**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 378QAQ and Core
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a string $s$ of length $n$. Define the core of a string as the substring$^\dagger$ with maximum lexicographic$^\ddagger$ order.

For example, the core of "$\mathtt{bazoka}$" is "$\mathtt{zoka}$", and the core of "$\mathtt{aaa}$" is "$\mathtt{aaa}$".

378QAQ wants to rearrange the string $s$ so that the core is lexicographically minimum. Find the lexicographically minimum possible core over all rearrangements of $s$.

$^\dagger$ A substring of string $s$ is a continuous segment of letters from $s$. For example, "$\mathtt{defor}$", "$\mathtt{code}$" and "$\mathtt{o}$" are all substrings of "$\mathtt{codeforces}$" while "$\mathtt{codes}$" and "$\mathtt{aaa}$" are not.

$^\ddagger$ A string $p$ is lexicographically smaller than a string $q$ if and only if one of the following holds:

  * $p$ is a prefix of $q$, but $p \ne q$; or    * in the first position where $p$ and $q$ differ, the string $p$ has a smaller element than the corresponding element in $q$ (when compared by their ASCII code).

For example, "$\mathtt{code}$" and "$\mathtt{coda}$" are both lexicographically smaller than "$\mathtt{codeforces}$" while "$\mathtt{codeforceston}$" and "$\mathtt{z}$" are not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\leq n\leq 10^6$) — the length of string $s$.

The next line of each test case contains the string $s$ of length $n$. The string $s$ consists of lowercase English letters.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, output the lexicographically minimum possible core over all rearrangements of $s$.

Examples:
input:
6
3
qaq
4
cccc
6
bazoka
6
zazzzz
7
ababbbb
7
ccbabcc
output:
qaq
cccc
z
zzz
bbababb
cbcacbc

Note: In the first test case, all possible rearrangements and their corresponding cores are as follows:

  * "$\mathtt{qaq}$", its core is "$\mathtt{qaq}$".    * "$\mathtt{aqq}$", its core is "$\mathtt{qq}$".    * "$\mathtt{qqa}$", its core is "$\mathtt{qqa}$".

So the core with the minimum lexicographic order in all rearrangement plans is "$\mathtt{qaq}$".

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
//#pragma GCC optimize(3)
#define DB double
#define LL long long
#define ULL unsigned long long
#define in128 __int128
#define cint const int
#define cLL const LL
#define For(z,e1,e2) for(int z=(e1);z<=(e2);z++)
#define Rof(z,e1,e2) for(int z=(e2);z>=(e1);z--)
#define For_(z,e1,e2) for(int z=(e1);z<(e2);z++)
#define Rof_(z,e1,e2) for(int z=(e2);z>(e1);z--)
#define inint(e) scanf("%d",&e)
#define inll(e) scanf("%lld",&e)
#define inpr(e1,e2) scanf("%d%d",&e1,&e2)
#define in3(e1,e2,e3) scanf("%d%d%d",&e1,&e2,&e3)
#define outint(e) printf("%d\n",e)
#define outint_(e) printf("%d%c",e," \n"[i==n])
#define outint2_(e,e1,e2) printf("%d%c",e," \n"[(e1)==(e2)])
#define outll(e) printf("%lld\n",e)
#define outll_(e) printf("%lld%c",e," \n"[i==n])
#define outll2_(e,e1,e2) printf("%lld%c",e," \n"[(e1)==(e2)])
#define exc(e) if(e) continue
#define stop(e) if(e) break
#define ret(e) if(e) return
#define ll(e) (1ll*(e))
#define pb push_back
#define ft first
#define sc second
#define pii pair<int,int>
#define pli pair<LL,int>
#define vct vector
#define clean(e) while(!e.empty()) e.pop()
#define all(ev) ev.begin(),ev.end()
#define sz(ev) ((int)ev.size())
#define debug(x) printf("%s=%d\n",#x,x)
#define x0 __xx00__
#define y1 __yy11__
#define ffo fflush(stdout)
// cLL mod=998244353,G=404;
cLL mod=1e9+7;
// cLL mod[2]={1686688681ll,1888686661ll},base[2]={166686661ll,188868881ll};
template <typename Type> void get_min(Type &w1,const Type w2) { if(w2<w1) w1=w2; } template <typename Type> void get_max(Type &w1,const Type w2) { if(w2>w1) w1=w2; }
template <typename Type> Type up_div(Type w1,Type w2) { return (w1/w2+(w1%w2?1:0)); }
template <typename Type> Type gcd(Type X_,Type Y_) { Type R_=X_%Y_; while(R_) { X_=Y_; Y_=R_; R_=X_%Y_; } return Y_; } template <typename Type> Type lcm(Type X_,Type Y_) { return (X_/gcd(X_,Y_)*Y_); }
template <typename Type> Type md(Type w1,const Type w2=mod) { w1%=w2; if(w1<0) w1+=w2; return w1; } template <typename Type> Type md_(Type w1,const Type w2=mod) { w1%=w2; if(w1<=0) w1+=w2; return w1; }
void ex_gcd(LL &X_,LL &Y_,LL A_,LL B_) { if(!B_) { X_=1ll; Y_=0ll; return ; } ex_gcd(Y_,X_,B_,A_%B_); X_=md(X_,B_); Y_=(1ll-X_*A_)/B_; } LL inv(LL A_,LL B_=mod) { LL X_=0ll,Y_=0ll; ex_gcd(X_,Y_,A_,B_); return X_; }
template <typename Type> void add(Type &w1,const Type w2,const Type M_=mod) { w1=md(w1+w2,M_); } void mul(LL &w1,cLL w2,cLL M_=mod) { w1=md(w1*md(w2,M_),M_); } template <typename Type> Type pw(Type X_,Type Y_,Type M_=mod) { Type S_=1; while(Y_) { if(Y_&1) mul(S_,X_,M_); Y_>>=1; mul(X_,X_,M_); } return S_; }
template <typename Type> Type bk(vector <Type> &V_) { auto T_=V_.back(); V_.pop_back(); return T_; } template <typename Type> Type tp(stack <Type> &V_) { auto T_=V_.top(); V_.pop(); return T_; } template <typename Type> Type frt(queue <Type> &V_) { auto T_=V_.front(); V_.pop(); return T_; }
template <typename Type> Type bg(set <Type> &V_) { auto T_=*V_.begin(); V_.erase(V_.begin()); return T_; } template <typename Type> Type bk(set <Type> &V_) { auto T_=*prev(V_.end()); V_.erase(*prev(V_.end())); return T_; }
mt19937 gen(time(NULL)); int rd() { return abs((int)gen()); }
void main_init()
{

}
cint N=1.02e6;
char s[N];
void solve(vct <int> &v)
{
    // for(auto i:v) printf("%d ",i);
    // printf("\n");
    // ffo;
    int n=sz(v),m=v.back();
    if(n==1||v[n-2]!=m)
    {
        v={v.back()};
        return ;
    }
    int cnt=n-1;
    while(cnt>0&&v[cnt-1]==m) cnt--;
    v.resize(cnt);
    cnt=n-cnt-1;
    n-=cnt+1;
    vct <vct<int>> g(1,(vct<int>){});
    vct <int> v2;
    int h=cnt,tot=0,lst=0,w=0;
    for(int l=0,r=0;l<n;l=(++r))
    {
        while(r+1<n&&v[r+1]==v[r]) r++;
        int len=r-l+1,c=v[l];
        Rof(k,1,len/h) g[w].pb(c);
        len%=h;
        exc(!len);
        tot++;
        lst+=len;
        g.pb(g.back());
        g[w++].pb(c);
        h-=len;
        while(len)
        {
            len--;
            v2.pb(tot);
        }
    }
    tot++;
    Rof(i,1,cnt-lst) v2.pb(tot);
    // reverse(all(g));
    v.clear(); //vct <int> emp; swap(v,emp);
    solve(v2);
    v.pb(m);
    for(auto i:v2)
    {
        i--;
        for(auto c:g[i])
            v.pb(c);
        v.pb(m);
    }
}
void main_solve()
{
    int n;
    scanf("%d%s",&n,s);
    vct <int> v(n,0);
    For_(i,0,n) v[i]=s[i]-'a'+1;
    sort(all(v));
    solve(v);
    for(auto i:v) printf("%c",'a'+i-1);
    printf("\n");
}
int main()
{
    // ios::sync_with_stdio(0); cin.tie(0);
    // freopen("in.txt","r",stdin);
    // freopen("out.txt","w",stdout);
    // srand(time(NULL));
    main_init();
    int _; inint(_); For(__,1,_) // T>1 ?
        // printf("\n------------\n\n"),
        main_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 expose the bug in the given code, we need to focus on edge cases and scenarios that are likely to cause issues in the algorithm. Here are some potential scenarios to consider:

1. **Strings with repeated characters**: The code might have issues handling strings where one character is repeated many times.
2. **Strings with a mix of characters**: The code might have issues handling strings where the characters are not uniformly distributed.
3. **Strings with a lexicographically small core**: The code might have issues when the core is expected to be a small lexicographical string.
4. **Strings with a lexicographically large core**: The code might have issues when the core is expected to be a large lexicographical string.
5. **Strings with a single character**: The code might have issues handling strings of length 1.

Let's consider a small-scale test input that combines these scenarios. Specifically, we will use a string with a mix of repeated and distinct characters, and a string with a single character.

### Test Input
```plaintext
2
5
aabbz
1
a
```

### Explanation
1. **First Test Case**:
   - **Input**: `aabbz`
   - **Expected Output**: `aabbz`
   - **Reason**: The lexicographically minimum core for any rearrangement of `aabbz` is `aabbz` itself, as `z` is the largest character and should be placed at the end.

2. **Second Test Case**:
   - **Input**: `a`
   - **Expected Output**: `a`
   - **Reason**: The string is already the lexicographically minimum core.

This test input is designed to check how the code handles strings with repeated characters and a single character, which are common edge cases that can expose bugs in the algorithm.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 378QAQ and Core
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a string $s$ of length $n$. Define the core of a string as the substring$^\dagger$ with maximum lexicographic$^\ddagger$ order.

For example, the core of "$\mathtt{bazoka}$" is "$\mathtt{zoka}$", and the core of "$\mathtt{aaa}$" is "$\mathtt{aaa}$".

378QAQ wants to rearrange the string $s$ so that the core is lexicographically minimum. Find the lexicographically minimum possible core over all rearrangements of $s$.

$^\dagger$ A substring of string $s$ is a continuous segment of letters from $s$. For example, "$\mathtt{defor}$", "$\mathtt{code}$" and "$\mathtt{o}$" are all substrings of "$\mathtt{codeforces}$" while "$\mathtt{codes}$" and "$\mathtt{aaa}$" are not.

$^\ddagger$ A string $p$ is lexicographically smaller than a string $q$ if and only if one of the following holds:

  * $p$ is a prefix of $q$, but $p \ne q$; or    * in the first position where $p$ and $q$ differ, the string $p$ has a smaller element than the corresponding element in $q$ (when compared by their ASCII code).

For example, "$\mathtt{code}$" and "$\mathtt{coda}$" are both lexicographically smaller than "$\mathtt{codeforces}$" while "$\mathtt{codeforceston}$" and "$\mathtt{z}$" are not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\leq n\leq 10^6$) — the length of string $s$.

The next line of each test case contains the string $s$ of length $n$. The string $s$ consists of lowercase English letters.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, output the lexicographically minimum possible core over all rearrangements of $s$.

Examples:
input:
6
3
qaq
4
cccc
6
bazoka
6
zazzzz
7
ababbbb
7
ccbabcc
output:
qaq
cccc
z
zzz
bbababb
cbcacbc

Note: In the first test case, all possible rearrangements and their corresponding cores are as follows:

  * "$\mathtt{qaq}$", its core is "$\mathtt{qaq}$".    * "$\mathtt{aqq}$", its core is "$\mathtt{qq}$".    * "$\mathtt{qqa}$", its core is "$\mathtt{qqa}$".

So the core with the minimum lexicographic order in all rearrangement plans is "$\mathtt{qaq}$".

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

int32_t main() {
    std::ios_base::sync_with_stdio(false), std::cin.tie(nullptr), std::cout.tie(nullptr);

    int tc; std::cin >> tc;

    std::function<void()> solve = [&]() {
        int n; std::cin >>n;
        std:: string s; std::cin >> s;

        std::vector<int> freq('z' - 'a' + 1);
        for(auto &it: s)
            freq[it - 'a'] ++;

        std::vector<std::pair<int, std::string>> v;
        for(int i = 'z' - 'a' ; i >= 0; i--){
            std::string t(1, i+'a');
            if(freq[i])
                v.emplace_back(freq[i], t);
        }

        std:: function<std::string(std::vector<std::pair<int, std::string>>)> rec =[&](std::vector<std::pair<int, std::string>> v) ->std::string{
            if(v.size() - 1 == 0){
                std::string s;
                for(int j = 0; j < v[0].first; j++){
                    s += v[0].second;
                }
                return s;
            }
            if(v[0].first == 1){
                return v[0].second;
            }
            int sz = v.size();

            int f = v[0].first - 1;
            std:: string dp = v[0].second;
            int ind = 0;

            int freq2[30] = {0};
            std::string ans[30];
            for(int i = sz - 1; i > 0; i--){
                if(v[i].first == 0) {  }
                else if(v[i].first < f){
                    ans[ind] = v[0].second + v[i].second;
                    freq2[ind++] = v[i].first;
                    f -= v[i].first;
                    continue;
                }else{
                    v[i].first -= f;
                    v[0].second += v[i].second;
                    i++;
                }
            }
            ans[ind] = v[0].second;
            freq2[ind++] = f;

            std::vector<std::pair<int, std::string>> subANS;
            for(int i = ind-1; i >= 0; i--){
                if(freq2[i])subANS.push_back({freq2[i], ans[i]});
            }
            return rec(subANS) + dp;
        };
        std::cout << rec(v) << '\n';
    };

    while(tc--) solve();
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
1
z
2
ab
3
abc
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 378QAQ and Core
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a string $s$ of length $n$. Define the core of a string as the substring$^\dagger$ with maximum lexicographic$^\ddagger$ order.

For example, the core of "$\mathtt{bazoka}$" is "$\mathtt{zoka}$", and the core of "$\mathtt{aaa}$" is "$\mathtt{aaa}$".

378QAQ wants to rearrange the string $s$ so that the core is lexicographically minimum. Find the lexicographically minimum possible core over all rearrangements of $s$.

$^\dagger$ A substring of string $s$ is a continuous segment of letters from $s$. For example, "$\mathtt{defor}$", "$\mathtt{code}$" and "$\mathtt{o}$" are all substrings of "$\mathtt{codeforces}$" while "$\mathtt{codes}$" and "$\mathtt{aaa}$" are not.

$^\ddagger$ A string $p$ is lexicographically smaller than a string $q$ if and only if one of the following holds:

  * $p$ is a prefix of $q$, but $p \ne q$; or    * in the first position where $p$ and $q$ differ, the string $p$ has a smaller element than the corresponding element in $q$ (when compared by their ASCII code).

For example, "$\mathtt{code}$" and "$\mathtt{coda}$" are both lexicographically smaller than "$\mathtt{codeforces}$" while "$\mathtt{codeforceston}$" and "$\mathtt{z}$" are not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\leq n\leq 10^6$) — the length of string $s$.

The next line of each test case contains the string $s$ of length $n$. The string $s$ consists of lowercase English letters.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, output the lexicographically minimum possible core over all rearrangements of $s$.

Examples:
input:
6
3
qaq
4
cccc
6
bazoka
6
zazzzz
7
ababbbb
7
ccbabcc
output:
qaq
cccc
z
zzz
bbababb
cbcacbc

Note: In the first test case, all possible rearrangements and their corresponding cores are as follows:

  * "$\mathtt{qaq}$", its core is "$\mathtt{qaq}$".    * "$\mathtt{aqq}$", its core is "$\mathtt{qq}$".    * "$\mathtt{qqa}$", its core is "$\mathtt{qqa}$".

So the core with the minimum lexicographic order in all rearrangement plans is "$\mathtt{qaq}$".

Buggy Code:
// Problem: H. 378QAQ and Core
// Contest: Codeforces - Codeforces Round 947 (Div. 1 + Div. 2)
// URL: https://codeforces.com/problemset/problem/1975/H
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mkp make_pair
#define uint unsigned
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef double db;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<ll, ll> pii;

const int maxn = 1000100;
const ll base = 131, mod1 = 998244853, mod2 = 1004535809;

int n, b[26];
char s[maxn], t[maxn];
ll pw1[maxn], pw2[maxn];

struct node {
	vector<int> S;
	ll x, y;
} a[maxn];

void solve() {
	scanf("%d%s", &n, s + 1);
	for (int i = 0; i <= n; ++i) {
		vector<int>().swap(a[i].S);
		a[i].x = 0;
	}
	mems(b, 0);
	pw1[0] = pw2[0] = 1;
	for (int i = 1; i <= n; ++i) {
		++b[s[i] - 'a'];
		pw1[i] = pw1[i - 1] * base % mod1;
		pw2[i] = pw2[i - 1] * base % mod2;
	}
	a[0].S.pb(0);
	a[0].x = a[0].y = 1;
	int m = 0;
	for (int i = 0; i < 26; ++i) {
		while (b[i]) {
			--b[i];
			a[++m].S.pb(i + 1);
			a[m].x = a[m].y = i + 2;
		}
	}
	for (int i = 0, j = m + 1; i <= m && a[m].S.back(); ++i) {
		if (j == m + 1) {
			while (j - 1 > i && a[m].x == a[j - 1].x && a[m].y == a[j - 1].y) {
				--j;
			}
		}
		for (int x : a[i].S) {
			a[j].S.pb(x);
		}
		a[j].x = (a[j].x * pw1[(int)a[i].S.size()] + a[i].x) % mod1;
		a[j].y = (a[j].y * pw2[(int)a[i].S.size()] + a[i].y) % mod2;
		vector<int>().swap(a[i].S);
		++j;
	}
	for (int x : a[m].S) {
		if (x) {
			putchar('a' + x - 1);
		}
	}
	putchar('\n');
}

int main() {
	int T = 1;
	scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
1
a
2
ab
3
abc
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 378QAQ and Core
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a string $s$ of length $n$. Define the core of a string as the substring$^\dagger$ with maximum lexicographic$^\ddagger$ order.

For example, the core of "$\mathtt{bazoka}$" is "$\mathtt{zoka}$", and the core of "$\mathtt{aaa}$" is "$\mathtt{aaa}$".

378QAQ wants to rearrange the string $s$ so that the core is lexicographically minimum. Find the lexicographically minimum possible core over all rearrangements of $s$.

$^\dagger$ A substring of string $s$ is a continuous segment of letters from $s$. For example, "$\mathtt{defor}$", "$\mathtt{code}$" and "$\mathtt{o}$" are all substrings of "$\mathtt{codeforces}$" while "$\mathtt{codes}$" and "$\mathtt{aaa}$" are not.

$^\ddagger$ A string $p$ is lexicographically smaller than a string $q$ if and only if one of the following holds:

  * $p$ is a prefix of $q$, but $p \ne q$; or    * in the first position where $p$ and $q$ differ, the string $p$ has a smaller element than the corresponding element in $q$ (when compared by their ASCII code).

For example, "$\mathtt{code}$" and "$\mathtt{coda}$" are both lexicographically smaller than "$\mathtt{codeforces}$" while "$\mathtt{codeforceston}$" and "$\mathtt{z}$" are not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\leq n\leq 10^6$) — the length of string $s$.

The next line of each test case contains the string $s$ of length $n$. The string $s$ consists of lowercase English letters.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, output the lexicographically minimum possible core over all rearrangements of $s$.

Examples:
input:
6
3
qaq
4
cccc
6
bazoka
6
zazzzz
7
ababbbb
7
ccbabcc
output:
qaq
cccc
z
zzz
bbababb
cbcacbc

Note: In the first test case, all possible rearrangements and their corresponding cores are as follows:

  * "$\mathtt{qaq}$", its core is "$\mathtt{qaq}$".    * "$\mathtt{aqq}$", its core is "$\mathtt{qq}$".    * "$\mathtt{qqa}$", its core is "$\mathtt{qqa}$".

So the core with the minimum lexicographic order in all rearrangement plans is "$\mathtt{qaq}$".

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

const int maxn = (int)1e6 + 9, maxc = 27;

int n, m, q, pL[maxc], pR[maxc];
char buf[maxn];
pair<int, int> que[maxc];
array<int, 3> seq[maxn];

void dfs(int u) {
    if(seq[u].at(0) == -1) {
        buf[n++] = u ? 'a' + u - 1 : '\0';
        return;
    }
    // assert(seq[u].at(1) != -1 && seq[u].at(2) > 0);
    dfs(seq[u].at(0));
    int sta = n;
    dfs(seq[u].at(1));
    int cnt = seq[u].at(2), ofs = n - sta, d = 0;
    for( ; (2 << d) <= cnt; ++d)
        memcpy(buf + sta + (ofs << d), buf + sta, (ofs << d) * sizeof(char));
    if((1 << d) < cnt)
        memcpy(buf + sta + (ofs << d), buf + sta, (cnt - (1 << d)) * ofs * sizeof(char));
    n += (cnt - 1) * ofs;
}

void solve() {
    scanf("%d%s", &n, buf);
    m = maxc;
    for(int i = 0; i < m; ++i)
        que[i] = {i, 0};
    ++que[0].second;
    for(int i = 0; i < n; ++i) {
        int o = buf[i] - 'a' + 1;
        ++que[o].second;
    }
    q = 0;
    for(int i = 0; i < m; ++i)
        if(que[i].second)
            que[q++] = que[i];
    --q;
    for(int i = 0; i < q; ++i) {
        pL[i] = i - 1;
        pR[i] = i + 1;
    }
    pL[0] = q - 1;
    pR[q - 1] = 0;
    for(int i = 0; que[0].second > 0; i = pR[i]) {
        if(que[i].second >= que[q].second) {
            int c = que[i].second / que[q].second;
            que[i].second -= c * que[q].second;
            seq[m] = {que[q].first, que[i].first, c};
            que[q].first = m++;
        }
        if(!que[i].second) {
            pL[pR[i]] = pL[i];
            pR[pL[i]] = pR[i];
        } else {
            que[q].second -= que[i].second;
            seq[m] = {que[q].first, que[i].first, 1};
            que[i].first = m++;
        }
    }
    // assert(que[q].second == 1);
    // int las = que[q].first;
    // for( ; las >= maxc; las = seq[las].at(1));
    // assert(!las);
    n = 0;
    dfs(que[q].first);
    // assert(n > 0 && buf[n - 1] == '\0');
    puts(buf);
}

int main() {
    for(int i = 0; i < maxc; ++i)
        seq[i] = {-1, -1, 0};

    int T = 1;
    scanf("%d", &T);
    for(int Case = 1; Case <= T; ++Case) {
        // printf("Case #%d: ", Case);
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
5
1
a
2
ab
3
aba
4
abac
6
abacde
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vova really loves the [XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation (denoted as $\oplus$). Recently, when he was going to sleep, he came up with a fun game.

At the beginning of the game, Vova chooses two binary sequences $s$ and $t$ of length $n$ and gives them to Vanya. A binary sequence is a sequence consisting only of the numbers $0$ and $1$. Vanya can choose integers $l, r$ such that $1 \leq l \leq r \leq n$, and for all $l \leq i \leq r$ simultaneously replace $s_i$ with $s_i \oplus s_{i - l + 1}$, where $s_i$ is the $i$-th element of the sequence $s$.

In order for the game to be interesting, there must be a possibility to win. Vanya wins if, with an unlimited number of actions, he can obtain the sequence $t$ from the sequence $s$. Determine if the game will be interesting for the sequences $s$ and $t$.

Input: Each test consists of multiple test cases. The first line contains an integer $q$ ($1 \le q \le 10^{4}$) — the number of test cases. Then follows the description of the test cases.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the length of the sequences $s$ and $t$.

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

The third line of each test case contains a binary sequence $t$ of length $n$.

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

Output: For each test case, output "Yes" if the game will be interesting, otherwise output "No".

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
1
0
1
7
0110100
0110100
9
100101010
101111110
4
0011
1011
4
0100
0001
8
10110111
01100000
output:
NO
YES
YES
NO
YES
YES

Note: In the first test case, Vanya will not be able to change the sequence $s$ with the only possible action of choosing $l = r = 1$.

In the second test case, the sequences $s$ and $t$ are already equal.

In the third test case, Vanya can act as follows:

  1. Choose $l = 3$ and $r = 5$, then $s$ will become $\mathtt{101101010}$.   2. Choose $l = 5$ and $r = 6$, then $s$ will become $\mathtt{101111010}$.   3. Choose $l = 7$ and $r = 7$, then $s$ will become $\mathtt{101111110}$.

Buggy Code:
#include <iostream>

// ** COULDN'T USE THIS BITS HEADER   ** //
//

#ifndef _GLIBCXX_NO_ASSERT
#include <cassert>
#endif
#include <cctype>
#include <cerrno>
#include <cfloat>
#include <ciso646>
#include <climits>
#include <clocale>
#include <cmath>
#include <csetjmp>
#include <csignal>
#include <cstdarg>
#include <cstddef>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>

#if __cplusplus >= 201103L
#include <ccomplex>
#include <cfenv>
#include <cinttypes>
//#include <cstdalign>
#include <cstdbool>
#include <cstdint>
#include <ctgmath>
#include <cwchar>
#include <cwctype>
#endif

// C++
#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <locale>
#include <map>
#include <memory>
#include <new>
#include <numeric>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <typeinfo>
#include <utility>
#include <valarray>
#include <vector>

#if __cplusplus >= 201103L
#include <array>
#include <atomic>
#include <chrono>
#include <condition_variable>
#include <forward_list>
#include <future>
#include <initializer_list>
#include <mutex>
#include <random>
#include <ratio>
#include <regex>
#include <scoped_allocator>
#include <system_error>
#include <thread>
#include <tuple>
#include <typeindex>
#include <type_traits>
#include <unordered_map>
#include <unordered_set>
#endif

// HO GYA ITNA HI THA //

using namespace std;

using ll= long long;
using lld= long double;
using ull= unsigned long long;
 // constant
const lld pi= 3.141592653589793238;
const ll mod=1e9+7;

//Sorting
bool sorta(const pair<int,int> &a,const pair<int,int> &b){return (a.second < b.second);}
bool sortd(const pair<int,int> &a,const pair<int,int> &b){return (a.second > b.second);}

//Bits
string decToBinary(int n){string s="";int i = 0;while (n > 0) {s =to_string(n % 2)+s;n = n / 2;i++;}return s;}
ll binaryToDecimal(string n){string num = n;ll dec_value = 0;int base = 1;int len = num.length();for(int i = len - 1; i >= 0; i--){if (num[i] == '1')dec_value += base;base = base * 2;}return dec_value;}

//Check
bool isPrime(ll n){if(n<=1)return false;if(n<=3)return true;if(n%2==0||n%3==0)return false;for(int i=5;i*i<=n;i=i+6)if(n%i==0||n%(i+2)==0)return false;return true;}
bool isPowerOfTwo(int n){if(n==0)return false;return (ceil(log2(n)) == floor(log2(n)));}
bool isPerfectSquare(ll x){if (x >= 0) {ll sr = sqrt(x);return (sr * sr == x);}return false;}
ll min(ll a,int b) { if (a<b) return a; return b; }
ll min(int a,ll b) { if (a<b) return a; return b; }
ll max(ll a,int b) { if (a>b) return a; return b; }
ll max(int a,ll b) { if (a>b) return a; return b; }
ll gcd(ll a,ll b) { if (b==0) return a; return gcd(b, a%b); }
ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }
string to_upper(string a) { for (int i=0;i<(int)a.size();++i) if (a[i]>='a' && a[i]<='z') a[i]-='a'-'A'; return a; }
string to_lower(string a) { for (int i=0;i<(int)a.size();++i) if (a[i]>='A' && a[i]<='Z') a[i]+='a'-'A'; return a; }

typedef pair<ll, ll> pll;
typedef vector<ll> vll;
typedef vector<pll> vpll;
typedef vector<string> vs;
typedef unordered_map<ll,ll> umll;
typedef map<ll,ll> mll;

// Macros

#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define lp(i,s,e) for(long long int i=s;i<e;i++)
#define flp(i,s,e) for(long long int i=s;i<=e;i++)
#define rp(i,e,s) for(long long int i=e;i>s;i--)
#define rfl(i,e,s) for(long long int i=e;i>=s;i--)
#define cy cout<<"YES\n";
#define cmi cout<<"-1\n";
#define cn cout<<"NO\n";
#define cl cout<<"\n";
#define cimp cout<<"IMPOSSIBLE\n";
#define vr(v) v.begin(),v.end()
#define rv(v) v.end(),v.begin()
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()

//Code

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

string s,t;
cin>>s>>t;
ll c=0;

lp(i,0,n) {
    if(s[i]=='1') c++;
}

// lp(i,0,n) {
//     if(s[i]!=t[i]){
//         if(s[0]==t[0] && s[0]!=1){
//             cn
//             return;
//         }
//     }
// }

if(s==t){
    cy
    return;
}

 if(s.substr(1,n)==t.substr(1,n) && s[0]!=t[0] ){
        cout<<"NO\n";
        return;
    }

// if(s[0]!=s[1] ){
//     cn
// }
// else if(s!=t && (c==n)){
//     cn
// }
// else{
//     cy
// }

cy

}

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

    ll t;
    cin>>t;
    lp(i,0,t)
    {
        CF();
    }
    return 0;
}
//END

**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 where the code might fail. The code appears to have some logical issues in handling specific scenarios, particularly when the sequences \( s \) and \( t \) have certain properties.

### Analysis of the Buggy Code
1. **Counting '1's in \( s \):** The code counts the number of '1's in \( s \) but does not use this count effectively.
2. **Initial Check:** The code checks if \( s \) and \( t \) are already equal and outputs "Yes" if they are.
3. **Special Case Check:** The code has a special case check for when the substrings \( s[1:] \) and \( t[1:] \) are equal but the first elements differ, in which case it outputs "No".

### Edge Cases to Consider
1. **Sequences with the same number of '1's but different patterns:**
2. **Sequences where the first element differs but the rest are the same:**
3. **Sequences where the first element is the same but the rest differ:**
4. **Sequences of length 1:**
5. **Sequences with all '0's or all '1's:**

### Challenging Test Input
Let's create a test case that highlights the issue with the special case check and the handling of the first element.

```plaintext
3
5
10101
00101
5
11111
01111
5
00000
10000
```

### Explanation
1. **First Test Case:**
   - \( s = "10101" \)
   - \( t = "00101" \)
   - The sequences differ only in the first element, but the rest are the same. The code should output "No" because the first element cannot be changed by the XOR operation.

2. **Second Test Case:**
   - \( s = "11111" \)
   - \( t = "01111" \)
   - The sequences differ only in the first element, but the rest are the same. The code should output "No" because the first element cannot be changed by the XOR operation.

3. **Third Test Case:**
   - \( s = "00000" \)
   - \( t = "10000" \)
   - The sequences differ only in the first element, but the rest are the same. The code should output "No" because the first element cannot be changed by the XOR operation.

This test input should help 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: Fun Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vova really loves the [XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation (denoted as $\oplus$). Recently, when he was going to sleep, he came up with a fun game.

At the beginning of the game, Vova chooses two binary sequences $s$ and $t$ of length $n$ and gives them to Vanya. A binary sequence is a sequence consisting only of the numbers $0$ and $1$. Vanya can choose integers $l, r$ such that $1 \leq l \leq r \leq n$, and for all $l \leq i \leq r$ simultaneously replace $s_i$ with $s_i \oplus s_{i - l + 1}$, where $s_i$ is the $i$-th element of the sequence $s$.

In order for the game to be interesting, there must be a possibility to win. Vanya wins if, with an unlimited number of actions, he can obtain the sequence $t$ from the sequence $s$. Determine if the game will be interesting for the sequences $s$ and $t$.

Input: Each test consists of multiple test cases. The first line contains an integer $q$ ($1 \le q \le 10^{4}$) — the number of test cases. Then follows the description of the test cases.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the length of the sequences $s$ and $t$.

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

The third line of each test case contains a binary sequence $t$ of length $n$.

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

Output: For each test case, output "Yes" if the game will be interesting, otherwise output "No".

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
1
0
1
7
0110100
0110100
9
100101010
101111110
4
0011
1011
4
0100
0001
8
10110111
01100000
output:
NO
YES
YES
NO
YES
YES

Note: In the first test case, Vanya will not be able to change the sequence $s$ with the only possible action of choosing $l = r = 1$.

In the second test case, the sequences $s$ and $t$ are already equal.

In the third test case, Vanya can act as follows:

  1. Choose $l = 3$ and $r = 5$, then $s$ will become $\mathtt{101101010}$.   2. Choose $l = 5$ and $r = 6$, then $s$ will become $\mathtt{101111010}$.   3. Choose $l = 7$ and $r = 7$, then $s$ will become $\mathtt{101111110}$.

Buggy Code:
#include<stdio.h>
const int N=2e5+5;
int t,n;
char a[N],b[N];
int main(){
	scanf("%d",&t);
	while(t--){
		scanf("%d%s%s",&n,a+1,b+1);
		if(a[1]=='1'){puts("YES");continue;}
		if(b[1]=='1'){puts("NO");continue;}
		puts("YES");
	}
	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 expose the bug in the given buggy code, we need to focus on the conditions it incorrectly handles. The buggy code has a flawed logic where it only checks the first element of the sequences `a` and `b` and makes a decision based on that. This is clearly insufficient to determine if the sequences can be transformed from `s` to `t` using the allowed operations.

### Step-by-Step Analysis:
1. **First Element Check:**
   - The code outputs "YES" if `a[1]` is '1'.
   - The code outputs "NO" if `b[1]` is '1'.
   - Otherwise, it outputs "YES".

2. **Flaw in Logic:**
   - The code does not consider the entire sequence, only the first element.
   - This means it will fail for cases where the first element is not '1' but the sequences still cannot be transformed.

### Test Case Generation:
We need a test case where the first element of `a` and `b` does not immediately trigger the incorrect output, but the sequences cannot be transformed into each other.

### Example Test Case:
- **Length of sequences (`n`):** 3
- **Sequence `a`:** "001"
- **Sequence `b`:** "010"

### Explanation:
- The first element of both `a` and `b` is '0', so the code will output "YES".
- However, it is impossible to transform "001" into "010" using the allowed operations.

### Output:
```plaintext
1
3
001
010
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vova really loves the [XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation (denoted as $\oplus$). Recently, when he was going to sleep, he came up with a fun game.

At the beginning of the game, Vova chooses two binary sequences $s$ and $t$ of length $n$ and gives them to Vanya. A binary sequence is a sequence consisting only of the numbers $0$ and $1$. Vanya can choose integers $l, r$ such that $1 \leq l \leq r \leq n$, and for all $l \leq i \leq r$ simultaneously replace $s_i$ with $s_i \oplus s_{i - l + 1}$, where $s_i$ is the $i$-th element of the sequence $s$.

In order for the game to be interesting, there must be a possibility to win. Vanya wins if, with an unlimited number of actions, he can obtain the sequence $t$ from the sequence $s$. Determine if the game will be interesting for the sequences $s$ and $t$.

Input: Each test consists of multiple test cases. The first line contains an integer $q$ ($1 \le q \le 10^{4}$) — the number of test cases. Then follows the description of the test cases.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the length of the sequences $s$ and $t$.

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

The third line of each test case contains a binary sequence $t$ of length $n$.

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

Output: For each test case, output "Yes" if the game will be interesting, otherwise output "No".

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
1
0
1
7
0110100
0110100
9
100101010
101111110
4
0011
1011
4
0100
0001
8
10110111
01100000
output:
NO
YES
YES
NO
YES
YES

Note: In the first test case, Vanya will not be able to change the sequence $s$ with the only possible action of choosing $l = r = 1$.

In the second test case, the sequences $s$ and $t$ are already equal.

In the third test case, Vanya can act as follows:

  1. Choose $l = 3$ and $r = 5$, then $s$ will become $\mathtt{101101010}$.   2. Choose $l = 5$ and $r = 6$, then $s$ will become $\mathtt{101111010}$.   3. Choose $l = 7$ and $r = 7$, then $s$ will become $\mathtt{101111110}$.

Buggy Code:
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <iostream>
using namespace std;
using namespace __gnu_pbds;
#define jack420 ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define endl "\n"
#define int long long int
#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>

int mod(int n){
    return n % 1000000007;
}

bool isprime(int n){
    if(n==1) return false;
    for(int i=2;i*i<=n;i++)
        if(n%i==0) return false;
    return true;
}

int isSquare(int n){
    int l=1,h=n;
    while(l<=h){
        int mid=(l+h)/2;
        if(mid*mid==n) return mid;
        if(mid*mid<n) l=mid+1;
        else h=mid-1;
    }
    return -1;
}

bool cmp(int a, int b){
    return a>b;
}

int power(int a,int b){
    int temp=1;
    while(b--)
        temp*=a;
    return temp;
}

void code(){
    int n; cin>>n;
    string s,c; cin>>s>>c;
    if(s[0]=='0'&&c[0]=='1') cout<<"NO";
    else cout<<"YES";
}

int32_t main(){
    #ifndef ONLINE_JUDGE
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);
    #endif
        jack420

    int t;
    cin>>t;
    while(t--)
    {
        code();
        cout<<endl;
    }
}

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

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

Think step by step.

Response:
```plaintext
6
1
0
1
7
0110100
0110100
9
100101010
101111110
4
0011
1011
4
0100
0001
8
10110111
01100000
```

### Explanation:

1. **Test Case 1**:
   - `n = 1`
   - `s = "0"`
   - `t = "1"`
   - **Expected Output**: `NO`
   - **Reason**: The only possible action is to choose `l = r = 1`, which does not change the sequence `s`.

2. **Test Case 2**:
   - `n = 7`
   - `s = "0110100"`
   - `t = "0110100"`
   - **Expected Output**: `YES`
   - **Reason**: The sequences `s` and `t` are already equal, so no operations are needed.

3. **Test Case 3**:
   - `n = 9`
   - `s = "100101010"`
   - `t = "101111110"`
   - **Expected Output**: `YES`
   - **Reason**: Vanya can perform the following operations to transform `s` into `t`:
     - Choose `l = 3` and `r = 5`, then `s` becomes `101101010`.
     - Choose `l = 5` and `r = 6`, then `s` becomes `101111010`.
     - Choose `l = 7` and `r = 7`, then `s` becomes `101111110`.

4. **Test Case 4**:
   - `n = 4`
   - `s = "0011"`
   - `t = "1011"`
   - **Expected Output**: `NO`
   - **Reason**: The first element of `s` is `0` and the first element of `t` is `1`. Since the first element cannot be changed, the sequences cannot be made equal.

5. **Test Case 5**:
   - `n = 4`
   - `s = "0100"`
   - `t = "0001"`
   - **Expected Output**: `YES`
   - **Reason**: Vanya can perform the following operations to transform `s` into `t`:
     - Choose `l = 1` and `r = 3`, then `s` becomes `0111`.
     - Choose `l = 2` and `r = 4`, then `s` becomes `0001`.

6. **Test Case 6**:
   - `n = 8`
   - `s = "10110111"`
   - `t = "01100000"`
   - **Expected Output**: `YES`
   - **Reason**: Vanya can perform the following operations to transform `s` into `t`:
     - Choose `l = 1` and `r = 8`, then `s` becomes `11011100`.
     - Choose `l = 2` and `r = 7`, then `s` becomes `01100000`.

These test cases cover a range of scenarios, including edge cases where the sequences are already equal, where the first elements differ, and where multiple operations are required to transform `s` into `t`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vova really loves the [XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation (denoted as $\oplus$). Recently, when he was going to sleep, he came up with a fun game.

At the beginning of the game, Vova chooses two binary sequences $s$ and $t$ of length $n$ and gives them to Vanya. A binary sequence is a sequence consisting only of the numbers $0$ and $1$. Vanya can choose integers $l, r$ such that $1 \leq l \leq r \leq n$, and for all $l \leq i \leq r$ simultaneously replace $s_i$ with $s_i \oplus s_{i - l + 1}$, where $s_i$ is the $i$-th element of the sequence $s$.

In order for the game to be interesting, there must be a possibility to win. Vanya wins if, with an unlimited number of actions, he can obtain the sequence $t$ from the sequence $s$. Determine if the game will be interesting for the sequences $s$ and $t$.

Input: Each test consists of multiple test cases. The first line contains an integer $q$ ($1 \le q \le 10^{4}$) — the number of test cases. Then follows the description of the test cases.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the length of the sequences $s$ and $t$.

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

The third line of each test case contains a binary sequence $t$ of length $n$.

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

Output: For each test case, output "Yes" if the game will be interesting, otherwise output "No".

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
1
0
1
7
0110100
0110100
9
100101010
101111110
4
0011
1011
4
0100
0001
8
10110111
01100000
output:
NO
YES
YES
NO
YES
YES

Note: In the first test case, Vanya will not be able to change the sequence $s$ with the only possible action of choosing $l = r = 1$.

In the second test case, the sequences $s$ and $t$ are already equal.

In the third test case, Vanya can act as follows:

  1. Choose $l = 3$ and $r = 5$, then $s$ will become $\mathtt{101101010}$.   2. Choose $l = 5$ and $r = 6$, then $s$ will become $\mathtt{101111010}$.   3. Choose $l = 7$ and $r = 7$, then $s$ will become $\mathtt{101111110}$.

Buggy Code:
def solution():
    n = int(input())
    s = input()
    t = input()
    for i in range(n):
        if s[i] == '1':
            return 'YES'
        if t[i] == '1':
            return "NO"

for _ in range(int(input())):
    print(solution())

**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 buggy code, we need to focus on edge cases where the logic of the code fails. The buggy code checks if there is a '1' in either sequence and immediately returns "YES" if there is a '1' in `s` or "NO" if there is a '1' in `t`. This logic is clearly flawed because it does not consider the actual transformation process described in the problem.

Let's consider a test case where both `s` and `t` contain '1's, but the transformation is still possible. This will highlight the bug in the code.

### Test Case Analysis
1. **Case 1: Both `s` and `t` contain '1's, but transformation is possible.**
   - `s = "0101"`
   - `t = "1010"`
   - Here, Vanya can transform `s` to `t` by choosing appropriate `l` and `r` values. For example:
     - Choose `l = 1` and `r = 2`, then `s` becomes `1101`.
     - Choose `l = 2` and `r = 3`, then `s` becomes `1011`.
     - Choose `l = 3` and `r = 4`, then `s` becomes `1010`.

2. **Case 2: Both `s` and `t` contain '1's, but transformation is not possible.**
   - `s = "0101"`
   - `t = "1110"`
   - Here, it is impossible to transform `s` to `t` because the transformation rules do not allow it.

### Challenging Test Input
Let's create a test input that includes both cases to maximize the failure probability:

```plaintext
2
4
0101
1010
4
0101
1110
```

### Explanation
- **First Test Case:**
  - `n = 4`
  - `s = "0101"`
  - `t = "1010"`
  - The transformation is possible, so the correct output should be "YES". The buggy code will return "YES" because it finds a '1' in `s`, but this does not prove the transformation is possible.

- **Second Test Case:**
  - `n = 4`
  - `s = "0101"`
  - `t = "1110"`
  - The transformation is not possible, so the correct output should be "NO". The buggy code will return "YES" because it finds a '1' in `s`, which is incorrect.

This test input will expose the bug in the buggy code by showing that it fails to correctly determine the possibility of transformation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vova really loves the [XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation (denoted as $\oplus$). Recently, when he was going to sleep, he came up with a fun game.

At the beginning of the game, Vova chooses two binary sequences $s$ and $t$ of length $n$ and gives them to Vanya. A binary sequence is a sequence consisting only of the numbers $0$ and $1$. Vanya can choose integers $l, r$ such that $1 \leq l \leq r \leq n$, and for all $l \leq i \leq r$ simultaneously replace $s_i$ with $s_i \oplus s_{i - l + 1}$, where $s_i$ is the $i$-th element of the sequence $s$.

In order for the game to be interesting, there must be a possibility to win. Vanya wins if, with an unlimited number of actions, he can obtain the sequence $t$ from the sequence $s$. Determine if the game will be interesting for the sequences $s$ and $t$.

Input: Each test consists of multiple test cases. The first line contains an integer $q$ ($1 \le q \le 10^{4}$) — the number of test cases. Then follows the description of the test cases.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the length of the sequences $s$ and $t$.

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

The third line of each test case contains a binary sequence $t$ of length $n$.

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

Output: For each test case, output "Yes" if the game will be interesting, otherwise output "No".

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
1
0
1
7
0110100
0110100
9
100101010
101111110
4
0011
1011
4
0100
0001
8
10110111
01100000
output:
NO
YES
YES
NO
YES
YES

Note: In the first test case, Vanya will not be able to change the sequence $s$ with the only possible action of choosing $l = r = 1$.

In the second test case, the sequences $s$ and $t$ are already equal.

In the third test case, Vanya can act as follows:

  1. Choose $l = 3$ and $r = 5$, then $s$ will become $\mathtt{101101010}$.   2. Choose $l = 5$ and $r = 6$, then $s$ will become $\mathtt{101111010}$.   3. Choose $l = 7$ and $r = 7$, then $s$ will become $\mathtt{101111110}$.

Buggy Code:
for _ in range(int(input())):
    n=int(input())
    s=input()
    t=input()
    if((s[0]!=t[0] and n==1) or (n>1 and s[0]!=t[0]  and s[0]==s[1])):
        print("NO")
    else:
        brr=[]
        if(s[0]=="0"):
            brr.append([1,0])
        else:
            brr.append([0,1])
        for i in range(1,n):
            if(s[i]=="0"):
                brr.append([1,brr[i-1][1]])
            else:
                brr.append([brr[i-1][1],1])
        flag=0
        for i in range(1,n-1):
            if(s[i]!=t[i] and s[i]=="0" and brr[i-1][1]!=1 and s[i]==s[i+1]):
                flag=1
                break
            if(s[i]!=t[i] and s[i]=="1" and brr[i-1][0]!=1 and s[i]==s[i+1]):
                flag=1
                break
        if(s[n-1]!=t[n-1]):
            if(s[n-1]=="1" and brr[n-2][0]!=1):
                flag=1
                break
            if(s[n-1]=="0" and brr[n-2][1]!=1):
                flag=1
                break

        if(flag==0):
            print("YES")
        else:
            print("NO")

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

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

Think step by step.

Response:
```plaintext
6
1
0
1
7
0110100
0110100
9
100101010
101111110
4
0011
1011
4
0100
0001
8
10110111
01100000
```

### Explanation:
1. **Test Case 1:**
   - **n = 1**
   - **s = 0**
   - **t = 1**
   - **Expected Output: NO**
   - **Reason:** With only one element, it is impossible to change it using the XOR operation.

2. **Test Case 2:**
   - **n = 7**
   - **s = 0110100**
   - **t = 0110100**
   - **Expected Output: YES**
   - **Reason:** The sequences are already equal, so no operations are needed.

3. **Test Case 3:**
   - **n = 9**
   - **s = 100101010**
   - **t = 101111110**
   - **Expected Output: YES**
   - **Reason:** Vanya can perform the following operations to transform `s` into `t`:
     1. Choose `l = 3` and `r = 5` to get `101101010`.
     2. Choose `l = 5` and `r = 6` to get `101111010`.
     3. Choose `l = 7` and `r = 7` to get `101111110`.

4. **Test Case 4:**
   - **n = 4**
   - **s = 0011**
   - **t = 1011**
   - **Expected Output: NO**
   - **Reason:** The first element of `s` is different from the first element of `t`, and there is no way to change it using the XOR operation without affecting the other elements.

5. **Test Case 5:**
   - **n = 4**
   - **s = 0100**
   - **t = 0001**
   - **Expected Output: NO**
   - **Reason:** The second element of `s` is different from the second element of `t`, and there is no way to change it using the XOR operation without affecting the other elements.

6. **Test Case 6:**
   - **n = 8**
   - **s = 10110111**
   - **t = 01100000**
   - **Expected Output: YES**
   - **Reason:** Vanya can perform the following operations to transform `s` into `t`:
     1. Choose `l = 1` and `r = 8` to get `01100110`.
     2. Choose `l = 6` and `r = 8` to get `01100000`.

These test cases are designed to cover various edge cases and scenarios that can expose the bug in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vova really loves the [XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation (denoted as $\oplus$). Recently, when he was going to sleep, he came up with a fun game.

At the beginning of the game, Vova chooses two binary sequences $s$ and $t$ of length $n$ and gives them to Vanya. A binary sequence is a sequence consisting only of the numbers $0$ and $1$. Vanya can choose integers $l, r$ such that $1 \leq l \leq r \leq n$, and for all $l \leq i \leq r$ simultaneously replace $s_i$ with $s_i \oplus s_{i - l + 1}$, where $s_i$ is the $i$-th element of the sequence $s$.

In order for the game to be interesting, there must be a possibility to win. Vanya wins if, with an unlimited number of actions, he can obtain the sequence $t$ from the sequence $s$. Determine if the game will be interesting for the sequences $s$ and $t$.

Input: Each test consists of multiple test cases. The first line contains an integer $q$ ($1 \le q \le 10^{4}$) — the number of test cases. Then follows the description of the test cases.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the length of the sequences $s$ and $t$.

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

The third line of each test case contains a binary sequence $t$ of length $n$.

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

Output: For each test case, output "Yes" if the game will be interesting, otherwise output "No".

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
1
0
1
7
0110100
0110100
9
100101010
101111110
4
0011
1011
4
0100
0001
8
10110111
01100000
output:
NO
YES
YES
NO
YES
YES

Note: In the first test case, Vanya will not be able to change the sequence $s$ with the only possible action of choosing $l = r = 1$.

In the second test case, the sequences $s$ and $t$ are already equal.

In the third test case, Vanya can act as follows:

  1. Choose $l = 3$ and $r = 5$, then $s$ will become $\mathtt{101101010}$.   2. Choose $l = 5$ and $r = 6$, then $s$ will become $\mathtt{101111010}$.   3. Choose $l = 7$ and $r = 7$, then $s$ will become $\mathtt{101111110}$.

Buggy Code:
#!/usr/bin/env python
from __future__ import division, print_function

import os
import sys
from io import BytesIO, IOBase

if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip

def main():
    for _ in range(int(input())):
        n = int(input())
        s = input()
        t = input()

        flag = False
        for i in range(n):
            if s[i] == "1":
                flag = True
                break

            if t[i] == "1":
                if i < n - 1 and s[i + 1] == "1":
                    flag = True
                break

        print("YES" if flag else "NO")

# region fastio

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

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

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

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

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

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

def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()

if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

input = lambda: sys.stdin.readline().rstrip("\r\n")

# endregion

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios where the code might fail. The bug in the code seems to be that it does not correctly handle the XOR operation and the transformation of the sequence `s` to `t`.

Here are the steps to generate a small-scale test input:

1. **Identify the Bug**: The code only checks if there is a '1' in `s` or if there is a '1' in `t` followed by a '1' in `s` at the next position. This is insufficient to determine if `s` can be transformed into `t` using the XOR operation.

2. **Create a Test Case**: We need a test case where `s` can be transformed into `t` through a series of valid XOR operations, but the code fails to recognize this.

3. **Example Test Case**:
   - **Test Case 1**: `s = "0101"` and `t = "0011"`. In this case, Vanya can perform the following operations:
     - Choose `l = 1` and `r = 2`, then `s` becomes `"0101"`.
     - Choose `l = 2` and `r = 3`, then `s` becomes `"0011"`.
   - **Test Case 2**: `s = "1010"` and `t = "1100"`. In this case, Vanya can perform the following operations:
     - Choose `l = 1` and `r = 2`, then `s` becomes `"1010"`.
     - Choose `l = 2` and `r = 3`, then `s` becomes `"1100"`.

4. **Combine Test Cases**: We will combine these test cases into a single input.

Here is the test input:

```plaintext
2
4
0101
0011
4
1010
1100
```

This input should expose the bug in the provided code, as the code will incorrectly output "NO" for both cases, while the correct output should be "YES" for both.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vova really loves the [XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation (denoted as $\oplus$). Recently, when he was going to sleep, he came up with a fun game.

At the beginning of the game, Vova chooses two binary sequences $s$ and $t$ of length $n$ and gives them to Vanya. A binary sequence is a sequence consisting only of the numbers $0$ and $1$. Vanya can choose integers $l, r$ such that $1 \leq l \leq r \leq n$, and for all $l \leq i \leq r$ simultaneously replace $s_i$ with $s_i \oplus s_{i - l + 1}$, where $s_i$ is the $i$-th element of the sequence $s$.

In order for the game to be interesting, there must be a possibility to win. Vanya wins if, with an unlimited number of actions, he can obtain the sequence $t$ from the sequence $s$. Determine if the game will be interesting for the sequences $s$ and $t$.

Input: Each test consists of multiple test cases. The first line contains an integer $q$ ($1 \le q \le 10^{4}$) — the number of test cases. Then follows the description of the test cases.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the length of the sequences $s$ and $t$.

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

The third line of each test case contains a binary sequence $t$ of length $n$.

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

Output: For each test case, output "Yes" if the game will be interesting, otherwise output "No".

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
1
0
1
7
0110100
0110100
9
100101010
101111110
4
0011
1011
4
0100
0001
8
10110111
01100000
output:
NO
YES
YES
NO
YES
YES

Note: In the first test case, Vanya will not be able to change the sequence $s$ with the only possible action of choosing $l = r = 1$.

In the second test case, the sequences $s$ and $t$ are already equal.

In the third test case, Vanya can act as follows:

  1. Choose $l = 3$ and $r = 5$, then $s$ will become $\mathtt{101101010}$.   2. Choose $l = 5$ and $r = 6$, then $s$ will become $\mathtt{101111010}$.   3. Choose $l = 7$ and $r = 7$, then $s$ will become $\mathtt{101111110}$.

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

public class B_959 {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc=new Scanner(System.in);
		PrintStream out=new PrintStream(System.out);
		int t=sc.nextInt();
		while(t-->0)
		{
			int n=sc.nextInt();
			String s=sc.next();
			String tx=sc.next();

			List<Integer> al=new ArrayList<>();

			for(int i=0;i<n;i++)
			{
				if(s.charAt(i)!=tx.charAt(i))
				{
					al.add(i);
				}

			}
			if(al.isEmpty())
			{
				out.println("Yes");

			}
			else
			{
			if(al.size()==1)
			{
				if(al.get(0)==0)
				{
					out.println("No");
				}
				else
				{
					if(s.charAt(0)!=s.charAt(al.get(0)))
					{
						out.println("Yes");

					}
					else
					{
						out.println("No");

					}
				}
			}
			else
			{
				int d=al.get(1)-al.get(0);
				boolean flag=true;

			if(flag)
			{
				out.println("Yes");
			}
			else
			{
				out.println("No");
			}
			}
			}

		}

	}

}

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

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

Think step by step.

Response:
```plaintext
3
5
00000
11111
5
10101
01010
5
10101
10101
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vova really loves the [XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation (denoted as $\oplus$). Recently, when he was going to sleep, he came up with a fun game.

At the beginning of the game, Vova chooses two binary sequences $s$ and $t$ of length $n$ and gives them to Vanya. A binary sequence is a sequence consisting only of the numbers $0$ and $1$. Vanya can choose integers $l, r$ such that $1 \leq l \leq r \leq n$, and for all $l \leq i \leq r$ simultaneously replace $s_i$ with $s_i \oplus s_{i - l + 1}$, where $s_i$ is the $i$-th element of the sequence $s$.

In order for the game to be interesting, there must be a possibility to win. Vanya wins if, with an unlimited number of actions, he can obtain the sequence $t$ from the sequence $s$. Determine if the game will be interesting for the sequences $s$ and $t$.

Input: Each test consists of multiple test cases. The first line contains an integer $q$ ($1 \le q \le 10^{4}$) — the number of test cases. Then follows the description of the test cases.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the length of the sequences $s$ and $t$.

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

The third line of each test case contains a binary sequence $t$ of length $n$.

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

Output: For each test case, output "Yes" if the game will be interesting, otherwise output "No".

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
1
0
1
7
0110100
0110100
9
100101010
101111110
4
0011
1011
4
0100
0001
8
10110111
01100000
output:
NO
YES
YES
NO
YES
YES

Note: In the first test case, Vanya will not be able to change the sequence $s$ with the only possible action of choosing $l = r = 1$.

In the second test case, the sequences $s$ and $t$ are already equal.

In the third test case, Vanya can act as follows:

  1. Choose $l = 3$ and $r = 5$, then $s$ will become $\mathtt{101101010}$.   2. Choose $l = 5$ and $r = 6$, then $s$ will become $\mathtt{101111010}$.   3. Choose $l = 7$ and $r = 7$, then $s$ will become $\mathtt{101111110}$.

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

public class FunGame {
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder output = new StringBuilder();
        int T = Integer.parseInt(reader.readLine());

        for (int j = 0; j < T; j++) {
            int length = Integer.parseInt(reader.readLine());
            String s = reader.readLine();
            String t = reader.readLine();

            if (s.equals(t)) {
                output.append("YES\n");
                continue;
            }

            int count0 = 0, count1 = 0;
            boolean possible = true;

            for (int i = 0; i < length; i++) {
                if (s.charAt(i) == '0') {
                    count0++;
                } else {
                    count1++;
                }

                if (s.charAt(i) != t.charAt(i)) {
                    if (t.charAt(i) == '1') {
                        if (count1 > 0) {
                            count1--;
                        } else {
                            possible = false;
                            break;
                        }
                    }
                }
            }

            if (possible) {
                output.append("YES\n");
            } else {
                output.append("NO\n");
            }
        }
        System.out.print(output);
    }
}

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

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

Think step by step.

Response:
```plaintext
3
1
0
1
2
01
10
4
0011
1100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vova really loves the [XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation (denoted as $\oplus$). Recently, when he was going to sleep, he came up with a fun game.

At the beginning of the game, Vova chooses two binary sequences $s$ and $t$ of length $n$ and gives them to Vanya. A binary sequence is a sequence consisting only of the numbers $0$ and $1$. Vanya can choose integers $l, r$ such that $1 \leq l \leq r \leq n$, and for all $l \leq i \leq r$ simultaneously replace $s_i$ with $s_i \oplus s_{i - l + 1}$, where $s_i$ is the $i$-th element of the sequence $s$.

In order for the game to be interesting, there must be a possibility to win. Vanya wins if, with an unlimited number of actions, he can obtain the sequence $t$ from the sequence $s$. Determine if the game will be interesting for the sequences $s$ and $t$.

Input: Each test consists of multiple test cases. The first line contains an integer $q$ ($1 \le q \le 10^{4}$) — the number of test cases. Then follows the description of the test cases.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the length of the sequences $s$ and $t$.

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

The third line of each test case contains a binary sequence $t$ of length $n$.

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

Output: For each test case, output "Yes" if the game will be interesting, otherwise output "No".

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
1
0
1
7
0110100
0110100
9
100101010
101111110
4
0011
1011
4
0100
0001
8
10110111
01100000
output:
NO
YES
YES
NO
YES
YES

Note: In the first test case, Vanya will not be able to change the sequence $s$ with the only possible action of choosing $l = r = 1$.

In the second test case, the sequences $s$ and $t$ are already equal.

In the third test case, Vanya can act as follows:

  1. Choose $l = 3$ and $r = 5$, then $s$ will become $\mathtt{101101010}$.   2. Choose $l = 5$ and $r = 6$, then $s$ will become $\mathtt{101111010}$.   3. Choose $l = 7$ and $r = 7$, then $s$ will become $\mathtt{101111110}$.

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

void sol(){
	int n;
	cin>>n;
	cin>>s>>t;
	s=' '+s;
	t=' '+t;
	if(s[1]=='1'){
		cout<<"YES";
		return;
	}else{
		int len=0;
		for(int i=1;i<=s.size();i++){
			if(s[i]=='0') len++;
			if(s[i]=='1') break;
		}
		//
//		cout<<"eln:"<<len<<endl;
		if(len==0){
			if(s==t) cout<<"YES";
			else cout<<"NO";
		}else{
			int yes=1;
			for(int i=1;i<=len;i++){
				if(s[i]!=t[i])
				yes=0;
				break;
			}
			if(yes) cout<<"YES";
			else cout<<"NO";
		}
	}
}
signed main(){
	int t;
	cin>>t;
	while(t--){
		sol();
		cout<<endl;
	}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to focus on edge cases and scenarios that the code might handle incorrectly. The buggy code has several issues, including incorrect handling of leading zeros and the logic for determining if the sequences can be transformed.

### Analysis of the Buggy Code:
1. **Leading Zeros Handling:**
   - The code incorrectly assumes that if the first character of `s` is '1', the answer is always "YES". This is not always true.
   - The code also incorrectly handles the case where `s` starts with a long sequence of zeros.

2. **Transformation Logic:**
   - The code does not correctly check if the sequences can be transformed from `s` to `t`.

### Challenging Test Cases:
1. **Test Case with Leading Zeros:**
   - A sequence `s` that starts with a long sequence of zeros and a sequence `t` that cannot be transformed from `s`.

2. **Test Case with Different Initial Values:**
   - A sequence `s` that starts with '0' and a sequence `t` that starts with '1', which should not be transformable.

3. **Test Case with Different Lengths of Leading Zeros:**
   - A sequence `s` that starts with a different number of leading zeros compared to `t`.

### Test Input:
```plaintext
3
10
0000000000
1000000000
5
00000
11111
10
0000000000
0000000001
```

### Explanation:
1. **First Test Case:**
   - `s = "0000000000"`
   - `t = "1000000000"`
   - The sequence `s` starts with a long sequence of zeros, and it is impossible to transform `s` to `t` using the allowed operations. The expected output is "NO".

2. **Second Test Case:**
   - `s = "00000"`
   - `t = "11111"`
   - The sequence `s` starts with '0', and it is impossible to transform `s` to `t` using the allowed operations. The expected output is "NO".

3. **Third Test Case:**
   - `s = "0000000000"`
   - `t = "0000000001"`
   - The sequence `s` starts with a long sequence of zeros, and it is impossible to transform `s` to `t` using the allowed operations. The expected output is "NO".

These test cases should expose the bugs in the given code and help identify the issues in the logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vova really loves the [XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation (denoted as $\oplus$). Recently, when he was going to sleep, he came up with a fun game.

At the beginning of the game, Vova chooses two binary sequences $s$ and $t$ of length $n$ and gives them to Vanya. A binary sequence is a sequence consisting only of the numbers $0$ and $1$. Vanya can choose integers $l, r$ such that $1 \leq l \leq r \leq n$, and for all $l \leq i \leq r$ simultaneously replace $s_i$ with $s_i \oplus s_{i - l + 1}$, where $s_i$ is the $i$-th element of the sequence $s$.

In order for the game to be interesting, there must be a possibility to win. Vanya wins if, with an unlimited number of actions, he can obtain the sequence $t$ from the sequence $s$. Determine if the game will be interesting for the sequences $s$ and $t$.

Input: Each test consists of multiple test cases. The first line contains an integer $q$ ($1 \le q \le 10^{4}$) — the number of test cases. Then follows the description of the test cases.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the length of the sequences $s$ and $t$.

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

The third line of each test case contains a binary sequence $t$ of length $n$.

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

Output: For each test case, output "Yes" if the game will be interesting, otherwise output "No".

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
1
0
1
7
0110100
0110100
9
100101010
101111110
4
0011
1011
4
0100
0001
8
10110111
01100000
output:
NO
YES
YES
NO
YES
YES

Note: In the first test case, Vanya will not be able to change the sequence $s$ with the only possible action of choosing $l = r = 1$.

In the second test case, the sequences $s$ and $t$ are already equal.

In the third test case, Vanya can act as follows:

  1. Choose $l = 3$ and $r = 5$, then $s$ will become $\mathtt{101101010}$.   2. Choose $l = 5$ and $r = 6$, then $s$ will become $\mathtt{101111010}$.   3. Choose $l = 7$ and $r = 7$, then $s$ will become $\mathtt{101111110}$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rp(i,a,b)        for(int i=a;i<b;i++)

void solvetask(){
    int n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    if(a[0] == '0' && b[0] == '1') cout << "NO\n";
    else{
        if(a[0] == '1'){
            rp(i, 0, n){
                if(a[i] == '1' && b[i] == '0') a[i] = '0';
                if(a[i] == '0' && b[i] == '1') a[i] = '1';
            }
            //cout << a << "\n" << b << "\n";
            if(b == a) cout << "YES\n";
            else cout << "NO\n";
        }
        else{
            bool flag = false;
            rp(i, 0, n) if(a[i] == '1') flag = true;
            if(flag) rp(i, 0, n){
                if(a[i] == '1' && b[i] == '0') a[i] = '0';
                if(a[i] == '0' && b[i] == '1') a[i] = '1';
            }
            rp(i, 0, n){
                if(b[i] == '1') break;
                else a[i] = '0';
            }
            if(b == a) cout << "YES\n";
            else cout << "NO\n";
        }

    }

}

int main(){
    cin.tie(0)->sync_with_stdio(0);
    ll t = 1;
    cin >> t;
    while(t--) solvetask();
}

**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 highlight the issues in the logic. Let's break down the potential problems:

1. **Initial Condition Check**: The code checks if `a[0] == '0'` and `b[0] == '1'` and immediately outputs "NO". This is a valid check but can be misleading if the rest of the sequence can be transformed.
2. **Transformation Logic**: The code attempts to transform `a` to match `b` by flipping bits. However, it does not correctly handle all possible transformations, especially when the sequences have complex patterns.

### Challenging Test Case
We can create a test case where the initial condition fails but the sequences can still be transformed to match each other. For example, consider the following test case:

```plaintext
1
4
0101
1010
```

### Explanation
- **Initial Condition**: `a[0] == '0'` and `b[0] == '1'`. According to the initial check in the code, this should output "NO".
- **Transformation**: However, this sequence can be transformed as follows:
  - Choose `l = 1` and `r = 2`: `a` becomes `1101`.
  - Choose `l = 2` and `r = 3`: `a` becomes `1011`.
  - Choose `l = 3` and `r = 4`: `a` becomes `1010`.

This test case should expose the bug in the initial condition check and show that the sequences can be transformed to match each other, even though the initial check fails.

### Output
```plaintext
1
4
0101
1010
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vova really loves the [XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation (denoted as $\oplus$). Recently, when he was going to sleep, he came up with a fun game.

At the beginning of the game, Vova chooses two binary sequences $s$ and $t$ of length $n$ and gives them to Vanya. A binary sequence is a sequence consisting only of the numbers $0$ and $1$. Vanya can choose integers $l, r$ such that $1 \leq l \leq r \leq n$, and for all $l \leq i \leq r$ simultaneously replace $s_i$ with $s_i \oplus s_{i - l + 1}$, where $s_i$ is the $i$-th element of the sequence $s$.

In order for the game to be interesting, there must be a possibility to win. Vanya wins if, with an unlimited number of actions, he can obtain the sequence $t$ from the sequence $s$. Determine if the game will be interesting for the sequences $s$ and $t$.

Input: Each test consists of multiple test cases. The first line contains an integer $q$ ($1 \le q \le 10^{4}$) — the number of test cases. Then follows the description of the test cases.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the length of the sequences $s$ and $t$.

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

The third line of each test case contains a binary sequence $t$ of length $n$.

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

Output: For each test case, output "Yes" if the game will be interesting, otherwise output "No".

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
1
0
1
7
0110100
0110100
9
100101010
101111110
4
0011
1011
4
0100
0001
8
10110111
01100000
output:
NO
YES
YES
NO
YES
YES

Note: In the first test case, Vanya will not be able to change the sequence $s$ with the only possible action of choosing $l = r = 1$.

In the second test case, the sequences $s$ and $t$ are already equal.

In the third test case, Vanya can act as follows:

  1. Choose $l = 3$ and $r = 5$, then $s$ will become $\mathtt{101101010}$.   2. Choose $l = 5$ and $r = 6$, then $s$ will become $\mathtt{101111010}$.   3. Choose $l = 7$ and $r = 7$, then $s$ will become $\mathtt{101111110}$.

Buggy Code:
def solve(n,s,t):
    if s == t: return "Yes"
    if (t.count('1') == 0): return "Yes"
    s= list(s)
    t= list(t)
    if s[0]=='1': return "Yes"
    if s[0]=='0' and t[0]=='1': return "No"

    if s.count('1')==0:
        if t.count('1')==0: return "Yes"
        else: return "Yes"

    if(s.index('1')>t.index('1')): return "No"

    return "Yes"

for _ in range(int(input())):
    n= int(input())
    s= input()
    t= input()
    print(solve(n,s,t))

# print("devfb".index('x'))

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases that the code might handle incorrectly. Here are the key points to consider:

1. **Initial Conditions**: The code checks if `s` and `t` are already equal, if `t` has no '1's, and if the first character of `s` is '1'. These are straightforward conditions and are likely to be handled correctly.

2. **First '1' Position**: The code checks if the position of the first '1' in `s` is greater than the position of the first '1' in `t`. This is a critical check because it can determine whether the transformation is possible.

3. **Count of '1's**: The code has some logic to handle cases where `s` or `t` has no '1's, but it might not handle other cases correctly.

4. **Transformation Logic**: The code does not correctly simulate the transformation process, which can lead to incorrect results.

Given these points, a challenging test input would be one where the first '1' in `s` is at a position greater than the first '1' in `t`, but the transformation should still be possible. This will expose the bug in the code.

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

```plaintext
1
5
00011
01000
```

Explanation:
- The first '1' in `s` is at position 4 (0-based index), and the first '1' in `t` is at position 1.
- Despite this, it is possible to transform `s` into `t` using the allowed operations:
  - Choose `l = 1` and `r = 4`, then `s` becomes `01011`.
  - Choose `l = 2` and `r = 4`, then `s` becomes `01000`.

This test case should expose the bug in the code because the code incorrectly returns "No" for this input.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vova really loves the [XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation (denoted as $\oplus$). Recently, when he was going to sleep, he came up with a fun game.

At the beginning of the game, Vova chooses two binary sequences $s$ and $t$ of length $n$ and gives them to Vanya. A binary sequence is a sequence consisting only of the numbers $0$ and $1$. Vanya can choose integers $l, r$ such that $1 \leq l \leq r \leq n$, and for all $l \leq i \leq r$ simultaneously replace $s_i$ with $s_i \oplus s_{i - l + 1}$, where $s_i$ is the $i$-th element of the sequence $s$.

In order for the game to be interesting, there must be a possibility to win. Vanya wins if, with an unlimited number of actions, he can obtain the sequence $t$ from the sequence $s$. Determine if the game will be interesting for the sequences $s$ and $t$.

Input: Each test consists of multiple test cases. The first line contains an integer $q$ ($1 \le q \le 10^{4}$) — the number of test cases. Then follows the description of the test cases.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the length of the sequences $s$ and $t$.

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

The third line of each test case contains a binary sequence $t$ of length $n$.

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

Output: For each test case, output "Yes" if the game will be interesting, otherwise output "No".

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
1
0
1
7
0110100
0110100
9
100101010
101111110
4
0011
1011
4
0100
0001
8
10110111
01100000
output:
NO
YES
YES
NO
YES
YES

Note: In the first test case, Vanya will not be able to change the sequence $s$ with the only possible action of choosing $l = r = 1$.

In the second test case, the sequences $s$ and $t$ are already equal.

In the third test case, Vanya can act as follows:

  1. Choose $l = 3$ and $r = 5$, then $s$ will become $\mathtt{101101010}$.   2. Choose $l = 5$ and $r = 6$, then $s$ will become $\mathtt{101111010}$.   3. Choose $l = 7$ and $r = 7$, then $s$ will become $\mathtt{101111110}$.

Buggy Code:
# # for _ in range(int(input())):
# # 	n = int(input())
# # 	a = list(map(int,input().split()))
# # 	b = list(map(int,input().split()))
# # 	if a == b:
# # 		print("YES")
# # 		continue
# # 	if len(a) == 1:
# # 		print('NO')
# # 		continue
# # 	try:
# # 		c1 = a.index(1,-1)
# # 	except ValueError:
# # 		c1 = -1
# # 	try:
# # 		c0 = a.index(0,-1)
# # 	except ValueError:
# # 		c0 = -1

# # 	m = b[max(c1,c0):]
# # 	if len(set(m)) == 1 and m[0] == 0:
# # 		print("YES")
# # 	else:
# # 		print("NO")
# # 	# d = list(reversed(a))
# # 	# k = d[0]
# # 	# ans = []
# # 	# for i in range(len(d)):
# # 	# 	if d[i] != k:
# # 	# 		ans = d[:i]

# # 	# 		break
# # 	# print(ans)
# n = int(input())
# li = list(map(int,input().split()))
# li.sort()
# c1 = 0
# c3 = 0
# c2 = 0
# c4 = 0
# for i in li:
# 	if i ==
for i in range(int(input())):
	x = int(input())
	s = list(str(input()))
	t = list(str(input()))
	if s[0] == '0' and t[0]=='1':
		print("NO")
		continue
	s1 = s.index('1')
	s2 = t.index('1')
	if s1 > s2:
		print("NO")
	else:
		print("YES")

**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
6
1
0
1
7
0110100
0110100
9
100101010
101111110
4
0011
1011
4
0100
0001
8
10110111
01100000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Fun Game
time_limit_ms: 1000
memory_limit_mb: 256
Description: Vova really loves the [XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation (denoted as $\oplus$). Recently, when he was going to sleep, he came up with a fun game.

At the beginning of the game, Vova chooses two binary sequences $s$ and $t$ of length $n$ and gives them to Vanya. A binary sequence is a sequence consisting only of the numbers $0$ and $1$. Vanya can choose integers $l, r$ such that $1 \leq l \leq r \leq n$, and for all $l \leq i \leq r$ simultaneously replace $s_i$ with $s_i \oplus s_{i - l + 1}$, where $s_i$ is the $i$-th element of the sequence $s$.

In order for the game to be interesting, there must be a possibility to win. Vanya wins if, with an unlimited number of actions, he can obtain the sequence $t$ from the sequence $s$. Determine if the game will be interesting for the sequences $s$ and $t$.

Input: Each test consists of multiple test cases. The first line contains an integer $q$ ($1 \le q \le 10^{4}$) — the number of test cases. Then follows the description of the test cases.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the length of the sequences $s$ and $t$.

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

The third line of each test case contains a binary sequence $t$ of length $n$.

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

Output: For each test case, output "Yes" if the game will be interesting, otherwise output "No".

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
6
1
0
1
7
0110100
0110100
9
100101010
101111110
4
0011
1011
4
0100
0001
8
10110111
01100000
output:
NO
YES
YES
NO
YES
YES

Note: In the first test case, Vanya will not be able to change the sequence $s$ with the only possible action of choosing $l = r = 1$.

In the second test case, the sequences $s$ and $t$ are already equal.

In the third test case, Vanya can act as follows:

  1. Choose $l = 3$ and $r = 5$, then $s$ will become $\mathtt{101101010}$.   2. Choose $l = 5$ and $r = 6$, then $s$ will become $\mathtt{101111010}$.   3. Choose $l = 7$ and $r = 7$, then $s$ will become $\mathtt{101111110}$.

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

public class cf2 {
	static final int MOD = 1000000000+7;
	static class FastScanner {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");

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

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

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

		void sort(int[] a){
			ArrayList<Integer> list = new ArrayList<>();
			for(int i : a) list.add(i);
			Collections.sort(list);
			for(int i = 0; i < a.length; i++) a[i] = list.get(i);
		}

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

	public static void main(String[] args) throws IOException {
		FastScanner fs = new FastScanner();
		PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
		int tcs = fs.nextInt();
		for(int tc = 0; tc < tcs; tc++) {
			int n = fs.nextInt();
			char[] s = fs.next().toCharArray();
			char[] t = fs.next().toCharArray();
			if(s[0]=='1'){
				pw.println("YES");
				continue;
			}
			if(s[0]=='0' && t[0]=='1'){
				pw.println("NO");
				continue;
			}
			int contains1 = -1;
			int firstdisagreement = n;
			for(int i = 0; i < n; i++) if(s[i]=='1') { contains1=i; break; }
			for(int i = 0; i < n; i++) if(s[i]!=t[i]) {firstdisagreement=i; break; }
			if(contains1==-1) { pw.println("NO"); continue; }
			if(contains1 <= firstdisagreement) pw.println("YES");
			else pw.println("NO");
		}
		pw.close();
	}
	public static class trip implements Comparable<trip>{
		long a, b, c;
		public trip(long x, long y, long z){
			a = x;
			b = y;
			c = z;
		}
		public int compareTo(trip p){
			if(a==p.a) return Long.compare(b, p.b);
			return Long.compare(a, p.a);
		}
	}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases where the code's logic might fail. The buggy code has a flaw in its logic for determining if the game will be interesting. Specifically, it only checks the first character of the sequences and some basic conditions, which can lead to incorrect results.

Here are the steps to create a challenging test input:

1. **Edge Case: All Zeros in Both Sequences**:
   - This case should return "NO" because there are no '1's to perform any XOR operations.

2. **Edge Case: First Character Mismatch**:
   - If the first character of `s` is '0' and the first character of `t` is '1', the code correctly returns "NO".

3. **Edge Case: First '1' in `s` After the First Disagreement**:
   - If the first '1' in `s` appears after the first disagreement with `t`, the code should return "NO".

4. **Edge Case: First '1' in `s` Before the First Disagreement**:
   - If the first '1' in `s` appears before the first disagreement with `t`, the code should return "YES".

5. **Complex Case: Multiple Disagreements**:
   - A more complex case where the sequences have multiple disagreements, but the first '1' in `s` is before the first disagreement.

Based on these considerations, here is a small-scale test input that covers these edge cases:

```plaintext
5
1
0
1
7
0110100
0110100
9
100101010
101111110
4
0011
1011
8
10110111
01100000
```

This test input includes:
- A simple case where `s` and `t` are both "0", which should return "NO".
- A case where `s` and `t` are already equal, which should return "YES".
- A case where the first '1' in `s` is before the first disagreement with `t`, which should return "YES".
- A case where the first '1' in `s` is after the first disagreement with `t`, which should return "NO".
- A more complex case with multiple disagreements, which should return "YES".

This input should help expose the bug in the given buggy code.