2000
memory_limit_mb: 256
Description: Maxim has an array $$$a$$$ of $$$n$$$ integers and an array $$$b$$$ of $$$m$$$ integers ($$$m \le n$$$).

Maxim considers an array $$$c$$$ of length $$$m$$$ to be good if the elements of array $$$c$$$ can be rearranged in such a way that at least $$$k$$$ of them match the elements of array $$$b$$$.

For example, if $$$b = [1, 2, 3, 4]$$$ and $$$k = 3$$$, then the arrays $$$[4, 1, 2, 3]$$$ and $$$[2, 3, 4, 5]$$$ are good (they can be reordered as follows: $$$[1, 2, 3, 4]$$$ and $$$[5, 2, 3, 4]$$$), while the arrays $$$[3, 4, 5, 6]$$$ and $$$[3, 4, 3, 4]$$$ are not good.

Maxim wants to choose every subsegment of array $$$a$$$ of length $$$m$$$ as the elements of array $$$c$$$. Help Maxim count how many selected arrays will be good.

In other words, find the number of positions $$$1 \le l \le n - m + 1$$$ such that the elements $$$a_l, a_{l+1}, \dots, a_{l + m - 1}$$$ form a good array.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \le k \le m \le n \le 2 \cdot 10^5$$$) — the number of elements in arrays $$$a$$$ and $$$b$$$, the required number of matching elements.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^6$$$) — the elements of array $$$a$$$. Elements of the array $$$a$$$ are not necessarily unique.

The third line of each test case contains $$$m$$$ integers $$$b_1, b_2, \dots, b_m$$$ ($$$1 \le b_i \le 10^6$$$) — the elements of array $$$b$$$. Elements of the array $$$b$$$ are not necessarily unique.

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

Output: For each test case, output the number of good subsegments of array $$$a$$$ on a separate line.

Examples:
input:
5
7 4 2
4 1 2 3 4 5 6
1 2 3 4
7 4 3
4 1 2 3 4 5 6
1 2 3 4
7 4 4
4 1 2 3 4 5 6
1 2 3 4
11 5 3
9 9 2 2 10 9 7 6 3 6 3
6 9 7 8 10
4 1 1
4 1 5 6
6
output:
4
3
2
4
1

Note: In the first example, all subsegments are good.

In the second example, good subsegments start at positions $$$1$$$, $$$2$$$, and $$$3$$$.

In the third example, good subsegments start at positions $$$1$$$ and $$$2$$$.

Buggy Code:
# # logankeede
# import sys
# from collections import Counter
# import bisect
# print = sys.stdout.write
# input = sys.stdin.readline
#
#
# def solve(st, m):
#     for j in range(n-m):
#         for i in range(m):
#             if arr[j+i]==1:
#
# t = int(input())
# for i in range(t):
#     n = int(input())
#     st = list(map(int, list(input())))
#     m = n
#     while True:
#         solve(m)
mod = 10**6+2
import sys
from collections import Counter
import bisect

print = sys.stdout.write
input = sys.stdin.readline

def check(prea, preb, k,ind,m):
    # print(prea, preb)
    z = 0
    l = set()
    for i in range(m):
        if a[i+ind] not in l:
            l.add(a[i+ind])
            i= a[i+ind]
            z += min(preb[i], prea[i])
        if z >= k:
            # print(l)
            return 1
    # print(l)
    return 0

t = int(input())
for i in range(t):
    n, m, k = map(int, input().split())
    a = list(map(int, input().split()))
    b = list(map(int, input().split()))
    prea = [0]*(mod)
    preb = [0]*(mod)
    count = 0
    for i in b:
        preb[i]+=1
    for i in range(m):
        prea[a[i]]+=1
    for i in range(m, n):
        # print(i)
        count += check(prea, preb, k,i-m,m)
        # print(a[i-m], a[i])
        prea[a[i - m]] -= 1
        prea[a[i]] += 1
    else:
        # print("here")
        count += check(prea, preb, k,n-m,m)
    print(f"{count}\n")

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

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

Think step by step.

Response:
```plaintext
1
5 3 3
1 2 3 4 5
1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Inaccurate Subsequence Search
time_limit_ms: 2000
memory_limit_mb: 256
Description: Maxim has an array $$$a$$$ of $$$n$$$ integers and an array $$$b$$$ of $$$m$$$ integers ($$$m \le n$$$).

Maxim considers an array $$$c$$$ of length $$$m$$$ to be good if the elements of array $$$c$$$ can be rearranged in such a way that at least $$$k$$$ of them match the elements of array $$$b$$$.

For example, if $$$b = [1, 2, 3, 4]$$$ and $$$k = 3$$$, then the arrays $$$[4, 1, 2, 3]$$$ and $$$[2, 3, 4, 5]$$$ are good (they can be reordered as follows: $$$[1, 2, 3, 4]$$$ and $$$[5, 2, 3, 4]$$$), while the arrays $$$[3, 4, 5, 6]$$$ and $$$[3, 4, 3, 4]$$$ are not good.

Maxim wants to choose every subsegment of array $$$a$$$ of length $$$m$$$ as the elements of array $$$c$$$. Help Maxim count how many selected arrays will be good.

In other words, find the number of positions $$$1 \le l \le n - m + 1$$$ such that the elements $$$a_l, a_{l+1}, \dots, a_{l + m - 1}$$$ form a good array.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \le k \le m \le n \le 2 \cdot 10^5$$$) — the number of elements in arrays $$$a$$$ and $$$b$$$, the required number of matching elements.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^6$$$) — the elements of array $$$a$$$. Elements of the array $$$a$$$ are not necessarily unique.

The third line of each test case contains $$$m$$$ integers $$$b_1, b_2, \dots, b_m$$$ ($$$1 \le b_i \le 10^6$$$) — the elements of array $$$b$$$. Elements of the array $$$b$$$ are not necessarily unique.

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

Output: For each test case, output the number of good subsegments of array $$$a$$$ on a separate line.

Examples:
input:
5
7 4 2
4 1 2 3 4 5 6
1 2 3 4
7 4 3
4 1 2 3 4 5 6
1 2 3 4
7 4 4
4 1 2 3 4 5 6
1 2 3 4
11 5 3
9 9 2 2 10 9 7 6 3 6 3
6 9 7 8 10
4 1 1
4 1 5 6
6
output:
4
3
2
4
1

Note: In the first example, all subsegments are good.

In the second example, good subsegments start at positions $$$1$$$, $$$2$$$, and $$$3$$$.

In the third example, good subsegments start at positions $$$1$$$ and $$$2$$$.

Buggy Code:
def good_subsegments(n, m, k, a, b):
    # Initialize frequency counts for array b
    b_freq = {}
    for num in b:
        b_freq[num] = b_freq.get(num, 0) + 1

    # Initialize variables
    good_segments = 0
    window_freq = {}

    # Process the first window
    for i in range(m):
        window_freq[a[i]] = window_freq.get(a[i], 0) + 1

    # Check if the first window is good
    matching_count = sum(min(window_freq.get(num, 0), b_freq[num]) for num in b)
    if matching_count >= k:
        good_segments += 1

    # Slide the window and process remaining segments
    for i in range(1, n - m + 1):
        # Remove the leftmost element from the window
        window_freq[a[i - 1]] -= 1
        if window_freq[a[i - 1]] == 0:
            del window_freq[a[i - 1]]

        # Add the rightmost element to the window
        window_freq[a[i + m - 1]] = window_freq.get(a[i + m - 1], 0) + 1

        # Check if the current window is good
        matching_count = sum(min(window_freq.get(num, 0), b_freq[num]) for num in b)
        if matching_count >= k:
            good_segments += 1

    return good_segments

if __name__ == "__main__":
    t = int(input().strip())
    for _ in range(t):
        n, m, k = map(int, input().split())
        a = list(map(int, input().split()))
        b = list(map(int, input().split()))
        print(good_subsegments(n, m, k, a, b))

**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
10 5 3
1 2 3 4 5 1 2 3 4 5
1 2 3 4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Inaccurate Subsequence Search
time_limit_ms: 2000
memory_limit_mb: 256
Description: Maxim has an array $$$a$$$ of $$$n$$$ integers and an array $$$b$$$ of $$$m$$$ integers ($$$m \le n$$$).

Maxim considers an array $$$c$$$ of length $$$m$$$ to be good if the elements of array $$$c$$$ can be rearranged in such a way that at least $$$k$$$ of them match the elements of array $$$b$$$.

For example, if $$$b = [1, 2, 3, 4]$$$ and $$$k = 3$$$, then the arrays $$$[4, 1, 2, 3]$$$ and $$$[2, 3, 4, 5]$$$ are good (they can be reordered as follows: $$$[1, 2, 3, 4]$$$ and $$$[5, 2, 3, 4]$$$), while the arrays $$$[3, 4, 5, 6]$$$ and $$$[3, 4, 3, 4]$$$ are not good.

Maxim wants to choose every subsegment of array $$$a$$$ of length $$$m$$$ as the elements of array $$$c$$$. Help Maxim count how many selected arrays will be good.

In other words, find the number of positions $$$1 \le l \le n - m + 1$$$ such that the elements $$$a_l, a_{l+1}, \dots, a_{l + m - 1}$$$ form a good array.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \le k \le m \le n \le 2 \cdot 10^5$$$) — the number of elements in arrays $$$a$$$ and $$$b$$$, the required number of matching elements.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^6$$$) — the elements of array $$$a$$$. Elements of the array $$$a$$$ are not necessarily unique.

The third line of each test case contains $$$m$$$ integers $$$b_1, b_2, \dots, b_m$$$ ($$$1 \le b_i \le 10^6$$$) — the elements of array $$$b$$$. Elements of the array $$$b$$$ are not necessarily unique.

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

Output: For each test case, output the number of good subsegments of array $$$a$$$ on a separate line.

Examples:
input:
5
7 4 2
4 1 2 3 4 5 6
1 2 3 4
7 4 3
4 1 2 3 4 5 6
1 2 3 4
7 4 4
4 1 2 3 4 5 6
1 2 3 4
11 5 3
9 9 2 2 10 9 7 6 3 6 3
6 9 7 8 10
4 1 1
4 1 5 6
6
output:
4
3
2
4
1

Note: In the first example, all subsegments are good.

In the second example, good subsegments start at positions $$$1$$$, $$$2$$$, and $$$3$$$.

In the third example, good subsegments start at positions $$$1$$$ and $$$2$$$.

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;

/**
 * @author YC
 * @version 1.0
 */
public class Main {
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static int n,m,k;
    static int N = (int) (2e5 + 10);
    static int M = (int) (1e6 + 10);
    static int[] a = new int[N];
    static int[] b = new int[N];
    static int[] cntB = new int[M];
    static int[] cntC = new int[M];
    public static void main(String[] args) throws IOException {
        int t = Integer.parseInt(br.readLine());
        while(t -- > 0) {
            String[] strs = br.readLine().split(" ");
            n = Integer.parseInt(strs[0]);
            m = Integer.parseInt(strs[1]);
            k = Integer.parseInt(strs[2]);

            strs = br.readLine().split(" ");
            for(int i = 1 ; i <= n ; i ++) {
                a[i] = Integer.parseInt(strs[i - 1]);
            }
            strs = br.readLine().split(" ");
            for(int i = 1 ; i <= m ; i ++) {
                b[i] = Integer.parseInt(strs[i - 1]);
                cntB[b[i]] ++;
            }

            // 统计有多少个数字出现了
            int cnt = 0;
            int ans = 0;
            for(int i = 1, j = 1; j <= n ; j ++) {
                int num = a[j];
                if(cntC[num] == 0 && cntB[num] != 0) {
                    cnt++;
//                    System.out.println(cnt + " " + a[i]);
                }
                cntC[num] ++;
                if(j - i + 1 > m) {
                    cntC[a[i]] --;
                    if(cntC[a[i]] == 0 && cntB[a[i]] != 0) cnt --;
                    i ++;
                }
                // 1. 长度为m 2.出现的元素个数要>=k
                if(j - i + 1 == m && cnt >= k) {
                    ans ++;
                }
//                System.out.println(cnt);
            }
            System.out.println(ans);
            for(int i = 1 ; i <= m ; i ++) cntB[b[i]] = 0;
            for(int i = 1 ; i <= n ; i ++) cntC[a[i]] = 0;

        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases that might cause the algorithm to fail. The primary issue in the buggy code seems to be the incorrect handling of the count of matching elements. Specifically, the code might not correctly track the number of unique elements that match the elements in array `b` when the window slides.

Here are the steps to generate a challenging test case:

1. **Choose a small `n` and `m`**: To ensure the test case is small and manageable, let's choose `n = 10` and `m = 5`.
2. **Set `k` to a value that is challenging**: Let's set `k = 3` to ensure that the subsegments need to have at least 3 matching elements.
3. **Create arrays `a` and `b` with elements that can cause the algorithm to fail**:
   - `a` should have elements that are repeated and some that are not in `b`.
   - `b` should have elements that are repeated and some that are not in `a`.

Here is a small-scale test input that meets these criteria:

```plaintext
1
10 5 3
1 2 3 4 5 1 2 3 4 5
1 2 3 6 7
```

Explanation:
- `n = 10`, `m = 5`, `k = 3`
- `a = [1, 2, 3, 4, 5, 1, 2, 3, 4, 5]`
- `b = [1, 2, 3, 6, 7]`

This test case is designed to:
- Have multiple subsegments of length 5 in `a` that match at least 3 elements in `b`.
- Include elements in `a` that are not in `b` to test the algorithm's handling of non-matching elements.
- Use repeated elements to test the algorithm's ability to correctly count unique matching elements.

This input should expose any issues in the buggy code related to the counting of matching elements and the sliding window mechanism.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Inaccurate Subsequence Search
time_limit_ms: 2000
memory_limit_mb: 256
Description: Maxim has an array $$$a$$$ of $$$n$$$ integers and an array $$$b$$$ of $$$m$$$ integers ($$$m \le n$$$).

Maxim considers an array $$$c$$$ of length $$$m$$$ to be good if the elements of array $$$c$$$ can be rearranged in such a way that at least $$$k$$$ of them match the elements of array $$$b$$$.

For example, if $$$b = [1, 2, 3, 4]$$$ and $$$k = 3$$$, then the arrays $$$[4, 1, 2, 3]$$$ and $$$[2, 3, 4, 5]$$$ are good (they can be reordered as follows: $$$[1, 2, 3, 4]$$$ and $$$[5, 2, 3, 4]$$$), while the arrays $$$[3, 4, 5, 6]$$$ and $$$[3, 4, 3, 4]$$$ are not good.

Maxim wants to choose every subsegment of array $$$a$$$ of length $$$m$$$ as the elements of array $$$c$$$. Help Maxim count how many selected arrays will be good.

In other words, find the number of positions $$$1 \le l \le n - m + 1$$$ such that the elements $$$a_l, a_{l+1}, \dots, a_{l + m - 1}$$$ form a good array.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \le k \le m \le n \le 2 \cdot 10^5$$$) — the number of elements in arrays $$$a$$$ and $$$b$$$, the required number of matching elements.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^6$$$) — the elements of array $$$a$$$. Elements of the array $$$a$$$ are not necessarily unique.

The third line of each test case contains $$$m$$$ integers $$$b_1, b_2, \dots, b_m$$$ ($$$1 \le b_i \le 10^6$$$) — the elements of array $$$b$$$. Elements of the array $$$b$$$ are not necessarily unique.

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

Output: For each test case, output the number of good subsegments of array $$$a$$$ on a separate line.

Examples:
input:
5
7 4 2
4 1 2 3 4 5 6
1 2 3 4
7 4 3
4 1 2 3 4 5 6
1 2 3 4
7 4 4
4 1 2 3 4 5 6
1 2 3 4
11 5 3
9 9 2 2 10 9 7 6 3 6 3
6 9 7 8 10
4 1 1
4 1 5 6
6
output:
4
3
2
4
1

Note: In the first example, all subsegments are good.

In the second example, good subsegments start at positions $$$1$$$, $$$2$$$, and $$$3$$$.

In the third example, good subsegments start at positions $$$1$$$ and $$$2$$$.

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

public class Main {
    static RealFastReader sc = new RealFastReader(System.in);
    static PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) {
        //int n=sc.ni();
        //double d=sc.nd();
        //String s=sc.ns();
        int T=sc.ni();
        while(T-->0){
            solve();
        }

        out.close();
    }

    public static void solve() {
        int n=sc.ni();
        int m=sc.ni();
        int k=sc.ni();
        int arr[]=new int[n];
        int ar[]=new int[m];
        int max=0;
        for(int i=0;i<n;i++){
            arr[i]=sc.ni();
            max=Math.max(max,arr[i]);
        }
        for(int i=0;i<m;i++){
            ar[i]=sc.ni();
            max=Math.max(ar[i],max);
        }

        int cnt[]=new int[max+2];
        int nums[]=new int[max+2];
        for(int a:ar) cnt[a]++;
        int ans=0;

        for(int i=0;i<m;i++){
            nums[arr[i]]++;
            if(cnt[arr[i]]>=nums[arr[i]]) ans+=1;
        }
        //System.out.println(ans);
        int res=ans>=k?1:0;

        int l=0;
        for(int i=m;i<n;i++){
            int a=arr[i];
            nums[arr[l]]--;
            if(cnt[arr[l]]!=0&&cnt[arr[l]]!=nums[arr[l]]) ans--;
            nums[a]++;
            if(cnt[a]!=0&&cnt[a]>=nums[a]) ans++;
            //System.out.println(ans);
            if(ans>=k) res++;
            l++;
        }
        System.out.println(res);

    }

    //快读快写---------------------------------------------------------------------------------
    public static class RealFastReader {
        InputStream is;

        public RealFastReader(final InputStream is) {
            this.is = is;
        }

        private byte[] inbuf = new byte[8192];
        public int lenbuf = 0, ptrbuf = 0;

        public int readByte() {
            if (lenbuf == -1) {
                throw new InputMismatchException();
            }
            if (ptrbuf >= lenbuf) {
                ptrbuf = 0;
                try {
                    lenbuf = is.read(inbuf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (lenbuf <= 0) {
                    return -1;
                }
            }
            return inbuf[ptrbuf++];
        }

        private boolean isSpaceChar(int c) {
            return !(c >= 33 && c <= 126);
        }

        private int skip() {
            int b;
            while ((b = readByte()) != -1 && isSpaceChar(b)) ;
            return b;
        }

        public double nd() {
            return Double.parseDouble(ns());
        }

        public char nc() {
            return (char) skip();
        }

        public String ns() {
            int b = skip();
            StringBuilder sb = new StringBuilder();
            while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != ' ')
                sb.appendCodePoint(b);
                b = readByte();
            }
            return sb.toString();
        }

        public char[] ns(int n) {
            char[] buf = new char[n];
            int b = skip(), p = 0;
            while (p < n && !(isSpaceChar(b))) {
                buf[p++] = (char) b;
                b = readByte();
            }
            return n == p ? buf : Arrays.copyOf(buf, p);
        }

        public int[] na(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = ni();
            }
            return a;
        }

        public long[] nal(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = nl();
            }
            return a;
        }

        public char[][] nm(int n, int m) {
            char[][] map = new char[n][];
            for (int i = 0; i < n; i++) {
                map[i] = ns(m);
            }
            return map;
        }

        public int[][] nmi(int n, int m) {
            int[][] map = new int[n][];
            for (int i = 0; i < n; i++) {
                map[i] = na(m);
            }
            return map;
        }

        public int ni() {
            int num = 0;
            int b;
            boolean minus = false;
            while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
                ;
            if (b == '-') {
                minus = true;
                b = readByte();
            }

            while (true) {
                if (b >= '0' && b <= '9') {
                    num = num * 10 + (b - '0');
                } else {
                    return minus ? -num : num;
                }
                b = readByte();
            }
        }

        public long nl() {
            long num = 0;
            int b;
            boolean minus = false;
            while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
                ;
            if (b == '-') {
                minus = true;
                b = readByte();
            }

            while (true) {
                if (b >= '0' && b <= '9') {
                    num = num * 10 + (b - '0');
                } else {
                    return minus ? -num : num;
                }
                b = readByte();
            }
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
1
5 3 2
1 2 3 4 5
3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Inaccurate Subsequence Search
time_limit_ms: 2000
memory_limit_mb: 256
Description: Maxim has an array $$$a$$$ of $$$n$$$ integers and an array $$$b$$$ of $$$m$$$ integers ($$$m \le n$$$).

Maxim considers an array $$$c$$$ of length $$$m$$$ to be good if the elements of array $$$c$$$ can be rearranged in such a way that at least $$$k$$$ of them match the elements of array $$$b$$$.

For example, if $$$b = [1, 2, 3, 4]$$$ and $$$k = 3$$$, then the arrays $$$[4, 1, 2, 3]$$$ and $$$[2, 3, 4, 5]$$$ are good (they can be reordered as follows: $$$[1, 2, 3, 4]$$$ and $$$[5, 2, 3, 4]$$$), while the arrays $$$[3, 4, 5, 6]$$$ and $$$[3, 4, 3, 4]$$$ are not good.

Maxim wants to choose every subsegment of array $$$a$$$ of length $$$m$$$ as the elements of array $$$c$$$. Help Maxim count how many selected arrays will be good.

In other words, find the number of positions $$$1 \le l \le n - m + 1$$$ such that the elements $$$a_l, a_{l+1}, \dots, a_{l + m - 1}$$$ form a good array.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \le k \le m \le n \le 2 \cdot 10^5$$$) — the number of elements in arrays $$$a$$$ and $$$b$$$, the required number of matching elements.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^6$$$) — the elements of array $$$a$$$. Elements of the array $$$a$$$ are not necessarily unique.

The third line of each test case contains $$$m$$$ integers $$$b_1, b_2, \dots, b_m$$$ ($$$1 \le b_i \le 10^6$$$) — the elements of array $$$b$$$. Elements of the array $$$b$$$ are not necessarily unique.

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

Output: For each test case, output the number of good subsegments of array $$$a$$$ on a separate line.

Examples:
input:
5
7 4 2
4 1 2 3 4 5 6
1 2 3 4
7 4 3
4 1 2 3 4 5 6
1 2 3 4
7 4 4
4 1 2 3 4 5 6
1 2 3 4
11 5 3
9 9 2 2 10 9 7 6 3 6 3
6 9 7 8 10
4 1 1
4 1 5 6
6
output:
4
3
2
4
1

Note: In the first example, all subsegments are good.

In the second example, good subsegments start at positions $$$1$$$, $$$2$$$, and $$$3$$$.

In the third example, good subsegments start at positions $$$1$$$ and $$$2$$$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long int
#define pb push_back
#define vi vector<int>
#define vvi vector<vector<int>>
#define pi pair<int, int>
#define vpi vector<pi>
#define vvpi vector<vector<pi>>
#define a_sort(v) sort(v.begin(), v.end())
#define d_sort(v) sort(v.rbegin(), v.rend())
#define ulta(v) reverse(v.begin(), v.end())
#define endl '\n'
#define read(x) int x; cin >> x
#define fast_io ios_base::sync_with_stdio(false); cin.tie(NULL)
#define mod 1000000007

signed main(){
    fast_io;
    read(t);
    while(t--){
        read(n);
        read(m);
        read(k);

        vi a(n);
        vi b(m);

        map<int, int> mpb;
        map<int, int> mpa;
        for(int i = 0; i < n; i++) cin >> a[i];
        for(int i = 0; i < m; i++){
            cin >> b[i];
            mpb[b[i]]++;
        }

        int i = 0;
        int l = 0;
        for(; i < m; i++){
            mpa[a[i]]++;
        }

        int num = 0;
        vi inc(1000001, 0);

        for(auto i : mpb){
            int val = i.first;
            int freq = i.second;
            if(mpa[val] >= freq){
                num += freq;
                inc[val] = 1;
            }
        }

        // cout << num << endl;

        int ans = 0;
        if(num >= k) ans++;

        for(; i < n; i++){
            mpa[a[l]]--;
            if(inc[a[l]] && mpa[a[l]] < mpb[a[l]]){
                inc[a[l]] = 0;
                num -= mpb[a[l]];
            }

            mpa[a[i]]++;
            if(mpb[a[i]] && mpa[a[i]] >= mpb[a[i]] && inc[a[i]] == 0){
                num += mpb[a[i]];
                inc[a[i]] = 1;
                // cout << a[i] << endl;
            }

            if(num >= k) ans++;
            l++;
        }

        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 3 3
1 2 3 4 5
1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Inaccurate Subsequence Search
time_limit_ms: 2000
memory_limit_mb: 256
Description: Maxim has an array $$$a$$$ of $$$n$$$ integers and an array $$$b$$$ of $$$m$$$ integers ($$$m \le n$$$).

Maxim considers an array $$$c$$$ of length $$$m$$$ to be good if the elements of array $$$c$$$ can be rearranged in such a way that at least $$$k$$$ of them match the elements of array $$$b$$$.

For example, if $$$b = [1, 2, 3, 4]$$$ and $$$k = 3$$$, then the arrays $$$[4, 1, 2, 3]$$$ and $$$[2, 3, 4, 5]$$$ are good (they can be reordered as follows: $$$[1, 2, 3, 4]$$$ and $$$[5, 2, 3, 4]$$$), while the arrays $$$[3, 4, 5, 6]$$$ and $$$[3, 4, 3, 4]$$$ are not good.

Maxim wants to choose every subsegment of array $$$a$$$ of length $$$m$$$ as the elements of array $$$c$$$. Help Maxim count how many selected arrays will be good.

In other words, find the number of positions $$$1 \le l \le n - m + 1$$$ such that the elements $$$a_l, a_{l+1}, \dots, a_{l + m - 1}$$$ form a good array.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \le k \le m \le n \le 2 \cdot 10^5$$$) — the number of elements in arrays $$$a$$$ and $$$b$$$, the required number of matching elements.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^6$$$) — the elements of array $$$a$$$. Elements of the array $$$a$$$ are not necessarily unique.

The third line of each test case contains $$$m$$$ integers $$$b_1, b_2, \dots, b_m$$$ ($$$1 \le b_i \le 10^6$$$) — the elements of array $$$b$$$. Elements of the array $$$b$$$ are not necessarily unique.

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

Output: For each test case, output the number of good subsegments of array $$$a$$$ on a separate line.

Examples:
input:
5
7 4 2
4 1 2 3 4 5 6
1 2 3 4
7 4 3
4 1 2 3 4 5 6
1 2 3 4
7 4 4
4 1 2 3 4 5 6
1 2 3 4
11 5 3
9 9 2 2 10 9 7 6 3 6 3
6 9 7 8 10
4 1 1
4 1 5 6
6
output:
4
3
2
4
1

Note: In the first example, all subsegments are good.

In the second example, good subsegments start at positions $$$1$$$, $$$2$$$, and $$$3$$$.

In the third example, good subsegments start at positions $$$1$$$ and $$$2$$$.

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

using namespace::std;
using namespace __gnu_pbds;

#define ll long long int
typedef set<ll> si;
typedef multiset<ll> mi;
typedef vector<ll> vi;
typedef pair<ll,ll> pi;
typedef vector<pair<ll,ll>> vp;
typedef tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update>  os;
#define all(v) v.begin(),v.end()
#define mod 1000000007
#define mod2 998244353
#define ub upper_bound
#define lb lower_bound
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define fast ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define fab(i,a,b) for(ll i = a; i<b; i++)
#define fn(i,n) for(ll i = 0; i<n; i++)
#define fabc(i,a,b,c) for(ll i = a; i<c*(b/c) || i>c*(b/c); i+=c)
#define fc(i,n,cond) for(ll i = 0; i<n && cond; i++)
#define fa(i,a) for(auto i : a)
#define no cout<<"NO";return;
#define yes cout<<"YES";return;
#define nl cout<<"\n"

ll gcd(ll a, ll b){return (ll)b == 0 ? (ll)a : gcd((ll)b, (ll)a % (ll)b);}
ll count(int arr[], ll n, ll ran){
    ll coun = 0;
    fn(i,ran){if(arr[i]==n){coun++;}}
    return coun;
}
ll power(ll a, ll b, ll modi){
    ll res = 1;
    while(b > 0){
    if(b&1)
        res = (res * a) % modi;
    a = (a * a) % modi;
    b >>= 1;
    }
    return res%modi;
}
ll modinv(ll x, ll modi){return power(x, modi - 2, modi);}
void arrinp(ll c[], ll n){fn(i,n){cin>>c[i];}}

void solve(){
    ll n, k, m;
    cin>>n>>m>>k;
    ll a[n];
    ll b[m];
    arrinp(a, n);
    arrinp(b, m);

    map<ll, ll> m1, m2;

    fn(i,m){m1[b[i]]++;m2[b[i]]++;}

    fn(i,m){m1[a[i]]--;}

    ll co = 0, d = 0;

    fn(i,m){d+=max((ll)0, m1[b[i]]);}

    // fn(i,m){cout<<b[i]<<" "<<m2[b[i]];nl;}nl;

    fn(i,n-m){
        if(d<=m-k){co++;}
        // cout<<d<<"  ";
        // fn(i,m){cout<<m1[b[i]]<<" ";}nl;
        if(a[i]==a[i+m]){continue;}
        if(m1[a[i]]==0){d++;}
        if(m1[a[m+i]]>=1){d--;}
        m1[a[i]]++;
        // m1[a[i]] = max(m1[a[i]], m2[a[i]]);
        m1[a[m+i]]--;
    }
    // cout<<d<<"  ";fn(i,m){cout<<m1[b[i]]<<" ";}nl;
    if(d<=m-k){co++;}

    cout<<co;

}

int main(){
    fast;
    // freopen("input.txt","r",stdin);freopen("output.txt","w",stdout);
    ll tc = 1;
    cin>>tc;
    while(tc--){solve();nl;}
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
5 3 2
1 2 3 4 5
3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Inaccurate Subsequence Search
time_limit_ms: 2000
memory_limit_mb: 256
Description: Maxim has an array $$$a$$$ of $$$n$$$ integers and an array $$$b$$$ of $$$m$$$ integers ($$$m \le n$$$).

Maxim considers an array $$$c$$$ of length $$$m$$$ to be good if the elements of array $$$c$$$ can be rearranged in such a way that at least $$$k$$$ of them match the elements of array $$$b$$$.

For example, if $$$b = [1, 2, 3, 4]$$$ and $$$k = 3$$$, then the arrays $$$[4, 1, 2, 3]$$$ and $$$[2, 3, 4, 5]$$$ are good (they can be reordered as follows: $$$[1, 2, 3, 4]$$$ and $$$[5, 2, 3, 4]$$$), while the arrays $$$[3, 4, 5, 6]$$$ and $$$[3, 4, 3, 4]$$$ are not good.

Maxim wants to choose every subsegment of array $$$a$$$ of length $$$m$$$ as the elements of array $$$c$$$. Help Maxim count how many selected arrays will be good.

In other words, find the number of positions $$$1 \le l \le n - m + 1$$$ such that the elements $$$a_l, a_{l+1}, \dots, a_{l + m - 1}$$$ form a good array.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \le k \le m \le n \le 2 \cdot 10^5$$$) — the number of elements in arrays $$$a$$$ and $$$b$$$, the required number of matching elements.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^6$$$) — the elements of array $$$a$$$. Elements of the array $$$a$$$ are not necessarily unique.

The third line of each test case contains $$$m$$$ integers $$$b_1, b_2, \dots, b_m$$$ ($$$1 \le b_i \le 10^6$$$) — the elements of array $$$b$$$. Elements of the array $$$b$$$ are not necessarily unique.

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

Output: For each test case, output the number of good subsegments of array $$$a$$$ on a separate line.

Examples:
input:
5
7 4 2
4 1 2 3 4 5 6
1 2 3 4
7 4 3
4 1 2 3 4 5 6
1 2 3 4
7 4 4
4 1 2 3 4 5 6
1 2 3 4
11 5 3
9 9 2 2 10 9 7 6 3 6 3
6 9 7 8 10
4 1 1
4 1 5 6
6
output:
4
3
2
4
1

Note: In the first example, all subsegments are good.

In the second example, good subsegments start at positions $$$1$$$, $$$2$$$, and $$$3$$$.

In the third example, good subsegments start at positions $$$1$$$ and $$$2$$$.

Buggy Code:
from collections import Counter

def goodSequences(left, right):
    count = 0
    inSet = 0
    freq = Counter()

    for i in range(left, right + 1):
        if a[i] in b:
            if freq[a[i]] == 0:
                inSet += 1
            freq[a[i]] += 1

    if inSet >= k:
        count += 1

    while right < n - 1:
        outgoing = a[left]
        incoming = a[right + 1]

        if outgoing in b:
            freq[outgoing] -= 1
            if freq[outgoing] == 0:
                inSet -= 1
        left += 1

        if incoming in b:
            if freq[incoming] == 0:
                inSet += 1
            freq[incoming] += 1
        right += 1

        if inSet >= k:
            count += 1

    return count

if __name__ == "__main__":
    t = int(input())
    results = []
    for _ in range(t):
        n, m, k = map(int, input().split())
        a = list(map(int, input().split()))
        b = set(map(int, input().split()))
        results.append(goodSequences(0, m - 1))

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

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

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

Think step by step.

Response:
```plaintext
3
5 3 3
1 2 3 4 5
1 2 3
5 3 2
1 2 3 4 5
1 2 3
5 3 1
1 2 3 4 5
1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Inaccurate Subsequence Search
time_limit_ms: 2000
memory_limit_mb: 256
Description: Maxim has an array $$$a$$$ of $$$n$$$ integers and an array $$$b$$$ of $$$m$$$ integers ($$$m \le n$$$).

Maxim considers an array $$$c$$$ of length $$$m$$$ to be good if the elements of array $$$c$$$ can be rearranged in such a way that at least $$$k$$$ of them match the elements of array $$$b$$$.

For example, if $$$b = [1, 2, 3, 4]$$$ and $$$k = 3$$$, then the arrays $$$[4, 1, 2, 3]$$$ and $$$[2, 3, 4, 5]$$$ are good (they can be reordered as follows: $$$[1, 2, 3, 4]$$$ and $$$[5, 2, 3, 4]$$$), while the arrays $$$[3, 4, 5, 6]$$$ and $$$[3, 4, 3, 4]$$$ are not good.

Maxim wants to choose every subsegment of array $$$a$$$ of length $$$m$$$ as the elements of array $$$c$$$. Help Maxim count how many selected arrays will be good.

In other words, find the number of positions $$$1 \le l \le n - m + 1$$$ such that the elements $$$a_l, a_{l+1}, \dots, a_{l + m - 1}$$$ form a good array.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \le k \le m \le n \le 2 \cdot 10^5$$$) — the number of elements in arrays $$$a$$$ and $$$b$$$, the required number of matching elements.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^6$$$) — the elements of array $$$a$$$. Elements of the array $$$a$$$ are not necessarily unique.

The third line of each test case contains $$$m$$$ integers $$$b_1, b_2, \dots, b_m$$$ ($$$1 \le b_i \le 10^6$$$) — the elements of array $$$b$$$. Elements of the array $$$b$$$ are not necessarily unique.

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

Output: For each test case, output the number of good subsegments of array $$$a$$$ on a separate line.

Examples:
input:
5
7 4 2
4 1 2 3 4 5 6
1 2 3 4
7 4 3
4 1 2 3 4 5 6
1 2 3 4
7 4 4
4 1 2 3 4 5 6
1 2 3 4
11 5 3
9 9 2 2 10 9 7 6 3 6 3
6 9 7 8 10
4 1 1
4 1 5 6
6
output:
4
3
2
4
1

Note: In the first example, all subsegments are good.

In the second example, good subsegments start at positions $$$1$$$, $$$2$$$, and $$$3$$$.

In the third example, good subsegments start at positions $$$1$$$ and $$$2$$$.

Buggy Code:
for _ in range(int(input())):
    n,m,k=map(int,input().split())
    a=list(map(int,input().split()))
    b=list(map(int,input().split()))
    d={}
    for i in b:
        if i in d:
            d[i]+=1
        else:
            d[i]=1
    d1={}
    for i in range(m):
        if a[i] in d1:
            d1[a[i]]+=1
        else:
            d1[a[i]]=1
    res=0
    count=0
    for i in d:
        if i in d1:
            if d1[i]>=d[i]:
                count+=d[i]
    if count>=k:
        res+=1
    for i in range(m,n):
        if a[i]!=a[i-m]:
            if d1[a[i-m]]==d.get(a[i-m],0):
                count-=1
            d1[a[i-m]]-=1
            if a[i] in d1:
                d1[a[i]]+=1
            else:
                d1[a[i]]=1
            if d1[a[i]]==d.get(a[i],0):
                count+=1
        if count>=k:
            res+=1
    print(res)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Inaccurate Subsequence Search
time_limit_ms: 2000
memory_limit_mb: 256
Description: Maxim has an array $$$a$$$ of $$$n$$$ integers and an array $$$b$$$ of $$$m$$$ integers ($$$m \le n$$$).

Maxim considers an array $$$c$$$ of length $$$m$$$ to be good if the elements of array $$$c$$$ can be rearranged in such a way that at least $$$k$$$ of them match the elements of array $$$b$$$.

For example, if $$$b = [1, 2, 3, 4]$$$ and $$$k = 3$$$, then the arrays $$$[4, 1, 2, 3]$$$ and $$$[2, 3, 4, 5]$$$ are good (they can be reordered as follows: $$$[1, 2, 3, 4]$$$ and $$$[5, 2, 3, 4]$$$), while the arrays $$$[3, 4, 5, 6]$$$ and $$$[3, 4, 3, 4]$$$ are not good.

Maxim wants to choose every subsegment of array $$$a$$$ of length $$$m$$$ as the elements of array $$$c$$$. Help Maxim count how many selected arrays will be good.

In other words, find the number of positions $$$1 \le l \le n - m + 1$$$ such that the elements $$$a_l, a_{l+1}, \dots, a_{l + m - 1}$$$ form a good array.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \le k \le m \le n \le 2 \cdot 10^5$$$) — the number of elements in arrays $$$a$$$ and $$$b$$$, the required number of matching elements.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^6$$$) — the elements of array $$$a$$$. Elements of the array $$$a$$$ are not necessarily unique.

The third line of each test case contains $$$m$$$ integers $$$b_1, b_2, \dots, b_m$$$ ($$$1 \le b_i \le 10^6$$$) — the elements of array $$$b$$$. Elements of the array $$$b$$$ are not necessarily unique.

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

Output: For each test case, output the number of good subsegments of array $$$a$$$ on a separate line.

Examples:
input:
5
7 4 2
4 1 2 3 4 5 6
1 2 3 4
7 4 3
4 1 2 3 4 5 6
1 2 3 4
7 4 4
4 1 2 3 4 5 6
1 2 3 4
11 5 3
9 9 2 2 10 9 7 6 3 6 3
6 9 7 8 10
4 1 1
4 1 5 6
6
output:
4
3
2
4
1

Note: In the first example, all subsegments are good.

In the second example, good subsegments start at positions $$$1$$$, $$$2$$$, and $$$3$$$.

In the third example, good subsegments start at positions $$$1$$$ and $$$2$$$.

Buggy Code:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.*;
import java.io.BufferedReader;
import java.io.InputStreamReader;

public class D {
    public static void main(String[] args) throws IOException {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Task solver = new Task();
        int testCount = Integer.parseInt(in.next());
        for (int i = 1; i <= testCount; i++)
            solver.solve(i, in, out);
        out.close();
    }

    static class Task {
        static final long MODULO = (long)(1e9+7);
        int sz = 1000001;
        int[] fa;
        int[] fb;
        public void solve(int testNumber, InputReader in, PrintWriter out) throws IOException {
            int n = in.nextInt(), m = in.nextInt(), k = in.nextInt();
            int[] a = in.readArray(n);
            int[] b = in.readArray(m);
            fa = new int[sz];
            fb = new int[sz];
            int cnt = 0;
            for (int i = 0; i < m; i++) {
                fb[b[i]]++;
            }
            for (int i = 0; i < m; i++) {
                fa[a[i]]++;
                if (fa[a[i]] <= fb[a[i]]) cnt++;
            }
            int ans = 0;
            if (cnt >= k) ans++;
            for (int i = m; i < n; i++) {
                if (--fa[a[i - m]] < fb[a[i - m]]) cnt--;
                fa[a[i]]++;
                if (fa[a[i]] <= fb[a[i]]) cnt++;
                if (cnt >= k) ans++;
            }
            out.println(ans);
        }
    }

    static int lowerBound(int[] a, int L, int R, int x) {
        int ans = R;
        while (L <= R) {
            int mid = (L + R) / 2;
            if (a[mid] >= x) {
                R = mid - 1;
                ans = mid;
            }
            else L = mid + 1;
        }
        return ans;
    }

    static int upperBound(int[] a, int L, int R, int x) {
        int ans = R;
        while (L <= R) {
            int mid = (L + R) / 2;
            if (a[mid] > x) {
                ans = mid;
                R = mid - 1;
            } else {
                L = mid + 1;
            }
        }
        return ans;
    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

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

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

        int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider edge cases and scenarios that maximize the failure probability. The bug in the code seems to be related to how it handles the matching of elements and the counting of good subsegments. Specifically, the code might not correctly handle cases where elements in `b` are repeated and the subsegments of `a` have multiple occurrences of elements that match `b`.

Here's a step-by-step approach to generating a challenging test input:

1. **Small-scale Test Case**: Since we need a small-scale test input, we will use a small number of test cases and arrays of moderate length.
2. **Edge Cases**:
   - **Repetitive Elements in `b`**: Ensure that `b` contains repeated elements to test how the code handles multiple occurrences.
   - **Repetitive Elements in `a`**: Ensure that `a` contains multiple occurrences of elements that match elements in `b` to test the counting mechanism.
   - ** borderline `k` values**: Test with `k` values that are just below or just above the number of matching elements to see if the code handles these cases correctly.

Let's create a test input based on these considerations:

```plaintext
3
5 3 2
1 2 3 2 1
1 2 3
5 3 3
1 2 3 2 1
1 2 3
6 4 2
1 2 3 4 2 1
1 2 3 4
```

### Explanation:
1. **Test Case 1**:
   - `n = 5`, `m = 3`, `k = 2`
   - `a = [1, 2, 3, 2, 1]`
   - `b = [1, 2, 3]`
   - **Expected Output**: `3`
   - **Explanation**: The subsegments `[1, 2, 3]`, `[2, 3, 2]`, and `[3, 2, 1]` are good.

2. **Test Case 2**:
   - `n = 5`, `m = 3`, `k = 3`
   - `a = [1, 2, 3, 2, 1]`
   - `b = [1, 2, 3]`
   - **Expected Output**: `1`
   - **Explanation**: Only the subsegment `[1, 2, 3]` is good.

3. **Test Case 3**:
   - `n = 6`, `m = 4`, `k = 2`
   - `a = [1, 2, 3, 4, 2, 1]`
   - `b = [1, 2, 3, 4]`
   - **Expected Output**: `3`
   - **Explanation**: The subsegments `[1, 2, 3, 4]`, `[2, 3, 4, 2]`, and `[3, 4, 2, 1]` are good.

This test input is designed to challenge the code by including repeated elements and testing the boundary conditions for `k`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Inaccurate Subsequence Search
time_limit_ms: 2000
memory_limit_mb: 256
Description: Maxim has an array $$$a$$$ of $$$n$$$ integers and an array $$$b$$$ of $$$m$$$ integers ($$$m \le n$$$).

Maxim considers an array $$$c$$$ of length $$$m$$$ to be good if the elements of array $$$c$$$ can be rearranged in such a way that at least $$$k$$$ of them match the elements of array $$$b$$$.

For example, if $$$b = [1, 2, 3, 4]$$$ and $$$k = 3$$$, then the arrays $$$[4, 1, 2, 3]$$$ and $$$[2, 3, 4, 5]$$$ are good (they can be reordered as follows: $$$[1, 2, 3, 4]$$$ and $$$[5, 2, 3, 4]$$$), while the arrays $$$[3, 4, 5, 6]$$$ and $$$[3, 4, 3, 4]$$$ are not good.

Maxim wants to choose every subsegment of array $$$a$$$ of length $$$m$$$ as the elements of array $$$c$$$. Help Maxim count how many selected arrays will be good.

In other words, find the number of positions $$$1 \le l \le n - m + 1$$$ such that the elements $$$a_l, a_{l+1}, \dots, a_{l + m - 1}$$$ form a good array.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \le k \le m \le n \le 2 \cdot 10^5$$$) — the number of elements in arrays $$$a$$$ and $$$b$$$, the required number of matching elements.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^6$$$) — the elements of array $$$a$$$. Elements of the array $$$a$$$ are not necessarily unique.

The third line of each test case contains $$$m$$$ integers $$$b_1, b_2, \dots, b_m$$$ ($$$1 \le b_i \le 10^6$$$) — the elements of array $$$b$$$. Elements of the array $$$b$$$ are not necessarily unique.

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

Output: For each test case, output the number of good subsegments of array $$$a$$$ on a separate line.

Examples:
input:
5
7 4 2
4 1 2 3 4 5 6
1 2 3 4
7 4 3
4 1 2 3 4 5 6
1 2 3 4
7 4 4
4 1 2 3 4 5 6
1 2 3 4
11 5 3
9 9 2 2 10 9 7 6 3 6 3
6 9 7 8 10
4 1 1
4 1 5 6
6
output:
4
3
2
4
1

Note: In the first example, all subsegments are good.

In the second example, good subsegments start at positions $$$1$$$, $$$2$$$, and $$$3$$$.

In the third example, good subsegments start at positions $$$1$$$ and $$$2$$$.

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
//import java.util.ArrayList;
//import java.util.HashMap;
//import java.util.StringTokenizer;
import java.util.*;

public class D {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

        String nextLine()
        {
            String str = "";
            try {
                if(st.hasMoreTokens()){
                    str = st.nextToken("\n");
                }
                else{
                    str = br.readLine();
                }
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
    public static void main(String[] args)
    {
        FastReader sc= new FastReader();
        int t = sc.nextInt();
        while (t != 0) {
            t--;
            int n = sc.nextInt();
            int m = sc.nextInt();
            int k = sc.nextInt();
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = sc.nextInt();
            }
            int[] b = new int[m];
            for (int i = 0; i < m; i++) {
                b[i] = sc.nextInt();
            }
            HashMap<Integer,Integer> hma=new HashMap<>();
            HashMap<Integer,Integer> hmb=new HashMap<>();
            ArrayList<Integer> vis=new ArrayList<>();
            for(int i = 0; i < m; i++) {
               if (hma.containsKey(a[i])==false)
               {
                   hma.put(a[i],1);
               }
               else{
                   int f1 = hma.get(a[i]);
                   ++f1;
                   hma.put(a[i], f1);
               }

                if(hmb.containsKey(b[i])==false)
                {
                    hmb.put(b[i],1);
                }
                else{
                    int f2=hmb.get(b[i]);
                    ++f2;
                    hmb.put(b[i],f2);
                }
            }

        int c=0,ans=0;
            for (int i = 0;i < m; i++) {
                if (vis.contains(a[i]) && hma.get(a[i])>hmb.get(a[i])){
                    continue;
                }
                if (hmb.get(a[i])!=null){
                    c+=Math.min(hma.get(a[i]),hmb.get(a[i])) ;
                    vis.add(a[i]);
                }
            }

   if (c>=k){
       ++ans;
   }
//         System.out.println(c);
//   System.out.println(ans);
           int idx=0;
            for (int i = 1; i <=n-m ; i++) {
//              if (idx<vis.size() && a[i-1]==vis.get(idx)){
//                       vis.remove(idx);
//                       ++idx;
//              }
                 int f1=hma.get(a[i-1]);
                 if (hmb.get(a[i-1])!= null && hmb.get(a[i-1])>=f1){
                                --c;
                 }
                 hma.put(a[i - 1], --f1);
       // System.out.println("first part"+" "+c);
       if (hma.containsKey(a[i+m-1])==false)
       {
         //  System.out.println("not present");
           hma.put(a[i+m-1],1);
           if (hmb.get(a[i+m-1])!=null){
           ++c;
          // vis.add(a[i+m-1]);
           }

       }else {
           //System.out.println("present");
           int f2 = hma.get(a[i + m - 1]);
           //  System.out.println(f2);
           hma.put(a[i + m - 1], f2+1);
           //System.out.println(hma.get(a[i+m-1]));
           //System.out.println(hmb.get(a[i+m-1]));
           if (hmb.get(a[i + m - 1]) != null && hma.get(a[i+m-1])>hmb.get(a[i+m-1])){
               if (c>=k){
                   ++ans;
               }
               continue;
           }

           if (hmb.get(a[i + m - 1]) != null) {
              c+=(Math.min(f2+1,hmb.get(a[i+m-1]))-f2);
              //vis.add(a[i+m-1]);
           }
       }

               // System.out.println("second part"+" "+c);
       if (c>=k){
      ++ans;
         }
     //  System.out.println(ans+" "+c+" "+i);
   }

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

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

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

Think step by step.

Response:
```plaintext
1
10 5 3
1 2 3 4 5 1 2 3 4 5
1 2 3 4 5
```