**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Simultaneous Coloring
time_limit_ms: 6000
memory_limit_mb: 512
Description: You are given a matrix, consisting of $n$ rows and $m$ columns.

You can perform two types of actions on it:

  * paint the entire column in blue;    * paint the entire row in red.

Note that you cannot choose which color to paint the row or column.

In one second, you can perform either one action or multiple actions at the same time. If you perform one action, it will be free. If you perform $k > 1$ actions at the same time, it will cost $k^2$ coins. When multiple actions are performed at the same time, for each cell affected by actions of both types, the color can be chosen independently.

You are asked to process $q$ queries. Before each query, all cells become colorless. Initially, there are no restrictions on the color of any cells. In the $i$-th query, a restriction of the following form is added:

  * $x_i~y_i~c_i$ — the cell in row $x_i$ in column $y_i$ should be painted in color $c_i$.

Thus, after $i$ queries, there are $i$ restrictions on the required colors of the matrix cells. After each query, output the minimum cost of painting the matrix according to the restrictions.

Input: The first line contains three integers $n, m$ and $q$ ($1 \le n, m, q \le 2 \cdot 10^5$) — the size of the matrix and the number of queries.

In the $i$-th of the next $q$ lines, two integers $x_i, y_i$ and a character $c_i$ ($1 \le x_i \le n$; $1 \le y_i \le m$; $c_i \in$ {'R', 'B'}, where 'R' means red, and 'B' means blue) — description of the $i$-th restriction. The cells in all queries are pairwise distinct.

Output: Print $q$ integers — after each query, output the minimum cost of painting the matrix according to the restrictions.

Examples:
input:
2 2 4
1 1 R
2 2 R
1 2 B
2 1 B
output:
0
0
0
16

input:
3 5 10
1 1 B
2 5 B
2 2 B
2 3 R
2 1 B
3 2 R
3 3 B
1 2 R
1 3 B
3 1 B
output:
0
0
0
0
0
0
16
16
25
25

Note:

Buggy Code:
from sys import stdin, stdout, setrecursionlimit
from collections import defaultdict, deque

setrecursionlimit(300000)

def solve():
    input = stdin.read
    data = input().split()

    idx = 0

    n = int(data[idx])
    idx += 1
    m = int(data[idx])
    idx += 1
    q = int(data[idx])
    idx += 1

    queries = []
    for _ in range(q):
        x = int(data[idx])
        idx += 1
        y = int(data[idx])
        idx += 1
        c = data[idx]
        idx += 1
        color = 0 if c == 'R' else 1
        queries.append((x, y, color))

    answers = [0] * q

    def ff(x):
        return 0 if x == 1 else x * x

    class UnionFind:
        def __init__(self, n):
            self.n = n
            self.m = n
            self.gids = list(range(n))
            self.sizes = [1] * n
            self.sum_of_squares = 0

        def union(self, i, j):
            ri = self.find(i)
            rj = self.find(j)
            ans = 0
            if ri != rj:
                id = min(ri, rj)
                ans -= ff(self.sizes[ri])
                ans -= ff(self.sizes[rj])
                self.gids[ri] = id
                self.gids[rj] = id
                self.sizes[id] = self.sizes[ri] + self.sizes[rj]
                ans += ff(self.sizes[id])
                self.m -= 1
            return ans

        def find(self, i):
            while i != self.gids[i]:
                self.gids[i] = self.gids[self.gids[i]]
                i = self.gids[i]
            return i

        def is_single_group(self):
            return self.m == 1

        def get_num_groups(self):
            return self.m

        def get_groups(self):
            groups = defaultdict(list)
            for i in range(self.n):
                id = self.find(i)
                groups[id].append(i)
            return groups

    def solve(n, m, queries):
        class TestCase:
            def __init__(self, n, m, queries):
                self.n = n
                self.m = m
                self.q = len(queries)

                self.dfn = [0] * (n + m + 1)
                self.low = [0] * (n + m + 1)
                self.scc = [0] * (n + m + 1)
                self.st = [0] * (n + m + 1)
                self.in_stack = [False] * (n + m + 1)

                self.uf = UnionFind(n + m + 1)
                self.anses = [0] * self.q
                self.edges = [(0, 0)] * self.q

                all_edges = []
                for i, (x, y, c) in enumerate(queries):
                    y += n
                    if c == 0:
                        self.edges[i] = (x, y)
                        all_edges.append((x, y, i))
                    else:
                        self.edges[i] = (y, x)
                        all_edges.append((y, x, i))

                self.G = [[] for _ in range(n + m + 1)]
                self.s = [[] for _ in range(self.q + 1)]

                self.solve(0, self.q, all_edges)

                self.ans = 0
                for i in range(self.q):
                    for id in self.s[i]:
                        self.ans += self.uf.union(self.edges[id][0], self.edges[id][1])
                    self.anses[i] = self.ans

            def solve(self, l, r, edges):
                if l == r:
                    for x, y, id in edges:
                        self.s[l].append(id)
                    return

                mid = (l + r) // 2
                self.clear(edges)

                self.tp = self.cnt = self.tot = 0
                for x, y, id in edges:
                    if id <= mid:
                        self.G[x].append(y)

                for x, y, id in edges:
                    if self.dfn[x] == 0:
                        self.tarjan(x)
                    if self.dfn[y] == 0:
                        self.tarjan(y)

                nl = []
                nr = []
                for x, y, id in edges:
                    if self.scc[x] == self.scc[y]:
                        nl.append((x, y, id))
                    else:
                        nr.append((self.scc[x], self.scc[y], id))

                self.solve(l, mid, nl)
                self.solve(mid + 1, r, nr)

            def clear(self, edges):
                for x, y, _ in edges:
                    self.dfn[x] = self.dfn[y] = 0
                    self.in_stack[x] = self.in_stack[y] = False
                    self.G[x].clear()
                    self.G[y].clear()

            def tarjan(self, x):
                self.dfn[x] = self.low[x] = self.cnt + 1
                self.cnt += 1
                self.in_stack[x] = True
                self.st[self.tp] = x
                self.tp += 1
                for y in self.G[x]:
                    if self.dfn[y] == 0:
                        self.tarjan(y)
                        self.low[x] = min(self.low[x], self.low[y])
                    elif self.in_stack[y]:
                        self.low[x] = min(self.low[x], self.dfn[y])

                if self.low[x] == self.dfn[x]:
                    self.tot += 1
                    while True:
                        p = self.st[self.tp - 1]
                        self.tp -= 1
                        self.in_stack[p] = False
                        self.scc[p] = self.tot
                        if p == x:
                            break

        return TestCase(n, m, queries).anses

    results = solve(n, m, queries)
    stdout.write('\n'.join(map(str, results)) + '\n')

if __name__ == "__main__":
    solve()

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

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

Think step by step.

Response:
```plaintext
3 3 9
1 1 R
1 2 R
1 3 R
2 1 B
2 2 B
2 3 B
3 1 R
3 2 B
3 3 R
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Simultaneous Coloring
time_limit_ms: 6000
memory_limit_mb: 512
Description: You are given a matrix, consisting of $n$ rows and $m$ columns.

You can perform two types of actions on it:

  * paint the entire column in blue;    * paint the entire row in red.

Note that you cannot choose which color to paint the row or column.

In one second, you can perform either one action or multiple actions at the same time. If you perform one action, it will be free. If you perform $k > 1$ actions at the same time, it will cost $k^2$ coins. When multiple actions are performed at the same time, for each cell affected by actions of both types, the color can be chosen independently.

You are asked to process $q$ queries. Before each query, all cells become colorless. Initially, there are no restrictions on the color of any cells. In the $i$-th query, a restriction of the following form is added:

  * $x_i~y_i~c_i$ — the cell in row $x_i$ in column $y_i$ should be painted in color $c_i$.

Thus, after $i$ queries, there are $i$ restrictions on the required colors of the matrix cells. After each query, output the minimum cost of painting the matrix according to the restrictions.

Input: The first line contains three integers $n, m$ and $q$ ($1 \le n, m, q \le 2 \cdot 10^5$) — the size of the matrix and the number of queries.

In the $i$-th of the next $q$ lines, two integers $x_i, y_i$ and a character $c_i$ ($1 \le x_i \le n$; $1 \le y_i \le m$; $c_i \in$ {'R', 'B'}, where 'R' means red, and 'B' means blue) — description of the $i$-th restriction. The cells in all queries are pairwise distinct.

Output: Print $q$ integers — after each query, output the minimum cost of painting the matrix according to the restrictions.

Examples:
input:
2 2 4
1 1 R
2 2 R
1 2 B
2 1 B
output:
0
0
0
16

input:
3 5 10
1 1 B
2 5 B
2 2 B
2 3 R
2 1 B
3 2 R
3 3 B
1 2 R
1 3 B
3 1 B
output:
0
0
0
0
0
0
16
16
25
25

Note:

Buggy Code:
//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
#include<bits/stdc++.h>
#include<cstring>
#include<vector>
#include<string>
#include<bitset>
#include<queue>
#include<stack>
#include<cmath>
#include<set>
#include<map>
#define O(x) cout<<__LINE__<<" " #x " : "<<(x)<<endl;
#define PII pair<LL,LL>
#define PDD pair<du,du>
#define fi first
#define se second
#define du double
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(b);i>=(a);i--)
#define SZ(t) ((LL)((t).size()))
#define eps (1e-3)
#define N 8000011
#define LL long long
//#define int LL
using namespace std;
int n,m,q;
LL ans[N];
struct edge{
	int u,v;
	int ti;
};
vector<int>to[N];
int scccnt;
int sccnum[N];
int sz[N];
int dfn[N];
int vi[N];
int cl;
int low[N];
int st[N],top;
void tarjan(int now){
	cl++;
	low[now]=dfn[now]=cl;
	top++;
	st[top]=now;
	for(auto v:to[now]){
		if(!dfn[v]){
			tarjan(v);
			low[now]=min(low[now],low[v]);
		}else if(!sccnum[v]){
			low[now]=min(low[now],dfn[v]);
		}
	}
	if(low[now]==dfn[now]){
		scccnt++;
		while(st[top]!=now){
			sccnum[st[top]]=scccnt;
			sz[scccnt]+=sz[st[top]];
			st[top]=0;
			top--;
		}
		sccnum[st[top]]=scccnt;
		sz[scccnt]+=sz[st[top]];
		st[top]=0;
		top--;
	}
	return ;
}
void init_(vector<edge>ee,int mid){
	cl=0;
	for(auto e:ee){
		sccnum[e.u]=sccnum[e.v]=0;
		dfn[e.u]=dfn[e.v]=0;
		low[e.u]=low[e.v]=0;
		to[e.u].clear();
		to[e.v].clear();
		vi[e.u]=0;
		vi[e.v]=0;
	}
	return ;
}
void solve(vector<edge>ee,int mid){
	for(auto e:ee){
		if(e.ti<=mid&&e.u!=e.v){
			to[e.u].push_back(e.v);
		}
	}
	for(auto e:ee){
		if(e.ti<=mid&&!dfn[e.u]){
			tarjan(e.u);
		}
	}
	return ;
}
int changeans(vector<edge> ee,LL aan,int mid){
	init_(ee,mid);
	solve(ee,mid);
	int nans=aan;
	for(auto e:ee){
		if(e.ti<=mid){
			if(!vi[e.v]&&sz[e.v]!=1){
				nans-=sz[e.v]*1ll*sz[e.v];
				vi[e.v]=1;
			}
			if(!vi[e.u]&&sz[e.u]!=1){
				nans-=sz[e.u]*1ll*sz[e.u];
				vi[e.u]=1;
			}
			vi[sccnum[e.u]]=0;
			vi[sccnum[e.v]]=0;
		}
	}
	for(auto e:ee){
		if(e.ti<=mid){
			if(!vi[sccnum[e.v]]&&sz[sccnum[e.v]]!=1){
				nans+=sz[sccnum[e.v]]*1ll*sz[sccnum[e.v]];
				vi[sccnum[e.v]]=1;
			}
			if(!vi[sccnum[e.u]]&&sz[sccnum[e.u]]!=1){
				nans+=sz[sccnum[e.u]]*1ll*sz[sccnum[e.u]];
				vi[sccnum[e.u]]=1;
			}
		}
	}
	return nans;
}
void div(int l,int r,vector<edge> ee,LL aan){
	if(l==r){
		ans[l]=changeans(ee,aan,l);
		return ;
	}
	int mid=(l+r)>>1;
	int ansr=changeans(ee,aan,mid);
	vector<edge>el,er;
	for(auto e:ee){
		int uu=sccnum[e.u];
		int vv=sccnum[e.v];
		if(!uu) uu=e.u;
		if(!vv) vv=e.v;
		if(e.ti>mid||uu!=vv){
			er.push_back({uu,vv,e.ti});
		}else{
			el.push_back(e);
		}
	}
	if(1){vector<edge>tmp;swap(ee,tmp);}
	div(mid+1,r,er,ansr);
	div(l,mid,el,aan);
	return ;
}
signed main(){
//	cout<<(sizeof(to))/1024./1024<<endl;
//	cout<<(sizeof(st))/1024./1024<<endl;
	scanf("%d%d%d",&n,&m,&q);
	vector<edge>ed;
	rep(i,1,q){
		int a,b;
		char s[11];
		scanf("%d%d%s",&a,&b,s);
		if(s[0]=='R'){
			ed.push_back((edge){b+n,a,i});
		}else{
			ed.push_back((edge){a,b+n,i});
		}
	}
	rep(i,1,n+m){
		sz[i]=1;
	}
	scccnt=n+m;
	div(1,q,ed,0);
	rep(i,1,q){
		printf("%lld\n",ans[i]);
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
2 2 4
1 1 R
1 2 B
2 1 B
2 2 R
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Simultaneous Coloring
time_limit_ms: 6000
memory_limit_mb: 512
Description: You are given a matrix, consisting of $n$ rows and $m$ columns.

You can perform two types of actions on it:

  * paint the entire column in blue;    * paint the entire row in red.

Note that you cannot choose which color to paint the row or column.

In one second, you can perform either one action or multiple actions at the same time. If you perform one action, it will be free. If you perform $k > 1$ actions at the same time, it will cost $k^2$ coins. When multiple actions are performed at the same time, for each cell affected by actions of both types, the color can be chosen independently.

You are asked to process $q$ queries. Before each query, all cells become colorless. Initially, there are no restrictions on the color of any cells. In the $i$-th query, a restriction of the following form is added:

  * $x_i~y_i~c_i$ — the cell in row $x_i$ in column $y_i$ should be painted in color $c_i$.

Thus, after $i$ queries, there are $i$ restrictions on the required colors of the matrix cells. After each query, output the minimum cost of painting the matrix according to the restrictions.

Input: The first line contains three integers $n, m$ and $q$ ($1 \le n, m, q \le 2 \cdot 10^5$) — the size of the matrix and the number of queries.

In the $i$-th of the next $q$ lines, two integers $x_i, y_i$ and a character $c_i$ ($1 \le x_i \le n$; $1 \le y_i \le m$; $c_i \in$ {'R', 'B'}, where 'R' means red, and 'B' means blue) — description of the $i$-th restriction. The cells in all queries are pairwise distinct.

Output: Print $q$ integers — after each query, output the minimum cost of painting the matrix according to the restrictions.

Examples:
input:
2 2 4
1 1 R
2 2 R
1 2 B
2 1 B
output:
0
0
0
16

input:
3 5 10
1 1 B
2 5 B
2 2 B
2 3 R
2 1 B
3 2 R
3 3 B
1 2 R
1 3 B
3 1 B
output:
0
0
0
0
0
0
16
16
25
25

Note:

Buggy Code:
#pragma gcc optimize("O3,Ofast,no-stack-protector,fast-math,section-anchors,inline,unroll-loops")
#pragma gcc target("abm,mmx,sse,sse2,sse3,sse4,ssse3,tune=native,lzcnt,popcnt,avx,avx2")
//#include<immintrin.h>
#include<iostream>
#include<iomanip>
#include<algorithm>
#include<array>
#include<vector>
#include<set>
#include<string>
#include<map>
#include<tuple>
#include<cmath>
#include<deque>
#include<ctime>
#include<random>
#include<bitset>
#include<unordered_map>
#include<unordered_set>
#include<chrono>
#include<queue>
#include<fstream>
#include<cassert>
#include<thread>
using namespace std;

/*
//#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using ordered_set = tree<int,null_type,std::less<int>,rb_tree_tag,tree_order_statistics_node_update>;
*/
using ll = long long;
using ld = long double;
//using li = __int128;
using ui = unsigned int;
using ul = unsigned long long;
#define ft first
#define sc second
#define all(x) x.begin(), x.end()
template <typename T1, typename T2> ostream& operator << (ostream &c, const pair<T1, T2> &v);
template <typename T> ostream& operator << (ostream &c, const vector<T> &v) {
    c << '{';
    for (int i = 0; i < v.size(); i++) {
        c << v[i];
        if (i+1 != v.size()) c << ", ";
    }
    c << '}';
    return c;
}
template <typename T1, typename T2> ostream& operator << (ostream &c, const pair<T1, T2> &v) {
    c << '<' << v.ft << ", " << v.sc << '>';
    return c;
}
#define debug(x) cout << (#x) << " = " << x << endl;
const long double pi = acos(-1);
const ll inf = 1e18, mod = 998244353, N = 1e6;

vector<vector<int>> rebra, orebra;
vector<int> used, out, color; int col;
void dfs(int i) {
    used[i] = 1;
    for (auto j : rebra[i]) {
        if (used[j] == 0) dfs(j);
    }
    out.push_back(i);
}
void dfs1(int i) {
    color[i] = col;
    used[i] = 1;
    for (auto j : orebra[i]) {
        if (!used[j]) dfs1(j);
    }
}
struct egor {
    int u, v, t;
    egor(int u_, int v_, int t_) {u = u_; v = v_; t = t_;}
};
struct dsu {
    vector<int> p, sz;
    ll sum;
    void init(int n) {
        p.resize(n); iota(all(p), 0); sz.assign(n, 1);
        sum = 0;
    }
    int get(int i) { return (i == p[i] ? i : p[i] = get(p[i])); }
    void merge(int i, int j) {
        i = get(i); j = get(j);
        if (i == j) return;
        if (sz[i] < sz[j]) swap(i, j);
        p[j] = i;
        if (sz[j] != 1) sum -= (ll)sz[j]*sz[j];
        if (sz[i] != 1) sum -= (ll)sz[i]*sz[i];
        sz[i] += sz[j];
        sum += (ll)sz[i]*sz[i];
    }
};
int hui;
dsu d;
vector<egor> tar(vector<egor> g) {
    vector<int> m;
    rebra.clear(); orebra.clear(); rebra.resize(hui); orebra.resize(hui); out.clear();
    for (auto [u, v, t] : g) {
        u = d.get(u); v = d.get(v);
        m.push_back(u); m.push_back(v);
        rebra[u].push_back(v);
        orebra[v].push_back(u);
        used[u] = used[v] = 0;
    }
    for (auto i : m) {
        if (used[i] == 0) dfs(i);
    }
    reverse(all(out));
    for (auto i : m) used[i] = 0;
    vector<egor> ret;
    col = 0;
    for (auto i : out) {
        if (!used[i]) {
            dfs1(i);
            col++;
        }
    }
    for (auto [u, v, t] : g) {
        if (color[d.get(u)] == color[d.get(v)]) ret.emplace_back(u, v, t);
    }
    return ret;
}
vector<ll> fuck;
void f(int l, int r, vector<egor> g) {
    for (auto &[u, v, t] : g) u = d.get(u), v = d.get(v);
    if (l+1 == r) {
        for (auto [u, v, t] : g) d.merge(u, v);
        cout << d.sum << "\n";
    } else {
        int k = (l+r)/2;
        vector<egor> gl, gr;
        for (auto [u, v, t] : g) {
            if (t < k) gl.emplace_back(u, v, t);
        }
        gl = tar(gl);
        for (auto [u, v, t] : gl) fuck[t] = l + (1ll << 32)*r;
        for (auto [u, v, t] : g) {
            if (fuck[t] != l + (1ll << 32)*r) gr.emplace_back(u, v, t);
        }
        f(l, k, gl);
        f(k, r, gr);
    }
}
signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    //freopen("1B/tests/152", "r", stdin);
    //freopen("macloren.txt", "w", stdout);
    mt19937 rnd(std::chrono::steady_clock::now().time_since_epoch().count());

    int n1, n2, m;
    cin >> n1 >> n2 >> m;
    int n = n1+n2; hui = n;
    vector<egor> g;
    rebra.resize(n); orebra.resize(n); used.resize(n); color.resize(n);
    fuck.assign(m, -1);
    for (int i = 0; i < m; i++) {
        int x, y; char c;
        cin >> x >> y >> c; x--; y--;
        if (c == 'R') g.emplace_back(x, y+n1, i);
        else g.emplace_back(y+n1, x, i);
    }
    d.init(n);
    f(0, m, tar(g));
}

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

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

Think step by step.

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

1. **Small Scale**: Since the output length is limited, we should use a small matrix size (e.g., \( n = 3 \) and \( m = 3 \)).
2. **Distinct Cells**: Ensure that the cells in all queries are pairwise distinct.
3. **Mixed Constraints**: Include a mix of red and blue constraints to test the algorithm's ability to handle different types of restrictions.
4. **Complex Interactions**: Create a scenario where the constraints interact in a non-trivial way, requiring the algorithm to make complex decisions.

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

```plaintext
3 3 9
1 1 R
1 2 B
1 3 R
2 1 B
2 2 R
2 3 B
3 1 R
3 2 B
3 3 R
```

This test input:
- Uses a \(3 \times 3\) matrix.
- Includes 9 queries, each with a distinct cell.
- Mixes red and blue constraints to ensure the algorithm must handle both types.
- Creates a pattern that requires the algorithm to make decisions about simultaneous actions and their costs.

This input should help expose any bugs in the algorithm, particularly in how it handles the constraints and calculates the minimum cost.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefiquence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two binary strings $$$a$$$ and $$$b$$$. A binary string is a string consisting of the characters '0' and '1'.

Your task is to determine the maximum possible number $$$k$$$ such that a prefix of string $$$a$$$ of length $$$k$$$ is a subsequence of string $$$b$$$.

A sequence $$$a$$$ is a subsequence of a sequence $$$b$$$ if $$$a$$$ can be obtained from $$$b$$$ by the deletion of several (possibly, zero or all) elements.

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

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1\le n,m \le 2 \cdot 10^5$$$) — the length of string $$$a$$$ and the length of string $$$b$$$, respectively.

The second line of each test case contains a binary string $$$a$$$ of length $$$n$$$.

The third line of each test case contains a binary string $$$b$$$ of length $$$m$$$.

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

Output: For each test case, output a single number — the maximum $$$k$$$, such that the first $$$k$$$ characters of $$$a$$$ form a subsequence of $$$b$$$.

Examples:
input:
6
5 4
10011
1110
3 3
100
110
1 3
1
111
4 4
1011
1111
3 5
100
11010
3 1
100
0
output:
2
2
1
1
3
0

Note: In the first example, the string '$$$10$$$' is a subsequence of '$$$1\color{red}11\color{red}0$$$' but the string '$$$100$$$' is not. So the answer is $$$2$$$.

In the fifth example, $$$a$$$='$$$100$$$', $$$b$$$='$$$1\color{red}{10}1\color{red}0$$$', whole string $$$a$$$ is a subsequence of string $$$b$$$. So the answer is $$$3$$$.

In the sixth example, string $$$b$$$ does not contain '$$$1$$$' so the answer is $$$0$$$.

Buggy Code:
#include<iostream>
#include<vector>
//#include<cstring>
using namespace std;
const int N = 2e5 + 5;
int t = 1, n, m;
bool a[N], b[N];
void solve()
{
	cin >> n >> m;
	int ans = 0;
	char tp;
	for (int i = 1; i <= n; ++i) cin >> tp, a[i] = tp == '1';
	for (int i = 1; i <= m; ++i) cin >> tp, b[i] = tp == '1';
	for (int i = 1, j = 1; i <= m;)
	{
		if (b[i] == a[j]) ++i, ++j, ++ans;
		else ++i;
	}
	cout << ans << endl;
}
int main()
{
	cin >> t;
	while (t--) solve();
	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider the following:

1. **Edge Cases**: Test cases where the input strings are at the boundaries of the constraints, such as very short or very long strings.
2. **Special Patterns**: Test cases where the strings have special patterns that might cause the algorithm to fail, such as all '0's or all '1's.
3. **Mismatched Lengths**: Test cases where the lengths of `a` and `b` are significantly different.
4. **Subsequence Properties**: Test cases where the prefix of `a` is a subsequence of `b` in a non-trivial way.

Given the constraints, we will create a small-scale test input that is complete and valid. Let's consider a scenario where the algorithm might fail:

- **Test Case 1**: `a` is "101010" and `b` is "111111". Here, the prefix "10" of `a` is a subsequence of `b`, but the algorithm might fail to find it correctly.
- **Test Case 2**: `a` is "000000" and `b` is "111111". Here, the prefix "0" of `a` is not a subsequence of `b`, so the answer should be 0.
- **Test Case 3**: `a` is "111111" and `b` is "000000". Here, the prefix "1" of `a` is not a subsequence of `b`, so the answer should be 0.
- **Test Case 4**: `a` is "101010" and `b` is "101010". Here, the entire string `a` is a subsequence of `b`, so the answer should be 6.

Let's write the test input:

```plaintext
4
6 6
101010
111111
6 6
000000
111111
6 6
111111
000000
6 6
101010
101010
```

This test input includes a variety of edge cases and special patterns to ensure the algorithm is thoroughly tested.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefiquence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two binary strings $$$a$$$ and $$$b$$$. A binary string is a string consisting of the characters '0' and '1'.

Your task is to determine the maximum possible number $$$k$$$ such that a prefix of string $$$a$$$ of length $$$k$$$ is a subsequence of string $$$b$$$.

A sequence $$$a$$$ is a subsequence of a sequence $$$b$$$ if $$$a$$$ can be obtained from $$$b$$$ by the deletion of several (possibly, zero or all) elements.

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

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1\le n,m \le 2 \cdot 10^5$$$) — the length of string $$$a$$$ and the length of string $$$b$$$, respectively.

The second line of each test case contains a binary string $$$a$$$ of length $$$n$$$.

The third line of each test case contains a binary string $$$b$$$ of length $$$m$$$.

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

Output: For each test case, output a single number — the maximum $$$k$$$, such that the first $$$k$$$ characters of $$$a$$$ form a subsequence of $$$b$$$.

Examples:
input:
6
5 4
10011
1110
3 3
100
110
1 3
1
111
4 4
1011
1111
3 5
100
11010
3 1
100
0
output:
2
2
1
1
3
0

Note: In the first example, the string '$$$10$$$' is a subsequence of '$$$1\color{red}11\color{red}0$$$' but the string '$$$100$$$' is not. So the answer is $$$2$$$.

In the fifth example, $$$a$$$='$$$100$$$', $$$b$$$='$$$1\color{red}{10}1\color{red}0$$$', whole string $$$a$$$ is a subsequence of string $$$b$$$. So the answer is $$$3$$$.

In the sixth example, string $$$b$$$ does not contain '$$$1$$$' so the answer is $$$0$$$.

Buggy Code:
#include <iostream>
using namespace std;
int main()
{
    int n;
    cin >> n;
    for(int i = 0; i < n; i++)
    {
        string a, b; int nr0 = 0, nr1 = 0;
        cin >> a >> a >> a >> b;
        for(int j = 0; j < b.size(); j++)
        {
            if(b[j] == '0')
                nr0++;
            else
                nr1++;
        }
        //cout << nr0 << " " << nr1 << "\n";
        int a1 = 0, a0 = 0, k = 0;
        while(k <= a.size() && a1 <= nr1 && a0 <= nr0)
        {
            k++;
            if(a[k-1] == '0')
                a0++;
            else
                a1++;
            //cout << a0 << " " << a1 << " " << k << "\n";
        }
        cout << k-1 << "\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 5
10101
01010
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefiquence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two binary strings $$$a$$$ and $$$b$$$. A binary string is a string consisting of the characters '0' and '1'.

Your task is to determine the maximum possible number $$$k$$$ such that a prefix of string $$$a$$$ of length $$$k$$$ is a subsequence of string $$$b$$$.

A sequence $$$a$$$ is a subsequence of a sequence $$$b$$$ if $$$a$$$ can be obtained from $$$b$$$ by the deletion of several (possibly, zero or all) elements.

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

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1\le n,m \le 2 \cdot 10^5$$$) — the length of string $$$a$$$ and the length of string $$$b$$$, respectively.

The second line of each test case contains a binary string $$$a$$$ of length $$$n$$$.

The third line of each test case contains a binary string $$$b$$$ of length $$$m$$$.

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

Output: For each test case, output a single number — the maximum $$$k$$$, such that the first $$$k$$$ characters of $$$a$$$ form a subsequence of $$$b$$$.

Examples:
input:
6
5 4
10011
1110
3 3
100
110
1 3
1
111
4 4
1011
1111
3 5
100
11010
3 1
100
0
output:
2
2
1
1
3
0

Note: In the first example, the string '$$$10$$$' is a subsequence of '$$$1\color{red}11\color{red}0$$$' but the string '$$$100$$$' is not. So the answer is $$$2$$$.

In the fifth example, $$$a$$$='$$$100$$$', $$$b$$$='$$$1\color{red}{10}1\color{red}0$$$', whole string $$$a$$$ is a subsequence of string $$$b$$$. So the answer is $$$3$$$.

In the sixth example, string $$$b$$$ does not contain '$$$1$$$' so the answer is $$$0$$$.

Buggy Code:
/*  xxxxxxXXXXXXXXXxxxxxxxxxxxXX$$$$$$$$$$$$$XX$$$$$$$X$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$xx$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
          xxxxxxxXXXXXXXXXx++;:;+++X$$$$$$$$$$$$$$$XX$$$$$$$X$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$X$$$$$$$$$$$$$$$$$X$$$$$$$$
          xxxxxxXXXXXXXXXXXx+++xxxxX$$X$$$$$$$$$$$$$X$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$X$$$$$$$$$$XXXXXXXXXxxXXXX
          xxxxXXXXXXXXXXXXXxxxxxxXX$$$XXX$$$$$$$$$$$X$$$$XXX$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$x;x$$XXXXXXXXxxxxxx++
          xxxXXXXXXXXXXXXXXx+++xxXX$$$XXX$$$$$$$$$$$X$$$$XXX$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$X+xX$XXXXXXXxxxx++++;
          xxxXXXXXXXXXXXXXXX++++xX$$$XXXX$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$XXXXXXXx+xxx++++;;
          xxxXXXXXXXXXXXXXXXxxxxXX$$XXXXX$$$$$$$$$$$$XX$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$XX$$$$$$$$XXXXXXXXXx;;+++++;;;
          xxxXXXXXXXXXXXXXXXxxxXXX$$$$XXX$$$$$$$$$$$xxX$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$XXXXXXXXXXXxxx+++;;;;
          xXXXXXXXXXXXXXXXXXxxxXX$$$$XXXX$$$$$$$$$$$XX$$$$$$$$$$$$$$$$$$$$XX$$$$$$$$$$$$$$$$$$$$$$$$$$$$$XXXXXXXXXXxxx++++;;;:
          xXX$$$$$$XXXXXXXXXXXXXX$$$$XXX$$$$$$$$X$$$$$$$$$$$$$$$$$$$$$$$$$Xx$$$$$$$$$$$$$$$$$$$$$$$$$$$$XXXXXXXXXXxxx++++;;;;:
          XXX$$$$$$$$$$$$$$$X$XXX$$$$XX$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$XXXXXXXXXXXxxx+++;;;;::
          XX$$$$$$$$$$$$$$$$$$$X$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$XXXXXXXXXXXxxx+++;;;:::
          XXXXXX$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$X$$$$$$$$$$$$$$$$$$$$$$$XXX$$$$$XXXXXXXXXXXXXxxx+++;;;;;;
          XXxxxxXXXXX$$$$$$$$$$$$$$$$XX$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$XXXXXXXXXXXxxxx+++;;;+xx
          XXxxxxxxxxXXXXX$$$$$$$$$$$XXXX$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$XXXXXXXXXXXxxxx+++;;;xXX
          XXxxxxxxxxX$$$$$$$$$$$$$$$$XXXX$$$$$X$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$&&&&&&&$$$$$$$$$$$$$$$$$$$$XXXXXXXxXXxxx+++;;+XXX
          XXXXXXXXXX$$$$$$$$$$$$$$$$$$$$$$X$XXX$$$$$$$$$$$$$$$$$$$$&&&&&&&&&&&&&&&&&&&&&&&$$$$$$$$$$$$$$$XXXXXXXXXxxx+++;;xXX+
          X$$$&&&&&$$$$$$$$$$$$$$$$$$$$$$$$XXX$$$$$$$$$$$$$$&&&&&&&&&&&&&&&$&&&&&&&$$&&&&&&&&&&$$$$$$$$$$$$XXXXXXXXxx++;;;+x+;
          X$$$&&&&&&&&&$$$$$$&&&&&$$$$$$$$$$$$$$$$$$$$$$$$$$&&&&&&&&$$$&&&$$$$&&$$&&&$$$&&&&$$$&$$$$$$$$$$$$$$$XXXxxx+++;;;:::
          X$$$$&&&&&$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$X$$$&&&&&$$XX$XX$$&&XXX$$$$$$$$X$$X$$$$$$$$$$$$$$$$$$$$$XXXxxx+++;;::::
          xX$$$$$$&&&$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$&&&&&$$XxxX$$$X$&$X$$$$X$XX$$$&$$$XX$$$$$$$$$$$$$$$$$XXXxxx+++;;;:::
          xX$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$&&&&&&&$$$$$$$$$XXXXXXXXX$$X$$$$$$&&$&&$$$$$$$$$$$$$$$XXXxxx+++;;;:::
          xX$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$X$$$$$$$$$$$$&&&&&&$Xx++++++;;;;;;;++xXXX$$$$$$$&&&&$$$$$$$$$$$$$$XXXxxx+++;;;;::
          XX$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$XXXXX$$$$$$$&&$Xxx+++;;;;;;;;;;;;;;+++xxXXX$$$&&&&$&$$$$$$$$$$$$$XXXXxx++;;;;::
          XX$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$XXXXX$$$$$$&$$Xx+++;;;;;;;;;;;;;;;;+++xxXXXX$&&&$$$$$$$$$$$$$$$$$$$XXXxx++;;;:
          XX$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$XXXX$$$$$$&$Xx+++;;;;;;;;;;;;;;;;;++++++xXX$$$$$$$$$$$$$$$$$$$$$$XXXXXxx++;;:
          X$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$XXXXX$$$$$$&$xx+++;;;;;;;;;;:::;;;;;;++++xxXX$$$$&$$&&&$$$$$$&&$$XXxxx+++++;;;
          X$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$XXXXXX$$$$$$&Xxx+++;;;;;;;;;;;;;;;;;;;;+++xxXX$$$$&&&&&&&$$$$$&&&&&$$XXxx+++;;;
          $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$XXXXXX$$$$$&$Xxx++;;;;;;;;;;;;;;;;;;;;;+++xxX$$$&&&&&&&&$$$$$&&&&&&&&&&$$$XXx++
          $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$XXXXXX$$$$$&$$$Xx+;;;;::::::::;;;;;;;;+++++xxX$$$$&&&&&&$$$$&&&&&&&&&&&&&&&$Xxx
          $$&&$$$$$$$$$$$$$$$$$XXXXXX$$$$$$$$$$XXXXXX$$$$$&$$$$$$XXx;;;;:;;;;++++xx+++++++xX$$$$$&&&&&$$$$&&&&&&&&&&&&&&&$x+;+
          $$&&&$$$$$$$$$$$$$$$$XXXXXXX$$$$$$$$$XXXXXX$$$$$X+++;;+xxXx++;;+xXX$$XXxXXXXxx++++xX$$$&&&&$$$$&&&&&&&&&&&&&&&&X++;;
          $$&&&&$$$$$$$$$$X$$$XXXXXXXXX$$$$$$$$XXXXXX$$$$Xxx$X$X;;;+++;:;;++++++;;;;+xXXx++++xX$$$&&&&$$&&&&&&&&&&&&&&&&$x++++
          $$&&&&$$$$$$$XXXX$$$XXXXXXXX$$X$$$$$$XXXXX$$$$Xxxxxx$X:::+++;;;;;;+xX$Xx++;;;++++++xX$$$&&&&$$&&&&&&&&&&&&&&&$Xx++++
          $$&&&&$$$$$$XXXXX$$$XXXXXXXX$$X$$$$$$XXXXX$$$$Xx++++;;;;;++;;;+;;;;++XX;;+x+;;;+++xX$$$x+++X$$$$$&&&&&&&&&&&&$Xx++++
          $$&&&&$$$$$$XXXXX$$$XXXXXXXX$$XX$$$$$XXXXX$$$$$x+;;;;:;;+x+;;;;;;;;;;;;;;;;;;;;;++xX$$+;::;;X$$$$$$$&&&&&&&&&$Xx++++
          $$&&&&$$$$$$XXXXX$$$XXXXXXXX$$$$$$$$XXXXXX$$$$$x;;;:::;;x+;;;+;;;::::::;;;;;;;;;++xXX+;:::;:x$$$$$$$$$$&&&&&$$Xxx+++
          $$&&&&$$$$$XXXXXX$$$XXXXXXX$X$XX$$$$XXXXXX$$$$$+;;;:;:;+;;;;;+++;:::::::::::;;;+++xXx+;;::;:x$$$$$$$$$$&&&&&$$Xxxx++
          $$&&&&$$$$$XXXXXX$$$XXXXXX$$X$XX$$$$XXXXX$$$$$$x;;;;:;++::::;;;;+;::::::::::;;+++xxX+;::::;;$$$$$$$$$$$&&&&&$$Xxxxx+
          $$&&&&$$$$$XXXXX$$$$XXXXXXX$$$X$$$$$XXXXX$$$$$$X+;;;;;+++;;;;;;;+;::::::::;;;+++xxXx;:::;;;X$$$$$$$$$$$&&&&&$$Xxxxxx
          $&&&&$$$$$$XXXXX$$$$XXXXXXX$X$X$$$$$XXXXX$$$$$$Xx+;;++;+x++;;;;;;:::::::::;;;+++xxXx+;;;;;X$$$$$$$$$$$$&&&&&$Xxxxxxx
          $&&&&$$$$$$XXXXX$$$XXXXXXXX$X$$$$$$$XXXXX$$$$$$Xx+++xxxXxx++xx+;;;;;;;:;;;;;++++xxx+::;;;X&$$$$$$$$$$$$&&&&$$Xxxxxxx
          $&&&$$$$$$$XXXXX$$$XXXXXXXX$XXX$$$$$XXXXX$$$$$$$Xx+Xx++++;;;;+++++x++;;;;;;;++++xxx::::+$&&$$$$$$$$$$$$&&&&$$XXxxxxx
          $$$$$$$$$$$XXXXX$$$XXXXXXXX$XXX$$$$$XXXXX$$$$$$$Xxx+xxxxx++++++xxxx+++;;;;;+++++xXXXXX$&&&&$$$$$$$$$$$$&&&$$$XXXxxxx
          $$$$$$$$$$$XXXXX$$$XXXXXXXXXXXX$$$$XXXXXX$$$$$$$$Xx+++++;;;;;;;+;;;;;;;;;;;;;++xXXX&&&&&&&&&$$$$$$$$$$$$&$$$$XXXXXXx
          $$$$$$$$$$$$XXXX$$$XXXXXXX$$$$$$$$$XXXXX$$$$$$$$$XXx++++++xx++;;;;;;;;;;++;;+++xXx$&&&&&&&&$$$$$$$$$$$$$$$$$$XXXXXXX
          $$$$$$$&XX$$XXXX$$$XXXXXXX$$$$$$$$$XXXXX$$$$$$$$$$XXx++;;;++;;;;;;;;;;;++;;+++xXxx&&&&&&&&&$$$$$$$$$$$$$$$$$$XXXXXXX
          $$$$$$$&XX$$XXXX$$$XXXXXXX$$$$$$$$$XXXXX$$$$$$$$$$$Xx++;+;;;;;;;;;;;;+++++++xxX++x&&&&&&&&&$$$$$$$$$$$$$$$$$$XXXXXXX
          $$$$$$$$XX$$XXX$$$$XXXXXXX$$$$X$$$$XXXXX$$$$$$$$$$$$$xx++++++;;;;;;+++++xxxxxx+++X&&&&&&&&&$$$$$$$$&$$$$$$$$$XXXXXXX
          $$$$$$$$$$$$$XX$$$$XXXXXXX$$XXX$XXXXXXXX$$$$$$$$$$$$$$XXxx+x+++++++xxxXXxxxx+;;++$&&&&&&&&&$$$$$$$$&&$$$$$$$$XXXXXXX
          $$$$$$$$$$$$$$$$$$XXXXXXXXXXXX$$$XXXXXX$$$$$$$$$$$$$$$x+$$XXXxXXXXXXx++++++;;;;+x$&&&&&&&&&$$$$$$$&&&&$&$$$$XXXXXXXX
          $$$$$$$$$$$$$$$$$$XXXXXXXXXXXX$$$$$$XXX$$$$$$$$$$$$$$$x+;;;+xx+;;;;:;;;+++;;;;;+x$$&&&&&&&&$$$$$$$&&&&&&$$$$XXXXXXXX
          XXXX$$$$$$$$$$$$$$XXXXXXXXXXXX$$$$$$XXX$$$$$$$$$$$$$$$x+;;;;+++;;;::;;;++;;;;;;+x$XX$&&&&&&$$$$$$$&&&&&$$$$$XXXXXXXX
          XXXXXXXXXXXXXXXXXXXXXXXXXXX$X$$$$$$$XXX$$$$$$$$$$$$$$Xx+;;;;;+;;;;;;;;;;;;;;;;++xXXX$$$&&&&$$$$$$$$$$$$$$$$$$XXXXXXX
          XXXXXXXXXXXXXXXXXXXXXXXXXXX$$$$$$$$$XXX$$$$$$$$$XX$$XX++;;;;;;;;;;;;;;;;;;;;;;++xx$$$X$$$&$$$$$$$$&&&&&$$$$$$XXXXXXX
          XXXXXXXXXXXXXXXXXXXXXXXXXX$$$$$$$$$$XXX$$$$$$$$XXX$Xxx+;;;;;;;;;;;;;;;;;;;;;;++++xxX$$X$X$$$$$$$$$$$$$$$$$$$$XXXXXXX
          XXXXXXXXXXXXXXXXXXXXXXXXX$$$$$$$$$$$XXX$$$$$XxXXXXXx++;;;;;;;;;;;;::::;;;;;;+++++++xXXXXXXX$$$$$$$$$$$$$$$$$$$XXXXXX
          XXXXXXXXXXXXXXXXXXXXXXXXXXXXX$$$$$$XXXX$XXXXXXXXXXXx;;;;;;;;;;:;;;;;;;;;;;;;+++;;++++xXXXXXXXX$$$$$$$$$$$$$$$$$$$XXX
          XXXXXXXXXXXXXXXXXXXXXXXXXXXXX$$$$$$XXXxxxxxxxXXXXXXx;;;;;::;;;;:::::;;;;;;;++;;;;;;;;+xXXXXXXXXXXX$$$$$$$$$$$$$$$$$$
          XXXXXXXXXXXXXXXXXXXXXXXXXXX$$$$$$XXXxxxxxxxxxXXXXXXx;;;;::::;;;::::::::::;;++;;;;;;;;+XXXXXXXX$XxXXXXX$$$$$$$$$$$$$$
          XXXXXXXXXXXXXXXXXXXXXXXXXX$$XXXxxxxxxxxxXXXXXXXXXXXX+::;:::::;;:::::::::::;+;:::;;;;+XXXXXXXXXX$$$XXXXxXXXX$$$$$$$$$
          XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXxxxxXXXXXxXXXXXXXXXXXXXx;;;;::::;::::::::::;+;;:::;;;+xxxXXXXXXXXXX$XXXXXXXXxxXX$$$$$$
          XXXXXXXXXXXXXXXXXXXXXXXxxxxxXXXXXXXXXXXXXXXXXXXXXXXxXXXX++;+++;;;::::::::;;;++++++xxxxxxXXXXXXXXX$XXXXXXXXXXXxxXX$$$
          XXXXXXXXXXXXXXXXXXXXXxxxxXXXXXXXXXXXXXXXXXxxxXXXXXXXXXXXXxx+;;;;;;;;;:;;:;;::;;;+xxxxxXXXXXXXXXXXXX$XX$$$$$$$$XXXXX$
          XXXXXXXXXXXXXXXXXXXxxxxXXXXXXXXXXXXXXXXXXXXXXXXXxxxXXXXXXXxxxx;:::;;;;;;;;:::::+xxxxxxXxxXXXXXXxXXX$$$$$$$$$$$$$$XXX
          $XXXXXXXXXXXXXXXXxxxxxXXXxxxxxXXXXXXXXXXXXXXXXXxxxxxxxx;XXXxxxxx;:::::::;;:::;+xxxxxxxXXxxXXXXXxxX$$$$$$$$$XXXXXXXXX
          $$$$XXXXXXXXXXXXxxxxxXxxxxxxxxxxXXXXXXXXXXXXXXXXxxxxxxxxxxxxxxxxx+;:::::::::;+xxxxxxxxxxXXXXXXXXXXX$XXXXXXXXXXXXXXXX
          $$$$$$XXXXXXXXxxxxxxxxxxxxxxxxxxxXXXXXXXXXXXXXXXXXXXxxxxxxxxxxxxxxx+;::::::;xxxxxxxxxxxxXXXXXXXXXXXXXXXXXXXXXXXXXXXX
          $$$$$$$XXXXXXxxxxxxxxxxxxxxxxxxxxxxxXXXXXXXxxxxXXXXXXXXXXxxxxxxxxxxxxx;:::;+xxxxxxxxxXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
          $$$$$$$$$XXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXXXXXXXxxxxxXXxxx;;+xxxxxxXXXXXXXXXXXXxxxXXXXXXXXXXXXXXXXXXXX
          $$$$$$$$$XXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXxxxxxxxXXXXXXXXXXXXxxxxxxxxXXXXXXXXXXXXXXXXXXX
          XXXX$$$$$$Xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx;+xxXXxxxxxxxxxxxxxxxxxxxxxxxXXxxXXXXXxXXXX
          XXX$$$$$$$xxxxxxxxxxxxXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXxxxXXXXxXXX
          $$$$$$$$$XxxxxxxxxxxxxxXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXxxxXXXxxXXXxxXXX
          XX$$$$$$$xxxxxxxxxxxxxxXxxxxxxxxxxxxxxxxXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXXxxXXxxxXXXxXXX
          X$$$$$$$XxxxxxxxxxxxxxxXxxXxxxxxxxxxxxXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXXXxxXXxxxXXxXXX
          X$$$$$$$xxxxxxxxxxxxxxxXXxxXxxxxxxXxxXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXXxXXXxxxXxXXX
          $$$$$$$XxxxxxxxxxxxxxxxXXxxxXxxxxXXxXxxxxxxxxXXXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXXXXXxxXXxXXX
          X$$$$$$XxxxxxxxxxxxxxxxxXxxxXxxxXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXXXXXXXXXXXX
          $$$$$$$XxxxxxxxxxxxxxxxxxXxxXXxXXXxxxXXXxxxXXXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXXXXXXXXX$XX
          $$$$$$$XxxxxxxxxxxxxxxXXXxXXXXX$XXXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXXXXXXXXXXXX
          $$$$$$$XxxxxxxxxxxxxxxxXXXxXXXXXXXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXXXXXXXXX
          &&&$$$$xxxxxxxxxxxxxxxxxxXXxXXXXXXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx+::;xxxxxxxxxxXX$$XXXXXXXXx
          &&&&$$XxxxxxxxxxxxxxxxxxxxxxxXXXXXXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx+.::xxxxxxxxxXxX$$XXXXXXXxx
          &&&&&&XxxxxxxxxxxxxxxxxxxXXXxXXXXXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx:.+xxxxxxxxxxxxx+;;;xxxxxxxxxXXX$XXXXXXXXxx
          &&&&&$xxxxxxxxxxxxxXxxxxxxxxxXXXxxxxxxxxxxxxxxxxXXXXXXxxxxxxxxxxxxxxxxxxxx;+xxxxxxxxxxxxxxxxxxxxxxxxxxXXX$XXXXxxxxxx
          &&&&&$xxxxxxxxxxxxxXxxxxxxxxXXXxxxxxxxxxxxxxxXXXXXXXXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXXXXXxxxxxxx
          &&&&&XxxxxxxxxxxxxxxXxxxxxxxXXXxxxxxxxxxxxxxXXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXxXXxxxxxxxx
          &&&&$xxxxxxxxxxxxxxxXxxxxxxxxXXxxxxxxxxxxxxxXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXXXxXxxxxxxxx
          &&&&$xxxxxxxxxxxxxxxxXxxxxxxxXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXXXxxXXxxxxxxx
          &&&&XxxxxxxxxxxxxxxxxXxxxxxxxXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXXxxxXXxxxxxxx
          &&&&XxxxxxxxxxxxxxxxxxXxxxxxXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXXXXXxxxXXxxxxxxx
          &&&&XxxxxxxxxxxxxxxxxxxXxxxxXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXXxxxxXXxxxxxxx
          &&&$xxxxxxxxxxxxxxxxxxxxXxxx$XxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXxxxxXXXxxxxxx
          &&&$xxxxxxxxxxxxxxxxxxxxxXxXXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXxxxxxXXxxxxxxx
          $$$XxxxxxxxxxxxxxxxxxxxxxxxXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXxxxxxXXXxxxxxx
          $$$XxxxxxxxxxxxxxxxxxxxxxxxX$xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXXxxxxxXXxxxxxxx
          $$$Xxxxxxxxxxxxxxxxxxxxxxxxx$xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx+.:xxxxxxxxxxxxxxxxxxxxxxxXXxxxXXx$Xxxxxxxx
          $$$xxxxxxxxxxxxxxxxxxxxxxxxx$xxxxxxxxxxxxxxxxxxxxxXXXXXxxxxxxxxxxxxxxxxxxx::xxxxxxxxxxxxxxxxxxxxxxXXxxxxxxX$XXxXXxxx
          $$XXxxxxxxxxxxxxxxxxxxxxxxxx$xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXxxxxxxxX$Xxxxxxxx
          $$XxxxxxxxxxxxxxxxxxxxxxxxxXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXxxxxxxxxXXXXxxxxxX
          $$XXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXxxxxxxxxx$XXXXxxxxx
          $$XXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXxxxxxxxxxx$XXxxxxXxx
          x$XxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXxxxxxxxxxxx$XXXXXXXXX
          :XXxxxxxxxxxxxxxxxxxxxxxXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXxxxxxxxxxxX$$XXXXXXXX
          ;XXxxxxxxxxxxxxxxxxxxxxXXXXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx$$XXXXXxxx
          XXXxxxxxxxxxxxxxxxxxxxX$XXXXXXXXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXxxxxxxxxxxxxX$XXXXXXXxx
          $XXxxxxxxxxxXXxxXxxxxxX$$XXXXXXXXXXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx++xxxxxxxxxxXxxxxxxxxxxxxxX$XXXXXXxxx
          $xXxxxxxxxxxxxxxxxxXXXXX$$$$XXXXXXXXXXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx+++xxxxxxxxXxxxxxxxxxxxxxxX$xxxXXXXXX
          +xXxxxxxxXXXxxxxxxxxxxXX$&$$XXXxXXXxxxXXXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx+++xxxxxxXxxxxxxxxxxxxxxX$$XXXXxxxxx
          xxxxxxxxxXXxxxxxxXXXxxXX$&$XXxxxxxxxxxxxxXXXXxxxxxxxxxxxxxxxxxxxxxxxxxxxx+;+xxxx++xxxxxxxxxxxxxxxxxxxxxx$$XXXXXXXxxx
          XxxxxxxxxxxxxxxxXXXxxxxx$$$XxxxxxxxxxxxxxxxxXXXXxxxxxxxxxxxxxXXXXXXxxxxxx+.:xxxxxxxxxxxxxxxxxxxxxxxxxxxX&$XXXXXXXXXX
          $XxxxxxxxxxxxxxxxxxXXXXX$$XXXxxxxxxxxxxxxxxxxxXXXXxxxxxxxxxxxxXXxxxxxxxxxxxxxxxxxxxxxxxXxxxxxxxxxxxxxxx$&$XXXXXXXXXX
          $XxxxxxxxxxxxxxxxXXXXXXX$XXXXXXXXxxxxxxxxxxxxxxxxXXXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXxxxxxxxxxxxxxxx$&&$XXXXXXXXXx
          $XXxxxxxxxxxxxxxxxXXXXXXXxxxxxxXXXXXXxxxxxxxxxxxxxxXXXXXXXXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxX&&&XXXXXXXXXXX
          $XxxxxxxxxxxxxxxxXXXXXXXXxxxxxxxxxxxxxxXxxxxxxxxxxxxxXXXXXXXXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxX$&&&$XXXXXXXXXX
          XxxxxxxxxxxxxxxxXXXXXXXXXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXX&&&$XXXXXXXXxx
          XxXxxxxxxxxxxxxxXXXXXXX$XXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXX$&&$XXXXXXXxxx
          XXXXxXxxxxxxxxxxxxXXXX$$XXxXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXxxxxxxxxxxxxxxxXxXX$&&&XXXXXXXxxx
          XXXxxxxxxxxxxxxxxXXXXXX$XXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx+xxxxXxxxxxxxxxxxxxxxXXxXX$&&$XXXXXxxxxx
          XXXxxxxxxxxxxxxxxxXXXXX$XxxxxxxxxxXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx++xxxxxxxxxxxxxxxxxxxxXxXXX$&&&$Xx+++;;;+
          XXxxxxxxxxxxxxxxxxXXXXX$XxxxxxxxxxxxxxXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx+xxxxxxxxxxxxxxxxxxxxXxxXXXX$&&&Xx+++;;;+
          XXXXXxxxxxxxxxxxxxxxxXXXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXXxxxxx+++xxxxxxxxxxxxxxxxxxxxxXxxXXXxxX&&&Xx++;;;;+
          XXXXXXXxxXx+;;;;;;;++xXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx++++xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx$&$Xx++;;;++
          &&$$$Xx+++;;;;;;;;;;+xXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx++:.;xxxxxxxxxxxxxxxxxxxxxxxxxxxxxX&$X+++;;+++
          &&&&$xx+++;;;;;;;;;;++Xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx+xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxX$&$x+++;++++
          &&&&&Xx+++;;;;;;;;;;++xXxxxxxxxxxxxxxxxxxxxxxxxxxxxXXXxxxxxxxxxxxxxxxx+xxxxxxxxxxxxxxxxxxxxxxxxxxxxXX$&&&&$x++;+++++
          $&&&&$x+++;;;;;;;;;;;+xXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXXXXXXXXXXxxXXXxxxxxxxxxxxxxxxxxxxxxxxxXXXXXxxXxX$&&X++++++++x
          $&&&&&Xx++;;;;;;;;;;;++XxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXXxxxxxxXxXX&$x+++++++xx
          $$&&&&$x+++;;;;;;;;;;++xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXxxxxxxxxXXXXXxX$x++++++xxx
          $$&&&&$X+++;;;;;;;;;;;+xXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXxxxxxxxxXXXXXXXxXXX+++++++xxx
          &&&&&&&$x++;;;;;;;;;;;+xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXxxxxxxxXXXxxxxxxXXXXx+++++++xxx
          &&&&&&&$x+++;;;;;;;;;;+xXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXxxxxxxxxxxxxxxxxxxXXxxx++;++++xxxX
          &&&&&&&&$x+++;;;;;;;;;++XxxxXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXXxxxxxxxxxxxxxxxxxxxXxXXx++++++xxxxxX
          &&&&&&&&&X+++;;;;;;;;;;+xXxxxxxxxXXXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXXXXXxxxxxxxxxxxxxxxxXXXxxxxxXx++++++xxxxX$
          */
         /*------------------------------------------------I AM THE ONLY ONE-------------------------------------------------*/
#include<iostream>
#include<string>
#include<vector>
#include<math.h>
#include<algorithm>
#include<map>
#include<set>
#include<unordered_set>
using namespace std;
typedef long double lld;
typedef unsigned long long ull;
typedef long long ll;
typedef int in;
typedef char ch;
typedef float fl;
typedef double dou;
typedef string ss;
#define ub upper_bound
#define lb lower_bound
#define AJ(i,a,b) for(int i=a;i<b;i++)
#define MOD 1000000007
#define fastio ios_base::sync_with_stdio(false);cin.tie(0);
#define test int t; cin>>t; while(t--)
#ifndef ONLINE_JUDGE
#define debug(x) cerr<<#x<<" "; _print(x); cerr<<endl;
#else
#define debug(x);
#endif

void _print(ll t) {cerr << t;}
void _print(int t) {cerr << t;}
void _print(string t) {cerr << t;}
void _print(char t) {cerr << t;}
void _print(lld t) {cerr << t;}
void _print(double t) {cerr << t;}
void _print(ull t) {cerr << t;}

template <class T, class V> void _print(pair <T, V> p);
template <class T> void _print(vector <T> v);
template <class T> void _print(set <T> v);
template <class T, class V> void _print(map <T, V> v);
template <class T> void _print(multiset <T> v);
template <class T, class V> void _print(pair <T, V> p) {cerr << "{"; _print(p.ff); cerr << ","; _print(p.ss); cerr << "}";}
template <class T> void _print(vector <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(set <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(multiset <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T, class V> void _print(map <T, V> v) {cerr << "[ "; for (auto i : v) {_print(i); cerr << " ";} cerr << "]";}

in main(){
    fastio
#ifndef ONLINE_JUDGE
        freopen("Error.txt", "w", stderr);
#endif
    test{
       ll a,b,c=0,d=0;
       cin>>a>>b;
       ss s1,s2;
       cin>>s1;
       cin>>s2;
       int g=0,f=1,m=0;
       AJ(i,0,a){
        if(g<=b-1){
         AJ(j,g,b){
           if(s1[i]==s2[j]){
            c++;
            m+=c;
            g=j+1;
            break;
           }
           c=0;
         }
         if(c==0) break;
        }
       }
       cout<<m<<endl;
    }
}

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

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

Think step by step.

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

1. **Understanding the Code Logic**:
   - The code iterates through the characters of string `a` and tries to find a matching subsequence in string `b`.
   - It uses nested loops to find the next matching character in `b` for each character in `a`.
   - If a match is found, it increments the count and moves to the next character in `b`.
   - If no match is found, it breaks out of the loop.

2. **Identifying Potential Bugs**:
   - The code might fail if it does not handle cases where the characters in `a` are not found in `b` efficiently.
   - It might also fail if it does not correctly handle the indices and breaks out of the loop prematurely.

3. **Generating Test Cases**:
   - **Edge Case 1**: Both strings are very short.
   - **Edge Case 2**: String `a` is much longer than string `b`.
   - **Edge Case 3**: String `b` contains only one character.
   - **Edge Case 4**: String `a` contains only one character.
   - **Edge Case 5**: String `a` and `b` are the same.
   - **Edge Case 6**: String `a` is a subsequence of `b` but not a prefix.

4. **Choosing the Most Challenging Test Case**:
   - A test case where `a` is a subsequence of `b` but not a prefix can be particularly challenging.

Here is a small-scale test input that fits the criteria:

```plaintext
6
5 4
10011
1110
3 3
100
110
1 3
1
111
4 4
1011
1111
3 5
100
11010
3 1
100
0
```

This test input includes various edge cases:
- The first test case (`10011` and `1110`) where the prefix `10` is a subsequence but not the entire string.
- The second test case (`100` and `110`) where the entire string `a` is a subsequence of `b`.
- The third test case (`1` and `111`) where `a` is a single character.
- The fourth test case (`1011` and `1111`) where `a` is a subsequence of `b`.
- The fifth test case (`100` and `11010`) where `a` is a subsequence of `b`.
- The sixth test case (`100` and `0`) where `a` is not a subsequence of `b`.

This should help expose any issues 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: Prefiquence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two binary strings $$$a$$$ and $$$b$$$. A binary string is a string consisting of the characters '0' and '1'.

Your task is to determine the maximum possible number $$$k$$$ such that a prefix of string $$$a$$$ of length $$$k$$$ is a subsequence of string $$$b$$$.

A sequence $$$a$$$ is a subsequence of a sequence $$$b$$$ if $$$a$$$ can be obtained from $$$b$$$ by the deletion of several (possibly, zero or all) elements.

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

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1\le n,m \le 2 \cdot 10^5$$$) — the length of string $$$a$$$ and the length of string $$$b$$$, respectively.

The second line of each test case contains a binary string $$$a$$$ of length $$$n$$$.

The third line of each test case contains a binary string $$$b$$$ of length $$$m$$$.

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

Output: For each test case, output a single number — the maximum $$$k$$$, such that the first $$$k$$$ characters of $$$a$$$ form a subsequence of $$$b$$$.

Examples:
input:
6
5 4
10011
1110
3 3
100
110
1 3
1
111
4 4
1011
1111
3 5
100
11010
3 1
100
0
output:
2
2
1
1
3
0

Note: In the first example, the string '$$$10$$$' is a subsequence of '$$$1\color{red}11\color{red}0$$$' but the string '$$$100$$$' is not. So the answer is $$$2$$$.

In the fifth example, $$$a$$$='$$$100$$$', $$$b$$$='$$$1\color{red}{10}1\color{red}0$$$', whole string $$$a$$$ is a subsequence of string $$$b$$$. So the answer is $$$3$$$.

In the sixth example, string $$$b$$$ does not contain '$$$1$$$' so the answer is $$$0$$$.

Buggy Code:
t = int(input())
ks = []
for i in range(t):
    n, m = map(int, input().split())
    a = input()
    b = input()
    prefix = ""
    subsequence = ""

    k = 0
    for j in range(len(a)):
        bs = []
        prefix += a[j]
        if(a[j] in b[j:]):
            subsequence += a[j]
            for v in b:
                bs.append(v)
            bs.remove(a[j])
            b = "".join(bs)
        elif(len(b) > j):
            for v in b:
                bs.append(v)
            bs.remove(b[j])
            b = "".join(bs)
    k = len(subsequence)
    ks.append(k)

for i in ks:
    print(i)

**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
10 10
1111111111
0000000000
5 5
10101
01010
6 6
110011
101010
8 8
10101010
01010101
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefiquence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two binary strings $$$a$$$ and $$$b$$$. A binary string is a string consisting of the characters '0' and '1'.

Your task is to determine the maximum possible number $$$k$$$ such that a prefix of string $$$a$$$ of length $$$k$$$ is a subsequence of string $$$b$$$.

A sequence $$$a$$$ is a subsequence of a sequence $$$b$$$ if $$$a$$$ can be obtained from $$$b$$$ by the deletion of several (possibly, zero or all) elements.

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

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1\le n,m \le 2 \cdot 10^5$$$) — the length of string $$$a$$$ and the length of string $$$b$$$, respectively.

The second line of each test case contains a binary string $$$a$$$ of length $$$n$$$.

The third line of each test case contains a binary string $$$b$$$ of length $$$m$$$.

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

Output: For each test case, output a single number — the maximum $$$k$$$, such that the first $$$k$$$ characters of $$$a$$$ form a subsequence of $$$b$$$.

Examples:
input:
6
5 4
10011
1110
3 3
100
110
1 3
1
111
4 4
1011
1111
3 5
100
11010
3 1
100
0
output:
2
2
1
1
3
0

Note: In the first example, the string '$$$10$$$' is a subsequence of '$$$1\color{red}11\color{red}0$$$' but the string '$$$100$$$' is not. So the answer is $$$2$$$.

In the fifth example, $$$a$$$='$$$100$$$', $$$b$$$='$$$1\color{red}{10}1\color{red}0$$$', whole string $$$a$$$ is a subsequence of string $$$b$$$. So the answer is $$$3$$$.

In the sixth example, string $$$b$$$ does not contain '$$$1$$$' so the answer is $$$0$$$.

Buggy Code:
t=int(input())
i=1
while(i<=t):
    nums=input().split()
    n=int(nums[0])
    m=int(nums[1])

    a=input()
    b=input()

    k=0

    P=b.count('1')
    Q=b.count('0')

    for g in range(1,n+1):
        subs_a=a[0:g]
        p=subs_a.count('1')
        q=subs_a.count('0')

        if (P>=p and Q>=q):
            k=k+1
            continue
        else:
            break
    print(k)

    i+=1

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider the following:

1. **Edge Cases**: Test cases where the lengths of `a` and `b` are at their limits, or where `b` contains very few or no '1's or '0's.
2. **Subsequence Conditions**: Test cases where the prefix of `a` is a subsequence of `b` but the conditions in the buggy code might fail to recognize it.
3. **Special Patterns**: Test cases with specific patterns that might cause the buggy code to fail, such as alternating '1's and '0's.

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

```plaintext
3
5 4
10011
1110
3 3
100
110
4 4
1011
0101
```

### Explanation:
1. **Test Case 1**:
   - `a = "10011"`, `b = "1110"`
   - The prefix "10" is a subsequence of "1110", but "100" is not. The correct answer should be 2.
   - The buggy code might fail to recognize that "100" is not a subsequence of "1110" because it only checks the counts of '1's and '0's.

2. **Test Case 2**:
   - `a = "100"`, `b = "110"`
   - The prefix "10" is a subsequence of "110", but "100" is not. The correct answer should be 2.
   - This test case is similar to the first one and should also expose the bug.

3. **Test Case 3**:
   - `a = "1011"`, `b = "0101"`
   - The prefix "101" is a subsequence of "0101", but "1011" is not. The correct answer should be 3.
   - This test case has alternating '1's and '0's, which might cause the buggy code to fail if it does not properly check the subsequence condition.

These test cases are designed to expose the bug in the buggy code by focusing on the conditions where the prefix of `a` is a subsequence of `b` but the counts of '1's and '0's might mislead the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefiquence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two binary strings $$$a$$$ and $$$b$$$. A binary string is a string consisting of the characters '0' and '1'.

Your task is to determine the maximum possible number $$$k$$$ such that a prefix of string $$$a$$$ of length $$$k$$$ is a subsequence of string $$$b$$$.

A sequence $$$a$$$ is a subsequence of a sequence $$$b$$$ if $$$a$$$ can be obtained from $$$b$$$ by the deletion of several (possibly, zero or all) elements.

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

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1\le n,m \le 2 \cdot 10^5$$$) — the length of string $$$a$$$ and the length of string $$$b$$$, respectively.

The second line of each test case contains a binary string $$$a$$$ of length $$$n$$$.

The third line of each test case contains a binary string $$$b$$$ of length $$$m$$$.

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

Output: For each test case, output a single number — the maximum $$$k$$$, such that the first $$$k$$$ characters of $$$a$$$ form a subsequence of $$$b$$$.

Examples:
input:
6
5 4
10011
1110
3 3
100
110
1 3
1
111
4 4
1011
1111
3 5
100
11010
3 1
100
0
output:
2
2
1
1
3
0

Note: In the first example, the string '$$$10$$$' is a subsequence of '$$$1\color{red}11\color{red}0$$$' but the string '$$$100$$$' is not. So the answer is $$$2$$$.

In the fifth example, $$$a$$$='$$$100$$$', $$$b$$$='$$$1\color{red}{10}1\color{red}0$$$', whole string $$$a$$$ is a subsequence of string $$$b$$$. So the answer is $$$3$$$.

In the sixth example, string $$$b$$$ does not contain '$$$1$$$' so the answer is $$$0$$$.

Buggy Code:
t=int(input())
for i in range(t):
    n,m=map(int,input().split())
    a=input()
    b=input()
    dic={}
    for i,x in enumerate(b):
        if x not in dic:
            dic[x] = [[i],[0]]
        else:
            dic[x][0].append(i)
    # print(dic)
    li=-1
    res=0
    flag=0
    for j,x in enumerate(a):
        # print("j",j)
        if x in dic:
            sind = dic[x][1][0]
            dic[x][1][0]+=1
            if sind < len(dic[x][0]):
                cind = dic[x][0][sind]
                if cind>li:
                    li = cind
                    continue
            else:
            	flag=1
            	res = j
            	break
        else:
            flag=1
            res=j
            break
    if flag==0:
    	print(len(a))
    else:
        print(res)

# your code goes here

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on scenarios where the code might fail to correctly determine the maximum prefix of `a` that is a subsequence of `b`. The key issues to consider are:

1. **Handling of repeated characters in `b`**: The code uses a dictionary to store indices of characters in `b`. If the code incorrectly handles the indices of repeated characters, it might skip valid indices or use invalid ones.
2. **Edge cases where `a` or `b` are very short**: The code might fail to handle cases where `a` or `b` are very short, especially if `b` is empty or contains only a single character.
3. **Cases where `a` is a subsequence of `b` but not as a prefix**: The code should correctly identify the longest prefix of `a` that is a subsequence of `b`.

Let's construct a test input that targets these issues:

```plaintext
3
5 5
10101
11011
4 4
1011
1111
3 3
100
101
```

### Explanation:
1. **Test Case 1:**
   - `a = "10101"`, `b = "11011"`
   - The prefix "101" of `a` is a subsequence of `b`, but "1010" is not. The correct answer is 3.
   - This test case checks if the code correctly handles multiple occurrences of the same character in `b`.

2. **Test Case 2:**
   - `a = "1011"`, `b = "1111"`
   - The prefix "10" of `a` is a subsequence of `b`, but "101" is not. The correct answer is 2.
   - This test case checks if the code correctly handles the case where `b` contains only one type of character.

3. **Test Case 3:**
   - `a = "100"`, `b = "101"`
   - The prefix "10" of `a` is a subsequence of `b`, but "100" is not. The correct answer is 2.
   - This test case checks if the code correctly handles the case where `b` does not contain all characters of `a`.

By using these test cases, we can effectively expose any issues in the code's handling of repeated characters and edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefiquence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two binary strings $$$a$$$ and $$$b$$$. A binary string is a string consisting of the characters '0' and '1'.

Your task is to determine the maximum possible number $$$k$$$ such that a prefix of string $$$a$$$ of length $$$k$$$ is a subsequence of string $$$b$$$.

A sequence $$$a$$$ is a subsequence of a sequence $$$b$$$ if $$$a$$$ can be obtained from $$$b$$$ by the deletion of several (possibly, zero or all) elements.

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

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1\le n,m \le 2 \cdot 10^5$$$) — the length of string $$$a$$$ and the length of string $$$b$$$, respectively.

The second line of each test case contains a binary string $$$a$$$ of length $$$n$$$.

The third line of each test case contains a binary string $$$b$$$ of length $$$m$$$.

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

Output: For each test case, output a single number — the maximum $$$k$$$, such that the first $$$k$$$ characters of $$$a$$$ form a subsequence of $$$b$$$.

Examples:
input:
6
5 4
10011
1110
3 3
100
110
1 3
1
111
4 4
1011
1111
3 5
100
11010
3 1
100
0
output:
2
2
1
1
3
0

Note: In the first example, the string '$$$10$$$' is a subsequence of '$$$1\color{red}11\color{red}0$$$' but the string '$$$100$$$' is not. So the answer is $$$2$$$.

In the fifth example, $$$a$$$='$$$100$$$', $$$b$$$='$$$1\color{red}{10}1\color{red}0$$$', whole string $$$a$$$ is a subsequence of string $$$b$$$. So the answer is $$$3$$$.

In the sixth example, string $$$b$$$ does not contain '$$$1$$$' so the answer is $$$0$$$.

Buggy Code:
import java.io.*;

public class Main {

    public static String s1;

    public static String s2;

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StreamTokenizer in = new StreamTokenizer(br);
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        in.nextToken();
        int t = (int) in.nval;
        while (t-- > 0) {
            in.nextToken();
            in.nextToken();
            in.nextToken();
            s1 = String.valueOf((int) in.nval);
            in.nextToken();
            s2 = String.valueOf((int) in.nval);
            out.println(solve(s1, s2));
        }
        out.flush();
        br.close();
        out.close();
    }

    public static int solve(String s1, String s2) {
        int n1 = s1.length(), n2 = s2.length();
        int p1 = 0, p2 = 0, res = 0;
        while (p1 < n1) {
            while (p2 < n2) {
                if (s1.charAt(p1) == s2.charAt(p2)) {
                    res++;
                    p2++;
                    break;
                } else {
                    p2++;
                }
            }
            if (p2 == n2) {
                break;
            }
            p1++;
        }
        return 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 5
10101
11111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefiquence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two binary strings $$$a$$$ and $$$b$$$. A binary string is a string consisting of the characters '0' and '1'.

Your task is to determine the maximum possible number $$$k$$$ such that a prefix of string $$$a$$$ of length $$$k$$$ is a subsequence of string $$$b$$$.

A sequence $$$a$$$ is a subsequence of a sequence $$$b$$$ if $$$a$$$ can be obtained from $$$b$$$ by the deletion of several (possibly, zero or all) elements.

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

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1\le n,m \le 2 \cdot 10^5$$$) — the length of string $$$a$$$ and the length of string $$$b$$$, respectively.

The second line of each test case contains a binary string $$$a$$$ of length $$$n$$$.

The third line of each test case contains a binary string $$$b$$$ of length $$$m$$$.

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

Output: For each test case, output a single number — the maximum $$$k$$$, such that the first $$$k$$$ characters of $$$a$$$ form a subsequence of $$$b$$$.

Examples:
input:
6
5 4
10011
1110
3 3
100
110
1 3
1
111
4 4
1011
1111
3 5
100
11010
3 1
100
0
output:
2
2
1
1
3
0

Note: In the first example, the string '$$$10$$$' is a subsequence of '$$$1\color{red}11\color{red}0$$$' but the string '$$$100$$$' is not. So the answer is $$$2$$$.

In the fifth example, $$$a$$$='$$$100$$$', $$$b$$$='$$$1\color{red}{10}1\color{red}0$$$', whole string $$$a$$$ is a subsequence of string $$$b$$$. So the answer is $$$3$$$.

In the sixth example, string $$$b$$$ does not contain '$$$1$$$' so the answer is $$$0$$$.

Buggy Code:
import java.util.*;

public class B {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- >= 1) {
            long n = sc.nextLong();
            long n1 = sc.nextLong();
            String a = sc.next();
            String b = sc.next();
            long zero = 0;
            long one = 0;
            long c = 0;
            for (long i = 0; i < n1; i++) {
                if (b.charAt((int)i) == '0'){
                    zero++;
                }else{
                    one++;
                }
            }
            for (long i = 0; i < n; i++) {
                if (a.charAt((int)i) == '0') {
                    if (zero > 0){
                        zero--;
                        c++;
                    }else{
                        break;
                    }
                }else {
                    if (one > 0){
                        one--;
                        c++;
                    }else{
                        break;
                    }
                }
            }
            System.out.println(c);
        }
    }

}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases where the code might fail. The bug in the provided code is that it only counts the total number of '0's and '1's in string `b` and then tries to match them with the characters in string `a`. This approach fails when the order of characters in `b` does not allow a subsequence match.

Let's consider the following edge cases:
1. **All characters in `a` are '0' and `b` has fewer '0's than `a`**: This will ensure that the code correctly stops when it runs out of '0's in `b`.
2. **All characters in `a` are '1' and `b` has fewer '1's than `a`**: This will ensure that the code correctly stops when it runs out of '1's in `b`.
3. **Mixed characters in `a` and `b` where the order in `b` does not allow a subsequence match**: This will expose the bug where the code does not correctly handle the order of characters.

Here is a small-scale test input that includes these edge cases:

```plaintext
3
5 4
10011
1110
3 3
100
110
4 4
1011
1111
```

Explanation:
1. **Test Case 1**: `a = "10011"`, `b = "1110"`
   - The prefix "10" is a subsequence of "1110".
   - The prefix "100" is not a subsequence of "1110".
   - Expected output: 2

2. **Test Case 2**: `a = "100"`, `b = "110"`
   - The prefix "10" is a subsequence of "110".
   - The prefix "100" is not a subsequence of "110".
   - Expected output: 2

3. **Test Case 3**: `a = "1011"`, `b = "1111"`
   - The prefix "10" is a subsequence of "1111".
   - The prefix "101" is a subsequence of "1111".
   - The prefix "1011" is a subsequence of "1111".
   - Expected output: 4

This test input should help expose the bug in the given code where it fails to correctly handle the order of characters in string `b`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefiquence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two binary strings $$$a$$$ and $$$b$$$. A binary string is a string consisting of the characters '0' and '1'.

Your task is to determine the maximum possible number $$$k$$$ such that a prefix of string $$$a$$$ of length $$$k$$$ is a subsequence of string $$$b$$$.

A sequence $$$a$$$ is a subsequence of a sequence $$$b$$$ if $$$a$$$ can be obtained from $$$b$$$ by the deletion of several (possibly, zero or all) elements.

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

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1\le n,m \le 2 \cdot 10^5$$$) — the length of string $$$a$$$ and the length of string $$$b$$$, respectively.

The second line of each test case contains a binary string $$$a$$$ of length $$$n$$$.

The third line of each test case contains a binary string $$$b$$$ of length $$$m$$$.

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

Output: For each test case, output a single number — the maximum $$$k$$$, such that the first $$$k$$$ characters of $$$a$$$ form a subsequence of $$$b$$$.

Examples:
input:
6
5 4
10011
1110
3 3
100
110
1 3
1
111
4 4
1011
1111
3 5
100
11010
3 1
100
0
output:
2
2
1
1
3
0

Note: In the first example, the string '$$$10$$$' is a subsequence of '$$$1\color{red}11\color{red}0$$$' but the string '$$$100$$$' is not. So the answer is $$$2$$$.

In the fifth example, $$$a$$$='$$$100$$$', $$$b$$$='$$$1\color{red}{10}1\color{red}0$$$', whole string $$$a$$$ is a subsequence of string $$$b$$$. So the answer is $$$3$$$.

In the sixth example, string $$$b$$$ does not contain '$$$1$$$' so the answer is $$$0$$$.

Buggy Code:
/*
	///////                //                    //   //
	//   //                //                     // //    //
	///////  //  //        //  //  //   //////     ///
	//       //  //   //   //  //  //   //  //     ///     //
	//       //  //   //   //  //  //   //  //    // //    //
	//       ///////  ///////  ///////  //  ///  //   //   //

 */

#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define YES cout<<"Yes"<<endl
#define NO cout<<"No"<<endl
#define fr1(a,b,c) for(int i=a;i<=b;i+=c)
#define fr2(a,b,c) for(int j=a;j<=b;j+=c)
#define fr3(a,b,c) for(int i=a;i>=b;i-=c)
#define fr4(a,b,c) for(int j=a;j>=n;j-=c)
#define fr5(a,b,c) for(a;b;c)
#define wh(a) while(a)
#define ioyh ios::sync_with_stdio(0)
#define bk break
#define con continue
#define uns using namespace std
#define rt return
uns;
int t,n,m;
string s1,s2;
signed main(){
	ioyh;
	cin>>t;
	wh(t--){
		cin>>n>>m>>s1>>s2;
		int a=0,b=0;
		fr1(0,m-1,1){
			if(s2[i]=='1') a++;
			else b++;
		}
		//cout<<a<<" "<<b<<endl;
		int x=0,y=0,k;
		bool f=0;
		fr1(0,n-1,1){
			if(s1[i]=='1') x++;
			else y++;
			k=i;
			if(x>a||y>b){
				f=1;
				bk;
			}
		}
		if(f) cout<<k<<endl;
		else cout<<k+1<<endl;
	}
	rt 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases that can highlight the issues in the algorithm. The code seems to be counting the number of '1's and '0's in both strings and then comparing these counts to determine the maximum prefix length that can be a subsequence. This approach is flawed because it does not correctly account for the order of characters in the subsequence.

### Key Observations:
1. **Order of Characters**: The code does not consider the order of characters when determining if a prefix is a subsequence.
2. **Edge Cases**: Test cases with strings that have a high concentration of '1's or '0's, or where the order of characters in `b` is significantly different from `a`.

### Challenging Test Input:
Let's create a test case where the order of characters in `b` is such that the prefix of `a` cannot be a subsequence, even though the counts of '1's and '0's are valid.

```plaintext
1
5 6
10101
111110
```

### Explanation:
- **Input**:
  - `t = 1` (one test case)
  - `n = 5`, `m = 6` (lengths of `a` and `b`)
  - `a = "10101"`
  - `b = "111110"`

- **Expected Output**: `3`
  - The prefix "101" of `a` is a subsequence of `b` (111110), but "1010" and "10101" are not.

### Why This Test Case is Challenging:
- The string `b` has a high concentration of '1's followed by a single '0'. This makes it difficult for the code to correctly determine the subsequence without considering the order of characters.
- The code will incorrectly count the number of '1's and '0's and might output a higher value than the correct answer.

This test input should effectively expose the bug 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: Prefiquence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two binary strings $$$a$$$ and $$$b$$$. A binary string is a string consisting of the characters '0' and '1'.

Your task is to determine the maximum possible number $$$k$$$ such that a prefix of string $$$a$$$ of length $$$k$$$ is a subsequence of string $$$b$$$.

A sequence $$$a$$$ is a subsequence of a sequence $$$b$$$ if $$$a$$$ can be obtained from $$$b$$$ by the deletion of several (possibly, zero or all) elements.

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

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1\le n,m \le 2 \cdot 10^5$$$) — the length of string $$$a$$$ and the length of string $$$b$$$, respectively.

The second line of each test case contains a binary string $$$a$$$ of length $$$n$$$.

The third line of each test case contains a binary string $$$b$$$ of length $$$m$$$.

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

Output: For each test case, output a single number — the maximum $$$k$$$, such that the first $$$k$$$ characters of $$$a$$$ form a subsequence of $$$b$$$.

Examples:
input:
6
5 4
10011
1110
3 3
100
110
1 3
1
111
4 4
1011
1111
3 5
100
11010
3 1
100
0
output:
2
2
1
1
3
0

Note: In the first example, the string '$$$10$$$' is a subsequence of '$$$1\color{red}11\color{red}0$$$' but the string '$$$100$$$' is not. So the answer is $$$2$$$.

In the fifth example, $$$a$$$='$$$100$$$', $$$b$$$='$$$1\color{red}{10}1\color{red}0$$$', whole string $$$a$$$ is a subsequence of string $$$b$$$. So the answer is $$$3$$$.

In the sixth example, string $$$b$$$ does not contain '$$$1$$$' so the answer is $$$0$$$.

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

using namespace std;
#define ll long long
#define int  long long

void Elzamly() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
}

template<typename T = int>
istream &operator>>(istream &in, vector<T> &v) {
    for (auto &x: v) in >> x;
    return in;
}

template<typename T = int>
ostream &operator<<(ostream &out, const vector<T> &v) {
    for (const T &x: v) out << x << " ";
    return out;
}

bool is_prime(int x) {
    if (x == 1 or x == 0) return false;
    for (int i = 2; i < x; i++) {
        if (x % i == 0) return false;
    }
    return true;
}

int bi_ex(int x, int n) {
    int r = 1;
    while (n > 0) {
        if (n % 2) r *= x;
        x *= x;
        n /= 2;
    }
    return r;
}

vector<int> factors(int n) {
    vector<int> v;
    for (int i = 1; i * i <= n; i++) {
        if (n % i == 0) v.push_back(i);
    }
    for (int i = sqrt(n); i * i >= 1; i--) {
        if (n % i == 0 and i * i != n) v.push_back(n / i);
    }
    return v;
}

vector<int> pfactors(int n) {
    vector<int> v;
    for (int i = 2; i * i <= n; i++) {
        while (n % i == 0) {
            n /= i;
            v.push_back(i);
        }
    }
    if (n != 1) v.push_back(n);
    return v;
}

int gcd(int a, int b) {
    while (b != 0) {
        int c = a;
        a = b;
        b = c % b;
    }
    return a;
}

int lcm(int a, int b) {
    return a / gcd(a, b) * b;
}
int32_t main() {
    Elzamly();
    int t ; cin >> t ;
    while (t--){
       int n ,m ; cin >> n >> m ;
       map<char,int>mp ;
       string a,b ; cin >> a >> b ;
       if(a == b){
           cout << n << endl ;
           continue;
       }
       bool f = false;
       for(auto it : b) mp[it]++ ;
       for(int i = 0 ; i < n ; i++){
           if(mp[a[i]] == 0) {
               cout << i << endl;
               f = true;
               break;
           }
           else {
               mp[a[i]]--;
           }
       }
       if(!f) cout << n << 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
6 5
111010
11010
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefiquence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two binary strings $$$a$$$ and $$$b$$$. A binary string is a string consisting of the characters '0' and '1'.

Your task is to determine the maximum possible number $$$k$$$ such that a prefix of string $$$a$$$ of length $$$k$$$ is a subsequence of string $$$b$$$.

A sequence $$$a$$$ is a subsequence of a sequence $$$b$$$ if $$$a$$$ can be obtained from $$$b$$$ by the deletion of several (possibly, zero or all) elements.

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

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1\le n,m \le 2 \cdot 10^5$$$) — the length of string $$$a$$$ and the length of string $$$b$$$, respectively.

The second line of each test case contains a binary string $$$a$$$ of length $$$n$$$.

The third line of each test case contains a binary string $$$b$$$ of length $$$m$$$.

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

Output: For each test case, output a single number — the maximum $$$k$$$, such that the first $$$k$$$ characters of $$$a$$$ form a subsequence of $$$b$$$.

Examples:
input:
6
5 4
10011
1110
3 3
100
110
1 3
1
111
4 4
1011
1111
3 5
100
11010
3 1
100
0
output:
2
2
1
1
3
0

Note: In the first example, the string '$$$10$$$' is a subsequence of '$$$1\color{red}11\color{red}0$$$' but the string '$$$100$$$' is not. So the answer is $$$2$$$.

In the fifth example, $$$a$$$='$$$100$$$', $$$b$$$='$$$1\color{red}{10}1\color{red}0$$$', whole string $$$a$$$ is a subsequence of string $$$b$$$. So the answer is $$$3$$$.

In the sixth example, string $$$b$$$ does not contain '$$$1$$$' so the answer is $$$0$$$.

Buggy Code:
for _ in range(int(input())):
    a,b=map(int,input().split())
    s=input()
    s1=input()
    d={"0":[],"1":[]}
    for i in range(len(s1)):

        d[s1[i]].append(i)
    m=0
    k,k1=0,0
    for i in s:
        if i=="0":
            if k<len(d["0"]):
                m+=1
                k+=1
            else:
                break
        else:
            if k1<len(d["1"]):
                m+=1
                k1+=1
            else:

                break
    print(m)

# print(int("111",2))

# print(2**100)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases that might cause the code to fail. The main issues in the buggy code are:

1. **Incorrect handling of the input format**: The code reads the lengths of the strings as integers but then reads the strings directly without considering the actual lengths.
2. **Subsequence checking**: The code uses a dictionary to store the indices of '0' and '1' in string `b` and then tries to match the characters of string `a` to these indices. However, it does not correctly handle the order of the subsequence.

To expose these issues, we can create a test case where:
- The lengths of `a` and `b` are provided correctly.
- The strings `a` and `b` are such that the subsequence check fails due to the incorrect handling of the indices.

Here is a small-scale test input that should expose the bug:

```plaintext
3
5 5
10011
11100
4 4
1011
1111
3 3
100
101
```

Explanation:
1. **First Test Case**:
   - `a = "10011"`, `b = "11100"`
   - The first prefix of `a` that is a subsequence of `b` is "100". The next character '1' cannot be matched in the remaining part of `b` after "100" is matched.
   - Expected output: 3
   - The buggy code will likely output 2 because it does not correctly handle the order of the subsequence.

2. **Second Test Case**:
   - `a = "1011"`, `b = "1111"`
   - The entire string `a` is a subsequence of `b`.
   - Expected output: 4
   - The buggy code should output 4, but it's important to ensure it handles this case correctly.

3. **Third Test Case**:
   - `a = "100"`, `b = "101"`
   - The first prefix of `a` that is a subsequence of `b` is "10". The next character '0' cannot be matched in the remaining part of `b` after "10" is matched.
   - Expected output: 2
   - The buggy code will likely output 2 because it does not correctly handle the order of the subsequence.

This test input should help in identifying the issues with the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefiquence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two binary strings $$$a$$$ and $$$b$$$. A binary string is a string consisting of the characters '0' and '1'.

Your task is to determine the maximum possible number $$$k$$$ such that a prefix of string $$$a$$$ of length $$$k$$$ is a subsequence of string $$$b$$$.

A sequence $$$a$$$ is a subsequence of a sequence $$$b$$$ if $$$a$$$ can be obtained from $$$b$$$ by the deletion of several (possibly, zero or all) elements.

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

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1\le n,m \le 2 \cdot 10^5$$$) — the length of string $$$a$$$ and the length of string $$$b$$$, respectively.

The second line of each test case contains a binary string $$$a$$$ of length $$$n$$$.

The third line of each test case contains a binary string $$$b$$$ of length $$$m$$$.

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

Output: For each test case, output a single number — the maximum $$$k$$$, such that the first $$$k$$$ characters of $$$a$$$ form a subsequence of $$$b$$$.

Examples:
input:
6
5 4
10011
1110
3 3
100
110
1 3
1
111
4 4
1011
1111
3 5
100
11010
3 1
100
0
output:
2
2
1
1
3
0

Note: In the first example, the string '$$$10$$$' is a subsequence of '$$$1\color{red}11\color{red}0$$$' but the string '$$$100$$$' is not. So the answer is $$$2$$$.

In the fifth example, $$$a$$$='$$$100$$$', $$$b$$$='$$$1\color{red}{10}1\color{red}0$$$', whole string $$$a$$$ is a subsequence of string $$$b$$$. So the answer is $$$3$$$.

In the sixth example, string $$$b$$$ does not contain '$$$1$$$' so the answer is $$$0$$$.

Buggy Code:
# contains random code for random problems i do on the fly
import math

n123 = int(input())

for i in range(n123):
    n,m=list(map(int,input().split()))
    a=input()
    b=input()
    ans=0
    for char in a:
        if char in b:
            ans+=1
            ind=b.find(char)
            b=b[:ind]+b[ind+1:]
        else:
            break
    print(ans)

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

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

Think step by step.

Response:
```plaintext
1
5 5
10101
11111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefiquence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two binary strings $$$a$$$ and $$$b$$$. A binary string is a string consisting of the characters '0' and '1'.

Your task is to determine the maximum possible number $$$k$$$ such that a prefix of string $$$a$$$ of length $$$k$$$ is a subsequence of string $$$b$$$.

A sequence $$$a$$$ is a subsequence of a sequence $$$b$$$ if $$$a$$$ can be obtained from $$$b$$$ by the deletion of several (possibly, zero or all) elements.

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

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1\le n,m \le 2 \cdot 10^5$$$) — the length of string $$$a$$$ and the length of string $$$b$$$, respectively.

The second line of each test case contains a binary string $$$a$$$ of length $$$n$$$.

The third line of each test case contains a binary string $$$b$$$ of length $$$m$$$.

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

Output: For each test case, output a single number — the maximum $$$k$$$, such that the first $$$k$$$ characters of $$$a$$$ form a subsequence of $$$b$$$.

Examples:
input:
6
5 4
10011
1110
3 3
100
110
1 3
1
111
4 4
1011
1111
3 5
100
11010
3 1
100
0
output:
2
2
1
1
3
0

Note: In the first example, the string '$$$10$$$' is a subsequence of '$$$1\color{red}11\color{red}0$$$' but the string '$$$100$$$' is not. So the answer is $$$2$$$.

In the fifth example, $$$a$$$='$$$100$$$', $$$b$$$='$$$1\color{red}{10}1\color{red}0$$$', whole string $$$a$$$ is a subsequence of string $$$b$$$. So the answer is $$$3$$$.

In the sixth example, string $$$b$$$ does not contain '$$$1$$$' so the answer is $$$0$$$.

Buggy Code:
import java.util.*;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.PrintWriter;
import java.lang.*;

public class Submission {

	public static void main(String[] args) throws FileNotFoundException {
		Scanner ip = new Scanner(System.in);

		int t = ip.nextInt();

			while(t-- > 0) {
				ip.nextInt();
				ip.nextInt();
				String a =  "" + ip.nextInt();
				String b = "" + ip.nextInt();

				int ans = 0;

				int i = 0;
				int j = 0;

				while(i < a.length() && j < b.length()) {
					while(j < b.length() && a.charAt(i) != b.charAt(j)) {
						j++;
					}
					if(j < b.length()) {
						ans++;
						i++;
						j++;
					}

				}

				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
5 5
10101
11111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefiquence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given two binary strings $$$a$$$ and $$$b$$$. A binary string is a string consisting of the characters '0' and '1'.

Your task is to determine the maximum possible number $$$k$$$ such that a prefix of string $$$a$$$ of length $$$k$$$ is a subsequence of string $$$b$$$.

A sequence $$$a$$$ is a subsequence of a sequence $$$b$$$ if $$$a$$$ can be obtained from $$$b$$$ by the deletion of several (possibly, zero or all) elements.

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

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1\le n,m \le 2 \cdot 10^5$$$) — the length of string $$$a$$$ and the length of string $$$b$$$, respectively.

The second line of each test case contains a binary string $$$a$$$ of length $$$n$$$.

The third line of each test case contains a binary string $$$b$$$ of length $$$m$$$.

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

Output: For each test case, output a single number — the maximum $$$k$$$, such that the first $$$k$$$ characters of $$$a$$$ form a subsequence of $$$b$$$.

Examples:
input:
6
5 4
10011
1110
3 3
100
110
1 3
1
111
4 4
1011
1111
3 5
100
11010
3 1
100
0
output:
2
2
1
1
3
0

Note: In the first example, the string '$$$10$$$' is a subsequence of '$$$1\color{red}11\color{red}0$$$' but the string '$$$100$$$' is not. So the answer is $$$2$$$.

In the fifth example, $$$a$$$='$$$100$$$', $$$b$$$='$$$1\color{red}{10}1\color{red}0$$$', whole string $$$a$$$ is a subsequence of string $$$b$$$. So the answer is $$$3$$$.

In the sixth example, string $$$b$$$ does not contain '$$$1$$$' so the answer is $$$0$$$.

Buggy Code:
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        for (int i = 0; i < t; i++) {
            int n = sc.nextInt();
            int m = sc.nextInt();
            char[] cha = sc.next().toCharArray();
            char[] chb = sc.next().toCharArray();
            int b0 = 0;
            int b1 = 0;
            for (int j = 0; j < m; j++) {
                if(chb[j] == '0') b0++;
                else b1++;
            }
            int a0 = 0;
            int a1 = 0;
            int val = 0;
            for (int j = 0; j < n; j++) {
                if(cha[j] == '0') a0++;
                else a1++;
                if(a0 > b0||a1 > b1){
                    System.out.println(j);
                    val = 1;
                    break;
                }
            }
            if(val == 0) System.out.println(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
6 6
101010
010101
```