**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Distance to Different
time_limit_ms: 2000
memory_limit_mb: 512
Description: Consider an array $a$ of $n$ integers, where every element is from $1$ to $k$, and every integer from $1$ to $k$ appears at least once.

Let the array $b$ be constructed as follows: for the $i$-th element of $a$, $b_i$ is the distance to the closest element in $a$ which is not equal to $a_i$. In other words, $b_i = \min \limits_{j \in [1, n], a_j \ne a_i} |i - j|$.

For example, if $a = [1, 1, 2, 3, 3, 3, 3, 1]$, then $b = [2, 1, 1, 1, 2, 2, 1, 1]$.

Calculate the number of different arrays $b$ you can obtain if you consider all possible arrays $a$, and print it modulo $998244353$.

Input: The only line of the input contains two integers $n$ and $k$ ($2 \le n \le 2 \cdot 10^5$; $2 \le k \le \min(n, 10)$).

Output: Print one integer — the number of different arrays $b$ you can obtain, taken modulo $998244353$.

Examples:
input:
2 2
output:
1

input:
4 3
output:
3

input:
6 2
output:
20

input:
6 5
output:
3

input:
133 7
output:
336975971

Note:

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

/**
 *
 * @author is2ac
 */
public class A_CF {

static final Random RAND = new Random();
   //static StringBuilder sb;
   static PrintWriter pw;
   static long mod = (long)(1e9+7);
   static TreeMap<Integer,Integer> map;
   static Long[] dp;
   public static void main(String[] args) throws IOException{
        FastScanner55 fs = new FastScanner55();
        //PrintWriter pw = new PrintWriter(System.out);
        pw = new PrintWriter(System.out);
        // int t = fs.ni();
        int t = 1;
        // pw.println("");
        long mod = 998244353L;
        StringBuilder sb =new StringBuilder();
        for (int tc = 1; tc <= t; tc++) {
            int n = fs.ni();
            int k = fs.ni();
            long[][] dp = new long[n+1][k+1];
            dp[0][k] = 1;
            long[]dpk = new long[k+1];
            dpk[k]++;
            for (int i = 1; i <= n; i++) {
                long[] newdp = new long[k+1];
                for (int j = 0; j <= k; j++) {
                    if (j!=k) dp[i][j] += dpk[j+1];
                    dp[i][j] %= mod;
                    if (i!=1 && i!=2 && i!=n) {
                        if (j!=k) dp[i][j] -= dp[i-2][j+1];
                    }
                    dp[i][j] %= mod;
                }
                dp[i][0] += dpk[0];
                if (i!=1 && i!=2 && i!=n) {
                    dp[i][0] -= dp[i-2][0];
                }
                for (int j = 0; j <= k; j++) {
                    dpk[j] += dp[i][j];
                    dpk[j] %= mod;
                }
            }
            // for (int i = 0; i <= n; i++) {
            //     pw.println(Arrays.toString(dp[i]));
            // }
            pw.println(dp[n][0]);
        }
        pw.println(sb);
        pw.close();

    }
    public static long moddiv(long a, long b, long m) {
        return (a%m * binpow(b,m-2,m)%m)%m;
    }
    public static long binpow(long a, long b, long m) {
        a %= m;
        long res = 1;
        while (b > 0) {
            if (b%2==1)
                res = res * a % m;
            a = a * a % m;
            b >>= 1;
        }
        return res;
    }
    static long dpow(long a,long n)
    {
        long r=1;
        while(n>0)
        {
            if((n&1)==1)
            {
                r=(r*a)%mod;
                n--;
            }
            else
            {
                a=(a*a)%mod;
                n=n/2;
            }
        }
        return r%mod;
    }
    public static long lcm(long a, long b) {
        return a*b/gcd(a,b);
    }
    public static long gcd(long a, long b) {
        if (b==0) return a;
        return gcd(Math.min(a,b),Math.max(a,b)%Math.min(a,b));
     }
    public static void sort(int[] arr) {
        for (int i = 0; i < arr.length; i++) {
            int r = RAND.nextInt(arr.length);
            int temp = arr[i];
            arr[i] = arr[r];
            arr[r] = temp;
        }
        Arrays.sort(arr);
    }
    public static void lsort(long[] arr) {
        for (int i = 0; i < arr.length; i++) {
            int r = RAND.nextInt(arr.length);
            long temp = arr[i];
            arr[i] = arr[r];
            arr[r] = temp;
        }
        Arrays.parallelSort(arr);
    }
}
class UnionFind {
    int[] id;
    int[] sz;
    public UnionFind (int n) {
        id = new int[n];
        sz = new int[n];
        for (int i = 0; i < n; i++) {
            id[i] = i;
            sz[i] = 1;
        }
    }
    public int find (int p) {
        int root = p;
        while (root != id[root]) {
            root = id[root];
        }
        while (p!=root) {
            int next = id[p];
            id[p] = root;
            p = next;
        }
        return root;
    }
    public int size(int p) {
        return sz[p];
    }
    public void union(int p, int q) {
        int a = find(p),b = find(q);
        if (a==b) return;
        int s = size(a) + size(b);
        id[b] = a;
        sz[a] = s;
        sz[b] = s;
    }
}

class MultiplicativeFunctionSieve {
    public int[] primes;
    public boolean[] isComp;
    public int primeLength;
    public int[] smallestPrimeFactor;

    public MultiplicativeFunctionSieve(int limit, boolean enableMobius, boolean enableEuler, boolean enableFactors) {
        isComp = new boolean[limit + 1];
        primes = new int[limit + 1];
        //numberOfSmallestPrimeFactor = new int[limit + 1];
        smallestPrimeFactor = new int[limit + 1];
        primeLength = 0;
        for (int i = 2; i <= limit; i++) {
            if (!isComp[i]) {
                primes[primeLength++] = i;
                smallestPrimeFactor[i] = i;
            }
            for (int j = 0, until = limit / i; j < primeLength && primes[j] <= until; j++) {
                int pi = primes[j] * i;
                smallestPrimeFactor[pi] = primes[j];
                isComp[pi] = true;
                if (i % primes[j] == 0) {
                    break;
                }
            }
        }

    }

}
class FastScanner55 {

    BufferedReader br;
    StringTokenizer st;

    public FastScanner55() {
        br = new BufferedReader(new InputStreamReader(System.in), 32768);
        st = null;
    }

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

    int[] intArray(int N) {
        int[] ret = new int[N];
        for (int i = 0; i < N; i++) {
            ret[i] = ni();
        }
        return ret;
    }

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

    long[] longArray(int N) {
        long[] ret = new long[N];
        for (int i = 0; i < N; i++) {
            ret[i] = nl();
        }
        return ret;
    }

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

    String nextLine() {
        String str = "";
        try {
            str = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }
}
class ST {
    int n;
    int[] t;

    public ST(int size) {
        n = size;
        t = new int[n * 2];
    }
    public ST(int[] a) {
        n = a.length;
        t = new int[n*2];
        for (int i = 0; i < a.length; i++) {
            t[n+i] = a[i];
        }
        build();
    }
    public void build() {
        for (int i = n-1; i > 0; i--) {
            t[i] = t[i>>1] + t[(i>>1)|1];
        }
    }
    public void update(int p, int u) {
        for (t[p+=n] += u; p > 1; p>>=1) {
            t[p>>1] = t[p] + t[p^1];
        }
    }
    public int query(int l, int r) {
        // take the sum of [l,r) -> inclusize of l, exclusive of r
        l += n;
        r += n+1; //
        int res = 0;
        while (l<r) {
            if ((l&1)==1) {
                res += t[l++];
            }
            if ((r&1)==1) {
                res += t[--r];
            }
            l>>=1;
            r>>=1;
        }
        return res;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider edge cases and scenarios that might cause the algorithm to fail. Here are the steps to think through this:

1. **Understand the Problem**: The problem requires calculating the number of different arrays `b` that can be obtained from all possible arrays `a` of length `n` with elements from 1 to `k`.

2. **Analyze the Code**: The provided code uses dynamic programming to solve the problem. It initializes a DP array and iterates through the elements to calculate the results. The code seems to have a complex logic for updating the DP values, which might be the source of the bug.

3. **Identify Edge Cases**:
   - **Small Values of `n` and `k`**: These can help in verifying the correctness of the base cases.
   - **Large Values of `n` and `k`**: These can help in identifying performance issues or logical errors in the DP transitions.
   - **Special Values**: Values where `n` and `k` are close to each other or where `k` is exactly 10 (the maximum allowed value).

4. **Generate Test Input**:
   - A small-scale test input that is complete and valid.
   - Choose values that are likely to expose the bug in the DP logic.

Given these considerations, let's generate a test input:

```plaintext
10 5
```

This input has `n = 10` and `k = 5`, which is a small enough scale to be manageable but large enough to potentially expose issues in the DP transitions and boundary conditions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A TRUE Battle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There is a list of $n$ booleans, each of which is either true or false, given as a binary string $^{\text{∗}}$ of length $n$ (where $\texttt{1}$ represents true, and $\texttt{0}$ represents false). Initially, there are no operators between the booleans.

Alice and Bob will take alternate turns placing and or or between the booleans, with Alice going first. Thus, the game will consist of $n-1$ turns since there are $n$ booleans. Alice aims for the final statement to evaluate to true, while Bob aims for it to evaluate to false. Given the list of boolean values, determine whether Alice will win if both players play optimally.

To evaluate the final expression, repeatedly perform the following steps until the statement consists of a single true or false:

  * If the statement contains an and operator, choose any one and replace the subexpression surrounding it with its evaluation.    * Otherwise, the statement contains an or operator. Choose any one and replace the subexpression surrounding the or with its evaluation.

For example, the expression true or false and false is evaluated as true or (false and false) $=$ true or false $=$ true. It can be shown that the result of any compound statement is unique.

$^{\text{∗}}$A binary string is a string that only consists of characters $\texttt{0}$ and $\texttt{1}$

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

The first line of each test case contains an integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the length of the string.

The second line contains a binary string of length $n$, consisting of characters $\texttt{0}$ and $\texttt{1}$ — the list of boolean values.

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

Output: For each testcase, output "YES" (without quotes) if Alice wins, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Examples:
input:
5
2
11
3
010
12
101111111100
10
0111111011
8
01000010
output:
YES
NO
YES
YES
NO

Note: In the first testcase, Alice can place and between the two booleans. The game ends as there are no other places to place operators, and Alice wins because true and true is true.

In the second testcase, Alice can place or between the middle true and the left false. Bob can place and between the middle true and the right false. The statement false or true and false is false.

Note that these examples may not be the best strategies for either Alice or Bob.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
int main() {
    int t;
    cin >>t;
    while (t--) {
        int n;
        string s;
        cin >>n>>s;

        int maxi=0;
        int curr=0;
        for (char c :s) {
            if(c=='1') {
                curr++;
                maxi=max(maxi, curr);
            } else {
                curr=0;
            }
        }
        if(maxi >1) cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
2
10
3
010
4
1010
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A TRUE Battle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There is a list of $n$ booleans, each of which is either true or false, given as a binary string $^{\text{∗}}$ of length $n$ (where $\texttt{1}$ represents true, and $\texttt{0}$ represents false). Initially, there are no operators between the booleans.

Alice and Bob will take alternate turns placing and or or between the booleans, with Alice going first. Thus, the game will consist of $n-1$ turns since there are $n$ booleans. Alice aims for the final statement to evaluate to true, while Bob aims for it to evaluate to false. Given the list of boolean values, determine whether Alice will win if both players play optimally.

To evaluate the final expression, repeatedly perform the following steps until the statement consists of a single true or false:

  * If the statement contains an and operator, choose any one and replace the subexpression surrounding it with its evaluation.    * Otherwise, the statement contains an or operator. Choose any one and replace the subexpression surrounding the or with its evaluation.

For example, the expression true or false and false is evaluated as true or (false and false) $=$ true or false $=$ true. It can be shown that the result of any compound statement is unique.

$^{\text{∗}}$A binary string is a string that only consists of characters $\texttt{0}$ and $\texttt{1}$

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

The first line of each test case contains an integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the length of the string.

The second line contains a binary string of length $n$, consisting of characters $\texttt{0}$ and $\texttt{1}$ — the list of boolean values.

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

Output: For each testcase, output "YES" (without quotes) if Alice wins, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Examples:
input:
5
2
11
3
010
12
101111111100
10
0111111011
8
01000010
output:
YES
NO
YES
YES
NO

Note: In the first testcase, Alice can place and between the two booleans. The game ends as there are no other places to place operators, and Alice wins because true and true is true.

In the second testcase, Alice can place or between the middle true and the left false. Bob can place and between the middle true and the right false. The statement false or true and false is false.

Note that these examples may not be the best strategies for either Alice or Bob.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pb push_back
#define all(v) v.begin(),v.end()
#define countsb(n) __builtin_popcount(n)
#define maxele_v(v) *max_element(v.begin(),v.end()) //return value of the pointer
#define desc_sortv(v) sort(v.rbegin(),v.rend()) //or use this sort(v.begin(),v.end(),greater<int>())
#define rev_v(v) reverse(v.begin(),v.end()) //same for string
#define erase_v(v,i) v.erase(v.begin()+i)
#define erase_range_v(v,i,j) v.erase(v.begin()+i,v.begin()+j)
#define insert_v(v,pos,val) v.insert(v.begin()+pos,val)
#define sum_v(v) accumulate(v.begin(),v.end(),0) // 0 -> initial value
#define unique_v(v) v.resize(unique(v.begin(),v.end())-v.end()) // (ON SORTED VECTOR)gives all the unique elements in vector. (same for string)
#define erasestring(s,pos,lengthtobe_deletedafterpos) s.erase(pos,lengthtobedeletedafterpos)
#define gcd(a,b) __gcd(a,b)

const int mod = 1e9+7;// prime number

//binary exponentiation -> used for finding power big integers
ll bin_expo(ll a, ll b, ll mod) {  // log b time -> iterative approach, prove it by example
    ll res = 1;
    while (b > 0) {
        if (b & 1) res = (res * a) % mod; // if b is odd
        a = (a * a) % mod; // b -> even
        b = b >> 1; //b = b/2;
    }
    return res;
}
ll inverse(ll b, ll m) {return bin_expo(b, m - 2, m);}  // b^-1 by fermats little theorem
ll mod_mul(ll a, ll b, ll m) {a = a % m; b = b % m; return (((a * b) % m) + m) % m;}  // +m to handle negative numbers
ll mod_sub(ll a, ll b, ll m) {a = a % m; b = b % m; return (((a - b) % m) + m) % m;}
ll mod_div(ll a, ll b, ll m) {a = a % m; b = b % m; return (mod_mul(a, inverse(b, m), m) + m) % m;}

vector<string> allpermutaions(string s){  // all permutations of s
    vector<string> ans;   // tc = n*n!
    sort(all(s));
    do{
        // if(s[0]!='0')
        ans.pb(s);
    }while(next_permutation(s.begin(),s.end()));
    return ans;
}

ll gcd(ll a,ll b){  // if a>b;
    if(b==0) return a;
    return gcd(b,a%b);
}

ll lcm(ll a, ll b){
    if(b==0) return a;
    return a*b/gcd(a,b);
}
// lamda function uses less parameter
// auto f = [&](ll n) -> bool{
//     return n*n<=num;
// };

class compare{
    public:
    bool operator()(pair<ll,ll> &p1, pair<ll,ll> &p2){
        // return true if p1 has less priority than p2
        if(p1.first!=p2.first) return p1.first > p2.first;
        return p1.second < p2.second;
    }
};

const int N = 1000100;
vector<ll>sieve(N,0);

//int idx=lower_bound(arr.begin(),arr.end(),target)-arr.begin();

void solve() {
    int n;
    cin >> n;
    // vector<int> v(n);
    // for(auto &it:v) cin >> it;
    string s;
    cin >> s;
    for (int i = 1; i < n; i++) {
        if (s[i] == '1' and s[i-1] == '1') {
            cout << "YES" << endl;
            return;
        }
    }
    cout << "NO" << endl;
}

int main() {
    // sieve[0]=1;
    // for(int i=2; i*i<=N; i++){
    //     if(!sieve[i]){
    //         for(int j = i*i; j<=N; j+=i){
    //             sieve[j]=1;
    //         }
    //     }
    // }
    int t = 1;
    cin >> t;
    while(t--){
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
2
10
4
1010
6
101010
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A TRUE Battle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There is a list of $n$ booleans, each of which is either true or false, given as a binary string $^{\text{∗}}$ of length $n$ (where $\texttt{1}$ represents true, and $\texttt{0}$ represents false). Initially, there are no operators between the booleans.

Alice and Bob will take alternate turns placing and or or between the booleans, with Alice going first. Thus, the game will consist of $n-1$ turns since there are $n$ booleans. Alice aims for the final statement to evaluate to true, while Bob aims for it to evaluate to false. Given the list of boolean values, determine whether Alice will win if both players play optimally.

To evaluate the final expression, repeatedly perform the following steps until the statement consists of a single true or false:

  * If the statement contains an and operator, choose any one and replace the subexpression surrounding it with its evaluation.    * Otherwise, the statement contains an or operator. Choose any one and replace the subexpression surrounding the or with its evaluation.

For example, the expression true or false and false is evaluated as true or (false and false) $=$ true or false $=$ true. It can be shown that the result of any compound statement is unique.

$^{\text{∗}}$A binary string is a string that only consists of characters $\texttt{0}$ and $\texttt{1}$

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

The first line of each test case contains an integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the length of the string.

The second line contains a binary string of length $n$, consisting of characters $\texttt{0}$ and $\texttt{1}$ — the list of boolean values.

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

Output: For each testcase, output "YES" (without quotes) if Alice wins, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Examples:
input:
5
2
11
3
010
12
101111111100
10
0111111011
8
01000010
output:
YES
NO
YES
YES
NO

Note: In the first testcase, Alice can place and between the two booleans. The game ends as there are no other places to place operators, and Alice wins because true and true is true.

In the second testcase, Alice can place or between the middle true and the left false. Bob can place and between the middle true and the right false. The statement false or true and false is false.

Note that these examples may not be the best strategies for either Alice or Bob.

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

int main() {
    long long t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        string s;
        cin>>s;
        if(s[0]=='1' || s[n-1]=='1')
        {
            cout<<"YES"<<endl;

        }
        else
        {
            cout<<"NO"<<endl;
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
3
3
000
4
1000
5
01010
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A TRUE Battle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There is a list of $n$ booleans, each of which is either true or false, given as a binary string $^{\text{∗}}$ of length $n$ (where $\texttt{1}$ represents true, and $\texttt{0}$ represents false). Initially, there are no operators between the booleans.

Alice and Bob will take alternate turns placing and or or between the booleans, with Alice going first. Thus, the game will consist of $n-1$ turns since there are $n$ booleans. Alice aims for the final statement to evaluate to true, while Bob aims for it to evaluate to false. Given the list of boolean values, determine whether Alice will win if both players play optimally.

To evaluate the final expression, repeatedly perform the following steps until the statement consists of a single true or false:

  * If the statement contains an and operator, choose any one and replace the subexpression surrounding it with its evaluation.    * Otherwise, the statement contains an or operator. Choose any one and replace the subexpression surrounding the or with its evaluation.

For example, the expression true or false and false is evaluated as true or (false and false) $=$ true or false $=$ true. It can be shown that the result of any compound statement is unique.

$^{\text{∗}}$A binary string is a string that only consists of characters $\texttt{0}$ and $\texttt{1}$

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

The first line of each test case contains an integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the length of the string.

The second line contains a binary string of length $n$, consisting of characters $\texttt{0}$ and $\texttt{1}$ — the list of boolean values.

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

Output: For each testcase, output "YES" (without quotes) if Alice wins, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Examples:
input:
5
2
11
3
010
12
101111111100
10
0111111011
8
01000010
output:
YES
NO
YES
YES
NO

Note: In the first testcase, Alice can place and between the two booleans. The game ends as there are no other places to place operators, and Alice wins because true and true is true.

In the second testcase, Alice can place or between the middle true and the left false. Bob can place and between the middle true and the right false. The statement false or true and false is false.

Note that these examples may not be the best strategies for either Alice or Bob.

Buggy Code:
n=int(input())
def boolian():
    m=int(input())
    z=input()
    x=list(z)
    if x[0]=='1':
        return True
    elif x[0]=='0' and x[1]=='0':
        return True
    elif x[0]=='0' and x[1]=='1':
        for i in range(0,len(x)-1):
            if x[i]==x[i+1] and x[i]=='1':
                return True
            elif x[i]==x[i+1] and x[i]=='0':
                return False
        if x[-1]=='0':
            return False
        else:
            return True
w=[]
for i in range(0,n):
    w.append(boolian())
for i in w:
    if i==True:
        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
3
2
10
4
1111
4
0101
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A TRUE Battle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There is a list of $n$ booleans, each of which is either true or false, given as a binary string $^{\text{∗}}$ of length $n$ (where $\texttt{1}$ represents true, and $\texttt{0}$ represents false). Initially, there are no operators between the booleans.

Alice and Bob will take alternate turns placing and or or between the booleans, with Alice going first. Thus, the game will consist of $n-1$ turns since there are $n$ booleans. Alice aims for the final statement to evaluate to true, while Bob aims for it to evaluate to false. Given the list of boolean values, determine whether Alice will win if both players play optimally.

To evaluate the final expression, repeatedly perform the following steps until the statement consists of a single true or false:

  * If the statement contains an and operator, choose any one and replace the subexpression surrounding it with its evaluation.    * Otherwise, the statement contains an or operator. Choose any one and replace the subexpression surrounding the or with its evaluation.

For example, the expression true or false and false is evaluated as true or (false and false) $=$ true or false $=$ true. It can be shown that the result of any compound statement is unique.

$^{\text{∗}}$A binary string is a string that only consists of characters $\texttt{0}$ and $\texttt{1}$

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

The first line of each test case contains an integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the length of the string.

The second line contains a binary string of length $n$, consisting of characters $\texttt{0}$ and $\texttt{1}$ — the list of boolean values.

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

Output: For each testcase, output "YES" (without quotes) if Alice wins, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Examples:
input:
5
2
11
3
010
12
101111111100
10
0111111011
8
01000010
output:
YES
NO
YES
YES
NO

Note: In the first testcase, Alice can place and between the two booleans. The game ends as there are no other places to place operators, and Alice wins because true and true is true.

In the second testcase, Alice can place or between the middle true and the left false. Bob can place and between the middle true and the right false. The statement false or true and false is false.

Note that these examples may not be the best strategies for either Alice or Bob.

Buggy Code:
def un(ch,x):
    s=0
    for i in range(len(ch)):
        if ch[i]==x:
            s=s+1
    return s
for _ in range(int(input())):
    n=int(input())
    s=input()
    if un(s,"1")>=un(s,"0") or (s[0]=="1") or s[len(s)-1]=="1":
            print("YES")
    elif un(s,"1")<un(s,"0"):
        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:
To expose the bug in the provided buggy code, we need to focus on edge cases that the code might not handle correctly. The code checks if the number of '1's is greater than or equal to the number of '0's, or if the first or last character is '1'. However, this logic is flawed because it doesn't consider the optimal play strategy for both players.

Let's consider a small-scale test input that could expose the bug:

1. A string where the number of '1's is equal to the number of '0's, but the optimal play strategy would still result in a win for Bob.
2. A string where the first and last characters are '0', but the number of '1's is greater than the number of '0's, and the optimal play strategy would still result in a win for Bob.

Here is a test input that fits these criteria:

```plaintext
3
4
1010
4
0101
4
1100
```

Explanation:
1. For the first test case "1010":
   - The number of '1's is equal to the number of '0's (2 each).
   - Alice can place an 'and' between the first and second characters, resulting in "1 and 0 or 1 or 0".
   - Bob can then place an 'or' between the second and third characters, resulting in "1 and (0 or 1) or 0 = 1 and 1 or 0 = 1 or 0 = 1".
   - Alice wins, but the buggy code might incorrectly output "NO".

2. For the second test case "0101":
   - The number of '1's is equal to the number of '0's (2 each).
   - Alice can place an 'or' between the second and third characters, resulting in "0 or 1 and 0 or 1".
   - Bob can then place an 'and' between the second and third characters, resulting in "0 or (1 and 0) or 1 = 0 or 0 or 1 = 1".
   - Alice wins, but the buggy code might incorrectly output "NO".

3. For the third test case "1100":
   - The number of '1's is greater than the number of '0's (2 '1's and 2 '0's).
   - Alice can place an 'and' between the first and second characters, resulting in "1 and 1 or 0 or 0".
   - Bob can then place an 'or' between the third and fourth characters, resulting in "1 and 1 or (0 or 0) = 1 and 1 or 0 = 1 or 0 = 1".
   - Alice wins, but the buggy code might incorrectly output "NO".

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: A TRUE Battle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There is a list of $n$ booleans, each of which is either true or false, given as a binary string $^{\text{∗}}$ of length $n$ (where $\texttt{1}$ represents true, and $\texttt{0}$ represents false). Initially, there are no operators between the booleans.

Alice and Bob will take alternate turns placing and or or between the booleans, with Alice going first. Thus, the game will consist of $n-1$ turns since there are $n$ booleans. Alice aims for the final statement to evaluate to true, while Bob aims for it to evaluate to false. Given the list of boolean values, determine whether Alice will win if both players play optimally.

To evaluate the final expression, repeatedly perform the following steps until the statement consists of a single true or false:

  * If the statement contains an and operator, choose any one and replace the subexpression surrounding it with its evaluation.    * Otherwise, the statement contains an or operator. Choose any one and replace the subexpression surrounding the or with its evaluation.

For example, the expression true or false and false is evaluated as true or (false and false) $=$ true or false $=$ true. It can be shown that the result of any compound statement is unique.

$^{\text{∗}}$A binary string is a string that only consists of characters $\texttt{0}$ and $\texttt{1}$

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

The first line of each test case contains an integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the length of the string.

The second line contains a binary string of length $n$, consisting of characters $\texttt{0}$ and $\texttt{1}$ — the list of boolean values.

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

Output: For each testcase, output "YES" (without quotes) if Alice wins, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Examples:
input:
5
2
11
3
010
12
101111111100
10
0111111011
8
01000010
output:
YES
NO
YES
YES
NO

Note: In the first testcase, Alice can place and between the two booleans. The game ends as there are no other places to place operators, and Alice wins because true and true is true.

In the second testcase, Alice can place or between the middle true and the left false. Bob can place and between the middle true and the right false. The statement false or true and false is false.

Note that these examples may not be the best strategies for either Alice or Bob.

Buggy Code:
t = int(input())
for _ in range(t):
    n = int(input())
    s = str(input())
    newS = s[0]
    for i in range(1, n):
        if s[i] != s[i - 1]:
            newS += s[i]
    if newS.count("1") >= newS.count("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
3
2
10
4
1010
12
101010101010
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A TRUE Battle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There is a list of $n$ booleans, each of which is either true or false, given as a binary string $^{\text{∗}}$ of length $n$ (where $\texttt{1}$ represents true, and $\texttt{0}$ represents false). Initially, there are no operators between the booleans.

Alice and Bob will take alternate turns placing and or or between the booleans, with Alice going first. Thus, the game will consist of $n-1$ turns since there are $n$ booleans. Alice aims for the final statement to evaluate to true, while Bob aims for it to evaluate to false. Given the list of boolean values, determine whether Alice will win if both players play optimally.

To evaluate the final expression, repeatedly perform the following steps until the statement consists of a single true or false:

  * If the statement contains an and operator, choose any one and replace the subexpression surrounding it with its evaluation.    * Otherwise, the statement contains an or operator. Choose any one and replace the subexpression surrounding the or with its evaluation.

For example, the expression true or false and false is evaluated as true or (false and false) $=$ true or false $=$ true. It can be shown that the result of any compound statement is unique.

$^{\text{∗}}$A binary string is a string that only consists of characters $\texttt{0}$ and $\texttt{1}$

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

The first line of each test case contains an integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the length of the string.

The second line contains a binary string of length $n$, consisting of characters $\texttt{0}$ and $\texttt{1}$ — the list of boolean values.

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

Output: For each testcase, output "YES" (without quotes) if Alice wins, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Examples:
input:
5
2
11
3
010
12
101111111100
10
0111111011
8
01000010
output:
YES
NO
YES
YES
NO

Note: In the first testcase, Alice can place and between the two booleans. The game ends as there are no other places to place operators, and Alice wins because true and true is true.

In the second testcase, Alice can place or between the middle true and the left false. Bob can place and between the middle true and the right false. The statement false or true and false is false.

Note that these examples may not be the best strategies for either Alice or Bob.

Buggy Code:
/**
 * C_A_TRUE_Battle
 */
import java.util.*;
public class C_A_TRUE_Battle {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int T = in.nextInt();
        while (T-->0) {
            int len = in.nextInt();
            char[] arr = in.next().toCharArray();
            if (arr.length<=2 ){
                if (new String(arr).equals("0")||new String(arr).equals("00") ) System.out.println("NO");
                else System.out.println("YES");
            }
            else{
                int i = 1;
                for (;i<=arr.length-1;i++){
                    if (arr[i]==arr[i-1] && arr[i]=='1') break;
                }
                if (i==arr.length) {
                    if ((arr[0]=='1'&& arr[2]=='1') || (arr[len-1]=='1'&& arr[len-3]=='1'))System.out.println("yes");
                    else System.out.println("NO");
                }
                else System.out.println("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
3
2
10
4
1111
5
10101
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A TRUE Battle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There is a list of $n$ booleans, each of which is either true or false, given as a binary string $^{\text{∗}}$ of length $n$ (where $\texttt{1}$ represents true, and $\texttt{0}$ represents false). Initially, there are no operators between the booleans.

Alice and Bob will take alternate turns placing and or or between the booleans, with Alice going first. Thus, the game will consist of $n-1$ turns since there are $n$ booleans. Alice aims for the final statement to evaluate to true, while Bob aims for it to evaluate to false. Given the list of boolean values, determine whether Alice will win if both players play optimally.

To evaluate the final expression, repeatedly perform the following steps until the statement consists of a single true or false:

  * If the statement contains an and operator, choose any one and replace the subexpression surrounding it with its evaluation.    * Otherwise, the statement contains an or operator. Choose any one and replace the subexpression surrounding the or with its evaluation.

For example, the expression true or false and false is evaluated as true or (false and false) $=$ true or false $=$ true. It can be shown that the result of any compound statement is unique.

$^{\text{∗}}$A binary string is a string that only consists of characters $\texttt{0}$ and $\texttt{1}$

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

The first line of each test case contains an integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the length of the string.

The second line contains a binary string of length $n$, consisting of characters $\texttt{0}$ and $\texttt{1}$ — the list of boolean values.

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

Output: For each testcase, output "YES" (without quotes) if Alice wins, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Examples:
input:
5
2
11
3
010
12
101111111100
10
0111111011
8
01000010
output:
YES
NO
YES
YES
NO

Note: In the first testcase, Alice can place and between the two booleans. The game ends as there are no other places to place operators, and Alice wins because true and true is true.

In the second testcase, Alice can place or between the middle true and the left false. Bob can place and between the middle true and the right false. The statement false or true and false is false.

Note that these examples may not be the best strategies for either Alice or Bob.

Buggy Code:
import java.util.*;

public class Solution {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        sc.nextLine();

        while (t-- > 0) {
            solve(sc);
        }
        sc.close();
    }

    public static void solve(Scanner sc) {
        int n = sc.nextInt();
        sc.nextLine();
        String s = sc.nextLine();

        int a = 0;
        int b = 0;

        for (int i = 0; i < n; i++) {
            if (s.charAt(i) == '1') {
                a++;
            } else if (s.charAt(i) == '0') {
                b++;
            }
        }

        if (a > b) {
            System.out.println("YES");
        } else {
            System.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
2
11
3
010
4
1010
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A TRUE Battle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There is a list of $n$ booleans, each of which is either true or false, given as a binary string $^{\text{∗}}$ of length $n$ (where $\texttt{1}$ represents true, and $\texttt{0}$ represents false). Initially, there are no operators between the booleans.

Alice and Bob will take alternate turns placing and or or between the booleans, with Alice going first. Thus, the game will consist of $n-1$ turns since there are $n$ booleans. Alice aims for the final statement to evaluate to true, while Bob aims for it to evaluate to false. Given the list of boolean values, determine whether Alice will win if both players play optimally.

To evaluate the final expression, repeatedly perform the following steps until the statement consists of a single true or false:

  * If the statement contains an and operator, choose any one and replace the subexpression surrounding it with its evaluation.    * Otherwise, the statement contains an or operator. Choose any one and replace the subexpression surrounding the or with its evaluation.

For example, the expression true or false and false is evaluated as true or (false and false) $=$ true or false $=$ true. It can be shown that the result of any compound statement is unique.

$^{\text{∗}}$A binary string is a string that only consists of characters $\texttt{0}$ and $\texttt{1}$

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

The first line of each test case contains an integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the length of the string.

The second line contains a binary string of length $n$, consisting of characters $\texttt{0}$ and $\texttt{1}$ — the list of boolean values.

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

Output: For each testcase, output "YES" (without quotes) if Alice wins, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Examples:
input:
5
2
11
3
010
12
101111111100
10
0111111011
8
01000010
output:
YES
NO
YES
YES
NO

Note: In the first testcase, Alice can place and between the two booleans. The game ends as there are no other places to place operators, and Alice wins because true and true is true.

In the second testcase, Alice can place or between the middle true and the left false. Bob can place and between the middle true and the right false. The statement false or true and false is false.

Note that these examples may not be the best strategies for either Alice or Bob.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define pii pair<int,int>
#define vpii vector<pair<int,int>>
#define vi vector<int>
#define vvi vector<vector<int>>
#define vl vector<long long>
#define vvl vector<vector<long long>>
#define pb push_back
#define mp make_pair
#define F first
#define unmp unordered_map<int,int>
#define S second
#define all(x) x.begin(),x.end()
#define sortasc(x) sort(all(x))
#define sortdes(x) sort(x.rbegin(),x.rend())
#define PI 3.1415926535897932384626
const int mod = 1000000007;

void solve(){
    int n;
    cin >> n;
    string s;
    cin >> s;
    int cnt =0;
    for(int i=0;i<n-1;i++){
        if((s[i]=='0' && s[i+1]=='1') || (s[i]=='1' && s[i+1]=='0')){
            cnt++;
        }
    }
    if(cnt==0){
        if(s[0]=='1'){
            cout << "YES" << endl;
            return;
        } else {
            cout << "NO" << endl;
            return;
        }
    }
    if(cnt%2==0){
        cout << "NO" << endl;
    } else {
        cout << "YES" << endl;
    }
}
#undef int
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    while(t--){
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that maximize the failure probability. The bug in the code is that it only counts transitions between `0` and `1` and uses this count to determine the outcome, which is not a correct strategy for determining the winner in the game.

### Step-by-Step Analysis:
1. **Understanding the Bug**:
   - The code counts the number of transitions between `0` and `1` and uses this count to determine the outcome.
   - If the count of transitions is even, it outputs "NO".
   - If the count of transitions is odd, it outputs "YES".

2. **Correct Strategy**:
   - The correct strategy involves considering the optimal moves for both players.
   - Alice wins if she can force the final expression to evaluate to `1` (true).
   - Bob wins if he can force the final expression to evaluate to `0` (false).

3. **Edge Cases**:
   - **All 1s**: If the string is all `1`s, Alice can always win by placing `and` operators.
   - **All 0s**: If the string is all `0`s, Bob can always win by placing `or` operators.
   - **Mixed 0s and 1s**: The optimal strategy for mixed strings is more complex and depends on the placement of operators.

4. **Challenging Test Input**:
   - A string with a mix of `0` and `1` that is not trivially solvable by the bug in the code.
   - A string that has an even number of transitions but still allows Alice to win.

### Chosen Test Input:
```plaintext
3
4
1010
5
01010
6
101010
```

### Explanation:
1. **Test Case 1**:
   - Input: `1010`
   - Transitions: `10`, `01`, `10` (3 transitions)
   - The code will output "YES" because the count is odd.
   - Correct output: "YES" (Alice can place `and` between the first and second booleans, and `or` between the second and third booleans, and `and` between the third and fourth booleans to win).

2. **Test Case 2**:
   - Input: `01010`
   - Transitions: `01`, `10`, `01`, `10` (4 transitions)
   - The code will output "NO" because the count is even.
   - Correct output: "YES" (Alice can place `or` between the first and second booleans, `and` between the second and third booleans, `or` between the third and fourth booleans, and `and` between the fourth and fifth booleans to win).

3. **Test Case 3**:
   - Input: `101010`
   - Transitions: `10`, `01`, `10`, `01`, `10` (5 transitions)
   - The code will output "YES" because the count is odd.
   - Correct output: "YES" (Alice can place `and` between the first and second booleans, `or` between the second and third booleans, `and` between the third and fourth booleans, `or` between the fourth and fifth booleans, and `and` between the fifth and sixth booleans to win).

This test input should expose the bug in the provided code by showing that the count of transitions is not a reliable strategy for determining the winner.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A TRUE Battle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There is a list of $n$ booleans, each of which is either true or false, given as a binary string $^{\text{∗}}$ of length $n$ (where $\texttt{1}$ represents true, and $\texttt{0}$ represents false). Initially, there are no operators between the booleans.

Alice and Bob will take alternate turns placing and or or between the booleans, with Alice going first. Thus, the game will consist of $n-1$ turns since there are $n$ booleans. Alice aims for the final statement to evaluate to true, while Bob aims for it to evaluate to false. Given the list of boolean values, determine whether Alice will win if both players play optimally.

To evaluate the final expression, repeatedly perform the following steps until the statement consists of a single true or false:

  * If the statement contains an and operator, choose any one and replace the subexpression surrounding it with its evaluation.    * Otherwise, the statement contains an or operator. Choose any one and replace the subexpression surrounding the or with its evaluation.

For example, the expression true or false and false is evaluated as true or (false and false) $=$ true or false $=$ true. It can be shown that the result of any compound statement is unique.

$^{\text{∗}}$A binary string is a string that only consists of characters $\texttt{0}$ and $\texttt{1}$

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

The first line of each test case contains an integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the length of the string.

The second line contains a binary string of length $n$, consisting of characters $\texttt{0}$ and $\texttt{1}$ — the list of boolean values.

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

Output: For each testcase, output "YES" (without quotes) if Alice wins, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Examples:
input:
5
2
11
3
010
12
101111111100
10
0111111011
8
01000010
output:
YES
NO
YES
YES
NO

Note: In the first testcase, Alice can place and between the two booleans. The game ends as there are no other places to place operators, and Alice wins because true and true is true.

In the second testcase, Alice can place or between the middle true and the left false. Bob can place and between the middle true and the right false. The statement false or true and false is false.

Note that these examples may not be the best strategies for either Alice or Bob.

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

using namespace std;

int n;

long long dif(int cnt1, int cnt0){
    return (abs((pow(2,n-1)*cnt1) - (pow(2,cnt0)-1)));
}

int main() {
    int t;
    cin >> t;
    while(t--){
        char in;
        int n;
        cin >> n;
        int rec=0;
        int cur=0;
        for(int i=0;i<n;i++){
            cin >> in;
            if (in=='1'){
                cur++;
            } else {
                cur = 0;
            }
            rec = max(rec,cur);

        }
        if (rec>n/2)
            cout << "YES" << '\n';
        else
            cout << "NO" << '\n';
    }
}

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

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

Think step by step.

Response:
```plaintext
5
2
11
3
010
12
101111111100
10
0111111011
8
01000010
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A TRUE Battle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There is a list of $n$ booleans, each of which is either true or false, given as a binary string $^{\text{∗}}$ of length $n$ (where $\texttt{1}$ represents true, and $\texttt{0}$ represents false). Initially, there are no operators between the booleans.

Alice and Bob will take alternate turns placing and or or between the booleans, with Alice going first. Thus, the game will consist of $n-1$ turns since there are $n$ booleans. Alice aims for the final statement to evaluate to true, while Bob aims for it to evaluate to false. Given the list of boolean values, determine whether Alice will win if both players play optimally.

To evaluate the final expression, repeatedly perform the following steps until the statement consists of a single true or false:

  * If the statement contains an and operator, choose any one and replace the subexpression surrounding it with its evaluation.    * Otherwise, the statement contains an or operator. Choose any one and replace the subexpression surrounding the or with its evaluation.

For example, the expression true or false and false is evaluated as true or (false and false) $=$ true or false $=$ true. It can be shown that the result of any compound statement is unique.

$^{\text{∗}}$A binary string is a string that only consists of characters $\texttt{0}$ and $\texttt{1}$

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

The first line of each test case contains an integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the length of the string.

The second line contains a binary string of length $n$, consisting of characters $\texttt{0}$ and $\texttt{1}$ — the list of boolean values.

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

Output: For each testcase, output "YES" (without quotes) if Alice wins, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Examples:
input:
5
2
11
3
010
12
101111111100
10
0111111011
8
01000010
output:
YES
NO
YES
YES
NO

Note: In the first testcase, Alice can place and between the two booleans. The game ends as there are no other places to place operators, and Alice wins because true and true is true.

In the second testcase, Alice can place or between the middle true and the left false. Bob can place and between the middle true and the right false. The statement false or true and false is false.

Note that these examples may not be the best strategies for either Alice or Bob.

Buggy Code:
for _ in range(int(input())):
    n = int(input())
    a = input()
    o = a.count('1')

    # for i in range(n):
    #     if a[i]=='1' :
    #         if i!=0 and i!= n-1:
    #             if a[i-1]=='0' or a[i+1]=='0' :
    #                 o = o+1
    #         elif i==0:
    #             if a[i+1]=='0':
    #                 o = o+1
    #         else :
    #             if a[i-1]=='0':
    #                 o = o+1
    # if o > a.count('1'):
    #     o = 1 + a.count('1')
    o = o + a.count('101')
    z = n-o
    if o>z+1:
        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
3
2
10
3
101
4
1010
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A TRUE Battle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There is a list of $n$ booleans, each of which is either true or false, given as a binary string $^{\text{∗}}$ of length $n$ (where $\texttt{1}$ represents true, and $\texttt{0}$ represents false). Initially, there are no operators between the booleans.

Alice and Bob will take alternate turns placing and or or between the booleans, with Alice going first. Thus, the game will consist of $n-1$ turns since there are $n$ booleans. Alice aims for the final statement to evaluate to true, while Bob aims for it to evaluate to false. Given the list of boolean values, determine whether Alice will win if both players play optimally.

To evaluate the final expression, repeatedly perform the following steps until the statement consists of a single true or false:

  * If the statement contains an and operator, choose any one and replace the subexpression surrounding it with its evaluation.    * Otherwise, the statement contains an or operator. Choose any one and replace the subexpression surrounding the or with its evaluation.

For example, the expression true or false and false is evaluated as true or (false and false) $=$ true or false $=$ true. It can be shown that the result of any compound statement is unique.

$^{\text{∗}}$A binary string is a string that only consists of characters $\texttt{0}$ and $\texttt{1}$

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

The first line of each test case contains an integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the length of the string.

The second line contains a binary string of length $n$, consisting of characters $\texttt{0}$ and $\texttt{1}$ — the list of boolean values.

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

Output: For each testcase, output "YES" (without quotes) if Alice wins, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Examples:
input:
5
2
11
3
010
12
101111111100
10
0111111011
8
01000010
output:
YES
NO
YES
YES
NO

Note: In the first testcase, Alice can place and between the two booleans. The game ends as there are no other places to place operators, and Alice wins because true and true is true.

In the second testcase, Alice can place or between the middle true and the left false. Bob can place and between the middle true and the right false. The statement false or true and false is false.

Note that these examples may not be the best strategies for either Alice or Bob.

Buggy Code:
for t in range(int(input())):
    n=int(input());
    b=input();
    d=b.count('0');
    e=n//2;
    if e>=d: 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
3
2
11
3
111
4
1010
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A TRUE Battle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There is a list of $n$ booleans, each of which is either true or false, given as a binary string $^{\text{∗}}$ of length $n$ (where $\texttt{1}$ represents true, and $\texttt{0}$ represents false). Initially, there are no operators between the booleans.

Alice and Bob will take alternate turns placing and or or between the booleans, with Alice going first. Thus, the game will consist of $n-1$ turns since there are $n$ booleans. Alice aims for the final statement to evaluate to true, while Bob aims for it to evaluate to false. Given the list of boolean values, determine whether Alice will win if both players play optimally.

To evaluate the final expression, repeatedly perform the following steps until the statement consists of a single true or false:

  * If the statement contains an and operator, choose any one and replace the subexpression surrounding it with its evaluation.    * Otherwise, the statement contains an or operator. Choose any one and replace the subexpression surrounding the or with its evaluation.

For example, the expression true or false and false is evaluated as true or (false and false) $=$ true or false $=$ true. It can be shown that the result of any compound statement is unique.

$^{\text{∗}}$A binary string is a string that only consists of characters $\texttt{0}$ and $\texttt{1}$

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

The first line of each test case contains an integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the length of the string.

The second line contains a binary string of length $n$, consisting of characters $\texttt{0}$ and $\texttt{1}$ — the list of boolean values.

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

Output: For each testcase, output "YES" (without quotes) if Alice wins, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Examples:
input:
5
2
11
3
010
12
101111111100
10
0111111011
8
01000010
output:
YES
NO
YES
YES
NO

Note: In the first testcase, Alice can place and between the two booleans. The game ends as there are no other places to place operators, and Alice wins because true and true is true.

In the second testcase, Alice can place or between the middle true and the left false. Bob can place and between the middle true and the right false. The statement false or true and false is false.

Note that these examples may not be the best strategies for either Alice or Bob.

Buggy Code:
import java.util.*;

public class hisfile{

    public static void main(String[] args) {
           Scanner input=new Scanner(System.in);
		   int test=input.nextInt();
		   for(int i=0;i<test;i++){
			int n=input.nextInt();
			String str=input.next();
			int no0=0;
			int no1=0;
			for(int j=0;j<str.length();j++){
				if(str.charAt(j)=='1'){
                      no1++;
				}
				else{
					no0++;}
			}

				if(no1>=no0){
					System.out.println("YES");
				}
				else{
					System.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:
To generate a test input that exposes the bug in the provided buggy code, we need to focus on scenarios where the code's logic of counting the number of '1's and '0's fails to correctly determine the winner. The buggy code assumes that if the number of '1's is greater than or equal to the number of '0's, Alice will always win. However, this assumption is incorrect because the placement of operators can significantly affect the outcome.

### Key Insights:
1. **Balanced Cases**: If the number of '1's and '0's are equal, the code will output "YES" because `no1 >= no0` evaluates to true. However, Alice might not always win in such cases.
2. **Edge Cases**: Cases where the number of '1's is just slightly more than the number of '0's but the placement of operators can still lead to Bob winning.

### Test Input:
Let's consider a test case where the number of '1's and '0's are balanced, but Bob can still force a win by placing operators strategically.

```plaintext
3
6
101010
4
1010
5
10101
```

### Explanation:
1. **Test Case 1: "101010"**
   - The number of '1's and '0's is equal (3 each).
   - Alice and Bob can place operators as follows:
     - Alice places an `and` between the first and second characters: `1 and 0 = 0`
     - The string becomes: `01010`
     - Bob places an `or` between the second and third characters: `0 or 1 = 1`
     - The string becomes: `0110`
     - Alice places an `and` between the second and third characters: `1 and 1 = 1`
     - The string becomes: `010`
     - Bob places an `or` between the second and third characters: `1 or 0 = 1`
     - The string becomes: `01`
     - Alice places an `and` between the remaining characters: `0 and 1 = 0`
     - The final result is `0`, and Bob wins.
   - The correct output should be "NO".

2. **Test Case 2: "1010"**
   - The number of '1's and '0's is equal (2 each).
   - Alice and Bob can place operators as follows:
     - Alice places an `and` between the first and second characters: `1 and 0 = 0`
     - The string becomes: `010`
     - Bob places an `or` between the second and third characters: `0 or 1 = 1`
     - The string becomes: `01`
     - Alice places an `and` between the remaining characters: `0 and 1 = 0`
     - The final result is `0`, and Bob wins.
   - The correct output should be "NO".

3. **Test Case 3: "10101"**
   - The number of '1's is one more than the number of '0's (3 '1's and 2 '0's).
   - Alice and Bob can place operators as follows:
     - Alice places an `and` between the first and second characters: `1 and 0 = 0`
     - The string becomes: `0101`
     - Bob places an `or` between the second and third characters: `0 or 1 = 1`
     - The string becomes: `011`
     - Alice places an `and` between the second and third characters: `1 and 1 = 1`
     - The string becomes: `01`
     - Bob places an `or` between the remaining characters: `0 or 1 = 1`
     - The final result is `1`, and Alice wins.
   - The correct output should be "YES".

This test input will expose the bug in the provided code because it shows that the simple count of '1's and '0's is not sufficient to determine the winner in all cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A TRUE Battle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There is a list of $n$ booleans, each of which is either true or false, given as a binary string $^{\text{∗}}$ of length $n$ (where $\texttt{1}$ represents true, and $\texttt{0}$ represents false). Initially, there are no operators between the booleans.

Alice and Bob will take alternate turns placing and or or between the booleans, with Alice going first. Thus, the game will consist of $n-1$ turns since there are $n$ booleans. Alice aims for the final statement to evaluate to true, while Bob aims for it to evaluate to false. Given the list of boolean values, determine whether Alice will win if both players play optimally.

To evaluate the final expression, repeatedly perform the following steps until the statement consists of a single true or false:

  * If the statement contains an and operator, choose any one and replace the subexpression surrounding it with its evaluation.    * Otherwise, the statement contains an or operator. Choose any one and replace the subexpression surrounding the or with its evaluation.

For example, the expression true or false and false is evaluated as true or (false and false) $=$ true or false $=$ true. It can be shown that the result of any compound statement is unique.

$^{\text{∗}}$A binary string is a string that only consists of characters $\texttt{0}$ and $\texttt{1}$

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

The first line of each test case contains an integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the length of the string.

The second line contains a binary string of length $n$, consisting of characters $\texttt{0}$ and $\texttt{1}$ — the list of boolean values.

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

Output: For each testcase, output "YES" (without quotes) if Alice wins, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Examples:
input:
5
2
11
3
010
12
101111111100
10
0111111011
8
01000010
output:
YES
NO
YES
YES
NO

Note: In the first testcase, Alice can place and between the two booleans. The game ends as there are no other places to place operators, and Alice wins because true and true is true.

In the second testcase, Alice can place or between the middle true and the left false. Bob can place and between the middle true and the right false. The statement false or true and false is false.

Note that these examples may not be the best strategies for either Alice or Bob.

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

public class C {

    public static void main(String[] args){
        FastReader fs=new FastReader();
        PrintWriter out=new PrintWriter(System.out);
        int N=fs.nextInt();
        final int MAX = Integer.MAX_VALUE;
        final int MIN = Integer.MIN_VALUE;
        final int MOD = 1000000007;

        for(int p = 0; p < N; p++){
            int len = fs.nextInt();
            String s = fs.nextLine();
            int zeroOnePairs = 0;
            int zeros = 0;
            int ones = 0;
            for (int i = 1; i < s.length(); i++) {
                if (s.charAt(i) != s.charAt(i - 1))
                    zeroOnePairs += 1;
                if (s.charAt(i) == '1')
                    ones += 1;
                if (s.charAt(i) == '0')
                    zeros += 1;
            }
            if (ones == 0) {
                out.println("No");
                continue;
            }

            if (zeroOnePairs == 0 && zeros == 0) {
                out.println("Yes");
                continue;
            }

            if (zeroOnePairs < ((len - 1) / 2)) {
                out.println("Yes");
            }
            else {
                out.println("No");
            }

        }
        out.close();
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

        String nextLine(){
            String str = "";
            try {
                str = br.readLine();
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}

**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
2
10
3
111
4
0110
5
10101
6
110110
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A TRUE Battle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There is a list of $n$ booleans, each of which is either true or false, given as a binary string $^{\text{∗}}$ of length $n$ (where $\texttt{1}$ represents true, and $\texttt{0}$ represents false). Initially, there are no operators between the booleans.

Alice and Bob will take alternate turns placing and or or between the booleans, with Alice going first. Thus, the game will consist of $n-1$ turns since there are $n$ booleans. Alice aims for the final statement to evaluate to true, while Bob aims for it to evaluate to false. Given the list of boolean values, determine whether Alice will win if both players play optimally.

To evaluate the final expression, repeatedly perform the following steps until the statement consists of a single true or false:

  * If the statement contains an and operator, choose any one and replace the subexpression surrounding it with its evaluation.    * Otherwise, the statement contains an or operator. Choose any one and replace the subexpression surrounding the or with its evaluation.

For example, the expression true or false and false is evaluated as true or (false and false) $=$ true or false $=$ true. It can be shown that the result of any compound statement is unique.

$^{\text{∗}}$A binary string is a string that only consists of characters $\texttt{0}$ and $\texttt{1}$

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

The first line of each test case contains an integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the length of the string.

The second line contains a binary string of length $n$, consisting of characters $\texttt{0}$ and $\texttt{1}$ — the list of boolean values.

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

Output: For each testcase, output "YES" (without quotes) if Alice wins, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Examples:
input:
5
2
11
3
010
12
101111111100
10
0111111011
8
01000010
output:
YES
NO
YES
YES
NO

Note: In the first testcase, Alice can place and between the two booleans. The game ends as there are no other places to place operators, and Alice wins because true and true is true.

In the second testcase, Alice can place or between the middle true and the left false. Bob can place and between the middle true and the right false. The statement false or true and false is false.

Note that these examples may not be the best strategies for either Alice or Bob.

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

#define int long long

const int mod = 1000000007;

int32_t main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int tt;
    cin >> tt;
    while (tt--)
    {
        int n, c0 = 0, c1 = 0;
        string s, ss = "";
        cin >> n >> s;
        ss += s[0];
        for (int i = 1; i < n; i++)
        {
            if (s[i] != s[i - 1])
                ss += s[i];
        }
        for (int i = 0; i < ss.length(); i++)
        {
            if (ss[i] == '0')
                c0++;
            else
                c1++;
        }
        if (c1 >= c0)
            cout << "yes\n";
        else
            cout << "NO\n";
    }
}

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

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

Think step by step.

Response:
```plaintext
3
2
10
4
1111
6
101010
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A TRUE Battle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There is a list of $n$ booleans, each of which is either true or false, given as a binary string $^{\text{∗}}$ of length $n$ (where $\texttt{1}$ represents true, and $\texttt{0}$ represents false). Initially, there are no operators between the booleans.

Alice and Bob will take alternate turns placing and or or between the booleans, with Alice going first. Thus, the game will consist of $n-1$ turns since there are $n$ booleans. Alice aims for the final statement to evaluate to true, while Bob aims for it to evaluate to false. Given the list of boolean values, determine whether Alice will win if both players play optimally.

To evaluate the final expression, repeatedly perform the following steps until the statement consists of a single true or false:

  * If the statement contains an and operator, choose any one and replace the subexpression surrounding it with its evaluation.    * Otherwise, the statement contains an or operator. Choose any one and replace the subexpression surrounding the or with its evaluation.

For example, the expression true or false and false is evaluated as true or (false and false) $=$ true or false $=$ true. It can be shown that the result of any compound statement is unique.

$^{\text{∗}}$A binary string is a string that only consists of characters $\texttt{0}$ and $\texttt{1}$

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

The first line of each test case contains an integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the length of the string.

The second line contains a binary string of length $n$, consisting of characters $\texttt{0}$ and $\texttt{1}$ — the list of boolean values.

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

Output: For each testcase, output "YES" (without quotes) if Alice wins, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Examples:
input:
5
2
11
3
010
12
101111111100
10
0111111011
8
01000010
output:
YES
NO
YES
YES
NO

Note: In the first testcase, Alice can place and between the two booleans. The game ends as there are no other places to place operators, and Alice wins because true and true is true.

In the second testcase, Alice can place or between the middle true and the left false. Bob can place and between the middle true and the right false. The statement false or true and false is false.

Note that these examples may not be the best strategies for either Alice or Bob.

Buggy Code:
#include <bits/stdc++.h>
#pragma GCC optimize(2)

// using namespace std;

using i64 = long long;
using u32 = unsigned int;
using u64 = unsigned long long;

void solve() {
    int n;
    std::cin >> n;
    std::string s;
    std::cin >> s;
    int a = 0, b = 0;

    for(int i = 0; i < n; i++){
        a += (s[i] == '1');
        if (i - 2 >= 0){
            if (s[i - 2] == '1' && s[i - 1] == '1' && s[i] == '1'){
                b = 1;
            }
        }
    }
    if (s[0] == '1' || s[n - 1] == '1'){
        std::cout << "YES\n";
        return;
    }
    if (b){
        std::cout << "YES\n";
        return;
    }
    n--;
    std::cout << (n / 2 < a ? "YES" : "NO") << "\n";
}

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    // std::cout << std::fixed << std::setprecision(10); // 固定输出精度
    int t = 1;
    std::cin >> t;

    while (t--) {
        solve();
    }

    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases where the logic might fail. The provided code has a few potential issues:

1. **Edge Case with Single '1' at the Beginning or End**: The code checks if the first or last character is '1' and immediately returns "YES" if so. However, this might not always be correct, especially if the length of the string is odd and the number of '1's is not sufficient for Alice to win.

2. **Three Consecutive '1's**: The code checks for three consecutive '1's and returns "YES" if found. This might not be sufficient to ensure Alice's win, especially in longer strings.

3. **General Strategy**: The code uses a simple count of '1's and a check to determine if Alice can win. This might fail in more complex scenarios where the distribution of '1's and '0's is not straightforward.

Let's construct a test case that exposes these issues:

- A string where the first and last characters are '1', but the number of '1's is not sufficient for Alice to win.
- A string with three consecutive '1's, but the overall distribution of '1's and '0's makes it difficult for Alice to win.
- A string where the number of '1's and the length of the string make it challenging for Alice to win.

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

```plaintext
5
2
11
3
010
12
101111111100
10
0111111011
8
01000010
```

This test input includes:
1. A simple case with two '1's where Alice can easily win.
2. A case with a '1' in the middle but surrounded by '0's, making it harder for Alice.
3. A longer string with multiple '1's but a tricky distribution.
4. A case with a '1' at the beginning and end but a challenging distribution of '1's and '0's.
5. A case with three consecutive '1's but a tricky overall distribution.

This should help expose the potential issues in the code's logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A TRUE Battle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There is a list of $n$ booleans, each of which is either true or false, given as a binary string $^{\text{∗}}$ of length $n$ (where $\texttt{1}$ represents true, and $\texttt{0}$ represents false). Initially, there are no operators between the booleans.

Alice and Bob will take alternate turns placing and or or between the booleans, with Alice going first. Thus, the game will consist of $n-1$ turns since there are $n$ booleans. Alice aims for the final statement to evaluate to true, while Bob aims for it to evaluate to false. Given the list of boolean values, determine whether Alice will win if both players play optimally.

To evaluate the final expression, repeatedly perform the following steps until the statement consists of a single true or false:

  * If the statement contains an and operator, choose any one and replace the subexpression surrounding it with its evaluation.    * Otherwise, the statement contains an or operator. Choose any one and replace the subexpression surrounding the or with its evaluation.

For example, the expression true or false and false is evaluated as true or (false and false) $=$ true or false $=$ true. It can be shown that the result of any compound statement is unique.

$^{\text{∗}}$A binary string is a string that only consists of characters $\texttt{0}$ and $\texttt{1}$

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

The first line of each test case contains an integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the length of the string.

The second line contains a binary string of length $n$, consisting of characters $\texttt{0}$ and $\texttt{1}$ — the list of boolean values.

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

Output: For each testcase, output "YES" (without quotes) if Alice wins, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Examples:
input:
5
2
11
3
010
12
101111111100
10
0111111011
8
01000010
output:
YES
NO
YES
YES
NO

Note: In the first testcase, Alice can place and between the two booleans. The game ends as there are no other places to place operators, and Alice wins because true and true is true.

In the second testcase, Alice can place or between the middle true and the left false. Bob can place and between the middle true and the right false. The statement false or true and false is false.

Note that these examples may not be the best strategies for either Alice or Bob.

Buggy Code:
t = int(input())
for _ in range(t):
    n = int(input())
    s = input()
    if s[0] == "1" or s[-1] == "1":
        print("YES")
    else:
        nz = 0
        no = 0
        for i in range(0,len(s)):
            if s[i] == "1":
                no += 1
            else:
                nz += 1
        if no >= nz:
            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:
To expose the bug in the provided code, we need to focus on scenarios where the code's logic fails to correctly determine the winner. The code's logic is flawed in the following ways:

1. It checks if the first or last character is "1" and immediately concludes that Alice wins if either is true. This is incorrect because the presence of a "1" at the start or end does not necessarily guarantee Alice's win.
2. It then counts the number of "1"s and "0"s and concludes that Alice wins if the number of "1"s is greater than or equal to the number of "0"s. This is also incorrect because the game's rules involve strategic placement of "and" and "or" operators, which can change the outcome.

To craft a test case that exposes these flaws, we need a binary string where:
- The first and last characters are "0".
- The number of "1"s is greater than or equal to the number of "0"s.
- Despite these conditions, Bob can still force a win by strategically placing operators.

A suitable test case is:
- A string of length 5: "01110"

Here's the reasoning:
- The first and last characters are "0".
- There are 3 "1"s and 2 "0"s, so the number of "1"s is greater than the number of "0"s.
- However, Bob can place the "and" operator between the "1"s to force a "0" result, and then place the "or" operators to ensure the final result is "0".

```plaintext
1
5
01110
```

This test case should 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: A TRUE Battle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There is a list of $n$ booleans, each of which is either true or false, given as a binary string $^{\text{∗}}$ of length $n$ (where $\texttt{1}$ represents true, and $\texttt{0}$ represents false). Initially, there are no operators between the booleans.

Alice and Bob will take alternate turns placing and or or between the booleans, with Alice going first. Thus, the game will consist of $n-1$ turns since there are $n$ booleans. Alice aims for the final statement to evaluate to true, while Bob aims for it to evaluate to false. Given the list of boolean values, determine whether Alice will win if both players play optimally.

To evaluate the final expression, repeatedly perform the following steps until the statement consists of a single true or false:

  * If the statement contains an and operator, choose any one and replace the subexpression surrounding it with its evaluation.    * Otherwise, the statement contains an or operator. Choose any one and replace the subexpression surrounding the or with its evaluation.

For example, the expression true or false and false is evaluated as true or (false and false) $=$ true or false $=$ true. It can be shown that the result of any compound statement is unique.

$^{\text{∗}}$A binary string is a string that only consists of characters $\texttt{0}$ and $\texttt{1}$

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

The first line of each test case contains an integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the length of the string.

The second line contains a binary string of length $n$, consisting of characters $\texttt{0}$ and $\texttt{1}$ — the list of boolean values.

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

Output: For each testcase, output "YES" (without quotes) if Alice wins, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Examples:
input:
5
2
11
3
010
12
101111111100
10
0111111011
8
01000010
output:
YES
NO
YES
YES
NO

Note: In the first testcase, Alice can place and between the two booleans. The game ends as there are no other places to place operators, and Alice wins because true and true is true.

In the second testcase, Alice can place or between the middle true and the left false. Bob can place and between the middle true and the right false. The statement false or true and false is false.

Note that these examples may not be the best strategies for either Alice or Bob.

Buggy Code:
import math
import sys
input = lambda : sys.stdin.readline().rstrip()

def solve():
    n = int(input())
    s = input()
    if s[0] == '1':return 'YES'
    flag = False
    cnt0 = 0
    cnt1 = 0
    for i in s:
        if i == '0' and flag == False:
            cnt0 += 1
        if flag == True and i == '0':
            break
        if i == '1':
            flag = True
            cnt1+=1
    if cnt1>cnt0 :return 'YES'
    flag = False
    cnt0 = 0
    cnt1 = 0
    for i in s[::-1]:
        if i == '0' and flag == False:
            cnt0 += 1
        if flag == True and i == '0':
            break
        if i == '1':
            flag = True
            cnt1+=1

    if cnt1 > cnt0:return 'YES'
    return 'NO'
for i in range(int(input())):print(solve())

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

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

Think step by step.

Response:
```plaintext
3
2
10
3
011
5
11111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: A TRUE Battle
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game. There is a list of $n$ booleans, each of which is either true or false, given as a binary string $^{\text{∗}}$ of length $n$ (where $\texttt{1}$ represents true, and $\texttt{0}$ represents false). Initially, there are no operators between the booleans.

Alice and Bob will take alternate turns placing and or or between the booleans, with Alice going first. Thus, the game will consist of $n-1$ turns since there are $n$ booleans. Alice aims for the final statement to evaluate to true, while Bob aims for it to evaluate to false. Given the list of boolean values, determine whether Alice will win if both players play optimally.

To evaluate the final expression, repeatedly perform the following steps until the statement consists of a single true or false:

  * If the statement contains an and operator, choose any one and replace the subexpression surrounding it with its evaluation.    * Otherwise, the statement contains an or operator. Choose any one and replace the subexpression surrounding the or with its evaluation.

For example, the expression true or false and false is evaluated as true or (false and false) $=$ true or false $=$ true. It can be shown that the result of any compound statement is unique.

$^{\text{∗}}$A binary string is a string that only consists of characters $\texttt{0}$ and $\texttt{1}$

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

The first line of each test case contains an integer $n$ ($2 \leq n \leq 2 \cdot 10^5$) — the length of the string.

The second line contains a binary string of length $n$, consisting of characters $\texttt{0}$ and $\texttt{1}$ — the list of boolean values.

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

Output: For each testcase, output "YES" (without quotes) if Alice wins, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Examples:
input:
5
2
11
3
010
12
101111111100
10
0111111011
8
01000010
output:
YES
NO
YES
YES
NO

Note: In the first testcase, Alice can place and between the two booleans. The game ends as there are no other places to place operators, and Alice wins because true and true is true.

In the second testcase, Alice can place or between the middle true and the left false. Bob can place and between the middle true and the right false. The statement false or true and false is false.

Note that these examples may not be the best strategies for either Alice or Bob.

Buggy Code:
import java.util.*;
public class Truebattle {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int a=sc.nextInt();
        while(a-->0){
            int x=sc.nextInt();
            String s=sc.next();
            if(s.equals("100")||s.equals("001")){
                System.out.println("YES");
                continue;
            }
            int c1=0;
            int c0=0;
            for(int i=0;i<x;i++){
               if(s.charAt(i)=='0'){
                  c0++;
               }else{c1++;}
                        }
                        if(c1>=c0){
                            System.out.println("YES");
                        }else{
                            System.out.println("NO");
                        }

        }
        sc.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
3
3
010
4
1010
5
11111
```