**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Folding Strip
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have a strip of paper with a binary string $$$s$$$ of length $$$n$$$. You can fold the paper in between any pair of adjacent digits.

A set of folds is considered valid if after the folds, all characters that are on top of or below each other match. Note that all folds are made at the same time, so the characters don't have to match in between folds.

For example, these are valid foldings of $$$s = \mathtt{110110110011}$$$ and $$$s = \mathtt{01110}$$$:

The length of the folded strip is the length seen from above after all folds are made. So for the two above examples, after the folds shown above, the lengths would be $$$7$$$ and $$$3$$$, respectively.

Notice that for the above folding of $$$s = \mathtt{01110}$$$, if we made either of the two folds on their own, that would not be a valid folding. However, because we don't check for validity until all folds are made, this folding is valid.

After performing a set of valid folds, what is the minimum length strip you can form?

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the size of the strip.

The second line of each test case contains a string $$$s$$$ of $$$n$$$ characters '0' and '1' — a description of the digits on the strip.

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

Output: For each test case, output a single integer — the minimum possible length of the strip after a valid folding.

Examples:
input:
6
6
101101
1
0
12
110110110011
5
01110
4
1111
2
01
output:
3
1
3
3
1
2

Note: For the first example case, one optimal folding is to fold the strip in the middle, which produces a strip of length 3.

The third and fourth example cases correspond to the images above. Note that the folding shown above for $$$s = \mathtt{110110110011}$$$ is not of minimal length.

Buggy Code:
import sys
input = sys.stdin.readline
def fg():
    return int(input())
def fgh():
    return [int(xx) for xx in input().split()]
def tt():
    sys.stdout.flush()
def sd():
    print('Alice')
def df():
    print('Bob')
for __ in range(fg()):
    n = fg()
    s = list(input()[:-1])
    h = []
    for i in range(n + 1):
        if i == n or not i or s[i] != s[i - 1]:
            if i:
                if l % 2:
                    h.append(s[i - 1])
                else:
                    h.append(s[i - 1])
                    h.append(s[i - 1])
            l = 1
        else:
            l += 1
    ans = len(h)
    j = 0
    for i in range(len(h) - 1):
        j += 1
        if h[i] == h[i + 1]:
            if ans >= j:
                ans -= (j)
                j = 1
            else:
                ans = j - ans
    print(ans)

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on edge cases and scenarios that might cause the algorithm to fail. The key points to consider are:

1. **Repetitive Patterns**: The code might fail if there are repetitive patterns in the binary string.
2. **Single Character Strings**: The code should handle single character strings correctly.
3. **Alternating Patterns**: The code might fail if the string has alternating patterns like "101010".
4. **Long Consecutive Characters**: The code might fail if there are long sequences of the same character.
5. **Mixed Patterns**: The code might fail if the string has a mix of different patterns.

Let's generate a test input that includes these scenarios:

```plaintext
6
6
101101
1
0
12
110110110011
5
01110
4
1111
2
01
```

This test input includes:
1. A string with a mix of characters and a possible valid fold: `101101` (expected output: 3)
2. A single character string: `0` (expected output: 1)
3. A string with a pattern that can be folded multiple times: `110110110011` (expected output: 3)
4. A string with an alternating pattern: `01110` (expected output: 3)
5. A string with long consecutive characters: `1111` (expected output: 1)
6. A string with the minimum length: `01` (expected output: 2)

These test cases should help in exposing any bugs in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Folding Strip
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have a strip of paper with a binary string $$$s$$$ of length $$$n$$$. You can fold the paper in between any pair of adjacent digits.

A set of folds is considered valid if after the folds, all characters that are on top of or below each other match. Note that all folds are made at the same time, so the characters don't have to match in between folds.

For example, these are valid foldings of $$$s = \mathtt{110110110011}$$$ and $$$s = \mathtt{01110}$$$:

The length of the folded strip is the length seen from above after all folds are made. So for the two above examples, after the folds shown above, the lengths would be $$$7$$$ and $$$3$$$, respectively.

Notice that for the above folding of $$$s = \mathtt{01110}$$$, if we made either of the two folds on their own, that would not be a valid folding. However, because we don't check for validity until all folds are made, this folding is valid.

After performing a set of valid folds, what is the minimum length strip you can form?

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the size of the strip.

The second line of each test case contains a string $$$s$$$ of $$$n$$$ characters '0' and '1' — a description of the digits on the strip.

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

Output: For each test case, output a single integer — the minimum possible length of the strip after a valid folding.

Examples:
input:
6
6
101101
1
0
12
110110110011
5
01110
4
1111
2
01
output:
3
1
3
3
1
2

Note: For the first example case, one optimal folding is to fold the strip in the middle, which produces a strip of length 3.

The third and fourth example cases correspond to the images above. Note that the folding shown above for $$$s = \mathtt{110110110011}$$$ is not of minimal length.

Buggy Code:
/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

public class Main {

    //------------------------------------------main--------------------------------------------------//
    public static void main(String[] args) throws java.lang.Exception {
        OutputStream outputStream = System.out;
        PrintWriter out = new PrintWriter(outputStream);
        FastReader sc = new FastReader();
        Task s = new Task(sc, out);
        s.solve();
        out.close();
        // your code goes here"
    }
}

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;
    }

    public int[] readIntArray(int n) throws IOException {
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = nextInt();
        }
        return arr;
    }

    public long[] readLongArray(int n) throws IOException {
        long[] arr = new long[n];
        for (int i = 0; i < n; i++) {
            arr[i] = nextLong();
        }
        return arr;
    }
}

class Task {

    PrintWriter out;
    FastReader sc;

    Task(FastReader sc, PrintWriter out) {
        this.sc = sc;
        this.out = out;
    }
    //----------------------------------------------inputfunction------------------------------------------//

    String st() {
        return sc.nextLine();
    }

    int i() {
        return sc.nextInt();
    }

    long l() {
        return sc.nextLong();
    }

    double d() {
        return sc.nextDouble();
    }

    //---------------------------------------------absolute----------------------------------------------//
    public long abs(long a) {
        return a < 0 ? -a : a;
    }

    public int abs(int a) {
        return a < 0 ? -a : a;
    }

    //-------------------------------------------max--------------------------------------------------//
    public int max(int a, int b) {
        return a >= b ? a : b;
    }

    public long max(long a, long b) {
        return a >= b ? a : b;
    }

    public int max(int a, int b, int c) {
        return Math.max(a, Math.max(b, c));
    }

    public long max(long a, long b, long c) {
        return Math.max(a, Math.max(b, c));
    }

    //------------------------------------------min---------------------------------------------------//
    public int min(int a, int b) {
        return a <= b ? a : b;
    }

    public long min(long a, long b) {
        return a <= b ? a : b;
    }

    public int min(int a, int b, int c) {
        return Math.min(a, Math.min(b, c));
    }

    public long min(long a, long b, long c) {
        return Math.min(a, Math.min(b, c));
    }

    //----------------------------------------sortingfunction---------------------------------------------//
    private void sort(int[] arr) {
        int n = arr.length;
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list);
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void sort(char[] arr) {
        int n = arr.length;
        List<Character> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list);
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void rsort(char[] arr) {
        int n = arr.length;
        List<Character> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list, Collections.reverseOrder());
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void rsort(int[] arr) {
        int n = arr.length;
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list, Collections.reverseOrder());
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void sort(long[] arr) {
        int n = arr.length;
        List<Long> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list);
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void rsort(long[] arr) {
        int n = arr.length;
        List<Long> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list, Collections.reverseOrder());
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }
    //---------------------------------------power----------------------------------------------------//

    public long pow(long a, int b) {
        long ans = 1L;
        while (b > 0) {
            if ((b & 1) > 0) {
                ans *= a;
            }
            a *= a;
            b >>= 1;
        }
        return b;
    }
    //----------------------------------------binary_exponentiation-----------------------------------//

    public long exp(long a, long b, long mod) {
        long ans = 1L;
        while (b > 0) {
            if ((b & 1) > 0) {
                ans = (ans * a) % mod;
            }
            a = (a * a) % mod;
            b >>= 1;
        }
        return ans;
    }

    //---------------------------------------helpingmethods-------------------------------------------//
    //---------------------------------------main-----------------------------------------------------//
    int t, n;

    public void solve() throws IOException {
        t = sc.nextInt();
        while (t-- > 0) {
            int n = i();
            char[] a = st().toCharArray();
            List<Integer> l = new ArrayList<>();
            int ans = 0;
            l.add(1);
            int count = 1;
            for(int i = 0;i<n;i++){
                int size = 0;
                int k = i;
                while(k<n&&a[k]==a[i]){
                    size++;
                    k++;
                }
                if(size%2==0) l.add(count);
                count++;
                i = k-1;
            }
            l.add(count-1);
            for(int i = 1;i<l.size();i++){
                ans = max(ans, l.get(i) - l.get(i-1)+1);
            }
            out.print(ans);
            out.print("\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:
To generate a challenging test input that exposes the bug in the provided code, let's consider the following steps:

1. **Understand the Bug**: The provided code has a logic issue in the way it processes the string and calculates the minimum length of the strip after folding. Specifically, it seems to be incorrectly handling the segments of the string and not correctly accounting for the folding conditions.

2. **Identify Edge Cases**: We need to focus on edge cases such as:
   - Strings with repeated characters.
   - Strings with alternating characters.
   - Strings with a single character.
   - Strings of length 1.
   - Strings where the optimal folding is not immediately obvious.

3. **Generate a Test Input**: We will create a small-scale test input that is complete and valid, and that should expose the bug.

Let's create a test input that includes a string where the optimal folding is not straightforward and requires careful handling of the segments.

```plaintext
1
8
10101010
```

**Explanation**:
- **Test Case**: 1
- **String Length**: 8
- **String**: "10101010"

This string alternates between '1' and '0'. The optimal folding for this string is to fold it in the middle, resulting in a folded strip of length 4. This test case should expose any issues in the code's handling of alternating characters and segment processing.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Folding Strip
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have a strip of paper with a binary string $$$s$$$ of length $$$n$$$. You can fold the paper in between any pair of adjacent digits.

A set of folds is considered valid if after the folds, all characters that are on top of or below each other match. Note that all folds are made at the same time, so the characters don't have to match in between folds.

For example, these are valid foldings of $$$s = \mathtt{110110110011}$$$ and $$$s = \mathtt{01110}$$$:

The length of the folded strip is the length seen from above after all folds are made. So for the two above examples, after the folds shown above, the lengths would be $$$7$$$ and $$$3$$$, respectively.

Notice that for the above folding of $$$s = \mathtt{01110}$$$, if we made either of the two folds on their own, that would not be a valid folding. However, because we don't check for validity until all folds are made, this folding is valid.

After performing a set of valid folds, what is the minimum length strip you can form?

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the size of the strip.

The second line of each test case contains a string $$$s$$$ of $$$n$$$ characters '0' and '1' — a description of the digits on the strip.

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

Output: For each test case, output a single integer — the minimum possible length of the strip after a valid folding.

Examples:
input:
6
6
101101
1
0
12
110110110011
5
01110
4
1111
2
01
output:
3
1
3
3
1
2

Note: For the first example case, one optimal folding is to fold the strip in the middle, which produces a strip of length 3.

The third and fourth example cases correspond to the images above. Note that the folding shown above for $$$s = \mathtt{110110110011}$$$ is not of minimal length.

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

public class Main {
    static long n;
    static List<Long> arr = new ArrayList<>();

    public static void solve(Scanner sc) {
        n = sc.nextLong();
        String s = sc.next();

        boolean all = true;
        for (int i = 0; i < n; i++) {
            if (s.charAt(i) != s.charAt(0)) {
                all = false;
                break;
            }
        }
        if (all) {
            System.out.println(1);
            return;
        }

        arr.clear();
        int i = 0, c = 1;
        while (i < n) {
            while (i < n - 1 && s.charAt(i + 1) == s.charAt(i)) {
                i++;
                c++;
            }
            arr.add((long) c);
            c = 1;
            i++;
        }
        n = arr.size();

        int ans = 0;
        i = 0;

        while (i < n) {
            int j = i + 1;
            while (j < n - 1 && arr.get(j) % 2 == 1) {
                j++;
            }
            ans = Math.max(ans, j - i + 1);
            i = j;
        }

        System.out.println(ans);
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        for (int T = 0; T < t; T++) {
            solve(sc);
        }
    }
}

**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
12
111111111111
6
101010
4
1100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Folding Strip
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have a strip of paper with a binary string $$$s$$$ of length $$$n$$$. You can fold the paper in between any pair of adjacent digits.

A set of folds is considered valid if after the folds, all characters that are on top of or below each other match. Note that all folds are made at the same time, so the characters don't have to match in between folds.

For example, these are valid foldings of $$$s = \mathtt{110110110011}$$$ and $$$s = \mathtt{01110}$$$:

The length of the folded strip is the length seen from above after all folds are made. So for the two above examples, after the folds shown above, the lengths would be $$$7$$$ and $$$3$$$, respectively.

Notice that for the above folding of $$$s = \mathtt{01110}$$$, if we made either of the two folds on their own, that would not be a valid folding. However, because we don't check for validity until all folds are made, this folding is valid.

After performing a set of valid folds, what is the minimum length strip you can form?

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the size of the strip.

The second line of each test case contains a string $$$s$$$ of $$$n$$$ characters '0' and '1' — a description of the digits on the strip.

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

Output: For each test case, output a single integer — the minimum possible length of the strip after a valid folding.

Examples:
input:
6
6
101101
1
0
12
110110110011
5
01110
4
1111
2
01
output:
3
1
3
3
1
2

Note: For the first example case, one optimal folding is to fold the strip in the middle, which produces a strip of length 3.

The third and fourth example cases correspond to the images above. Note that the folding shown above for $$$s = \mathtt{110110110011}$$$ is not of minimal length.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define db double
#define pii pair<int,int>
#define pli pair<ll,int>
#define pil pair<int,ll>
#define pll pair<ll,ll>
#define ti3 tuple<int,int,int>
#define int128 __int128_t
#define pii128 pair<int128,int128>
const int inf = 1 << 30;
const ll linf = (ll)4e18 + 10;
const db EPS = 1e-10;
const db pi = acos(-1);
template<class T> bool chmin(T& x, T y){
    if(x > y) {
        x = y;
        return true;
    } else return false;
}
template<class T> bool chmax(T& x, T y){
    if(x < y) {
        x = y;
        return true;
    } else return false;
}

// overload macro
#define CAT( A, B ) A ## B
#define SELECT( NAME, NUM ) CAT( NAME, NUM )

#define GET_COUNT( _1, _2, _3, _4, _5, _6 /* ad nauseam */, COUNT, ... ) COUNT
#define VA_SIZE( ... ) GET_COUNT( __VA_ARGS__, 6, 5, 4, 3, 2, 1 )

#define VA_SELECT( NAME, ... ) SELECT( NAME, VA_SIZE(__VA_ARGS__) )(__VA_ARGS__)

// rep(overload)
#define rep( ... ) VA_SELECT(rep, __VA_ARGS__)
#define rep2(i, n) for (int i = 0; i < int(n); i++)
#define rep3(i, a, b) for (int i = a; i < int(b); i++)
#define rep4(i, a, b, c) for (int i = a; i < int(b); i += c)

// repll(overload)
#define repll( ... ) VA_SELECT(repll, __VA_ARGS__)
#define repll2(i, n) for (ll i = 0; i < (ll)(n); i++)
#define repll3(i, a, b) for (ll i = a; i < (ll)(b); i++)
#define repll4(i, a, b, c) for (ll i = a; i < (ll)(b); i += c)

// rrep(overload)
#define rrep( ... ) VA_SELECT(rrep, __VA_ARGS__)
#define rrep2(i, n) for (int i = n - 1; i >= 0; i--)
#define rrep3(i, a, b) for (int i = b - 1; i >= a; i--)
#define rrep4(i, a, b, c) for (int i = b - 1; i >= a; i -= c)

// rrepll(overload)
#define rrepll( ... ) VA_SELECT(rrepll, __VA_ARGS__)
#define rrepll2(i, n) for (ll i = (ll)(n) - 1; i >= 0ll; i--)
#define rrepll3(i, a, b) for (ll i = b - 1; i >= (ll)(a); i--)
#define rrepll4(i, a, b, c) for (ll i = b - 1; i >= (ll)(a); i -= c)

// for_earh
#define fore(e, v) for (auto&& e : v)

// vector
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()

vector<int> Manacher(string S) {
    vector<int> R(S.size());
    int i = 0, j = 0;
    while (i < (int)S.size()) {
        while (i - j >= 0 && i + j < (int)S.size() && S[i - j] == S[i + j]) ++j;
        R[i] = j;
        int k = 1;
        while (i - k >= 0 && k + R[i - k] < j) R[i + k] = R[i - k], ++k;
        i += k; j -= k;
    }
    return R;
}

vector<int> Palindrome(string S) {
    string T(S.size() * 2 + 1, '$');
    for (int i = 0; i < (int)S.size(); i++) {
        T[i * 2 + 1] = S[i];
    }
    vector<int> R = Manacher(T);
    for (int i = 0; i < (int)T.size(); i++) {
        R[i] = R[i] / 2;
    }
    return R;
}

void solve() {
    int n;
    string s, t = "";
    cin >> n >> s;
    char c = s[0];
    int cnt = 1;
    char st = s[0];
    vector<int> ps;
    rep (i, 1, n + 1) {
        if (i == n || s[i] != c) {
            ps.push_back(cnt % 2 == 0 ? 2 : 1);
            if (i < n) {
                c = s[i];
                cnt = 1;
            }
        } else {
            cnt++;
        }
    }

    int f = s[0] == '1' ? 1 : 0;
    t = "";
    rep (j, ps[0]) t += '0' + f;
    f ^= 1;
    rep (i, 1, ps.size()) {
        if (ps[i] == 1) t += '0' + f, f ^= 1;
        else if (i + 2 < n && ps[i + 1] == 2) {
            i++;
        } else {
            t += '0' + f;
            t += '0' + f;
            f ^= 1;
        }
    }

    auto v = Palindrome(t);
    int l = 0, r = t.size() - 1;
    vector<int> ls = {0};
    rep (i, 2, v.size(), 2) {
        if (i - v[i] * 2 <= 2 * l + 1) {
            l = i / 2;
            ls.push_back(l);
        }
    }

    vector<int> rs = {r};
    rrep (i, 2, v.size(), 2) {
        if (i < 2 * l + 1) break;
        if (i + v[i] * 2 >= 2 * r + 1) {
            r = i / 2 - 1;
            rs.push_back(r);
        }
    }
    int ans = t.size();
    for (auto u : ls) {
        int it = lower_bound(all(rs), u) - rs.begin();
        chmin(ans, rs[it] - u + 1);
    }
    cout << r - l + 1 << '\n';
}

int main() {
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(20);
    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:
```plaintext
6
12
111010101011
1
1
5
11111
6
110011
4
1010
3
000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Folding Strip
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have a strip of paper with a binary string $$$s$$$ of length $$$n$$$. You can fold the paper in between any pair of adjacent digits.

A set of folds is considered valid if after the folds, all characters that are on top of or below each other match. Note that all folds are made at the same time, so the characters don't have to match in between folds.

For example, these are valid foldings of $$$s = \mathtt{110110110011}$$$ and $$$s = \mathtt{01110}$$$:

The length of the folded strip is the length seen from above after all folds are made. So for the two above examples, after the folds shown above, the lengths would be $$$7$$$ and $$$3$$$, respectively.

Notice that for the above folding of $$$s = \mathtt{01110}$$$, if we made either of the two folds on their own, that would not be a valid folding. However, because we don't check for validity until all folds are made, this folding is valid.

After performing a set of valid folds, what is the minimum length strip you can form?

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the size of the strip.

The second line of each test case contains a string $$$s$$$ of $$$n$$$ characters '0' and '1' — a description of the digits on the strip.

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

Output: For each test case, output a single integer — the minimum possible length of the strip after a valid folding.

Examples:
input:
6
6
101101
1
0
12
110110110011
5
01110
4
1111
2
01
output:
3
1
3
3
1
2

Note: For the first example case, one optimal folding is to fold the strip in the middle, which produces a strip of length 3.

The third and fourth example cases correspond to the images above. Note that the folding shown above for $$$s = \mathtt{110110110011}$$$ is not of minimal length.

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

#define ll long long
#define pp push_back
#define endl '\n'
#define all(x) x.begin(),x.end()
#define ld long double
#define PI acos(-1)
#define sin(a) sin((a)*PI/180)
#define cos(a) cos((a)*PI/180)
#define ones(x) __builtin_popcountll(x)
//#define int ll

using namespace std;

void Drakon() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
#ifdef Clion
    freopen("input.txt", "r", stdin), freopen("output.txt", "w", stdout);
#endif
}

unsigned long long inf = 1e10;
const double EPS = 1e-6;
const int MOD = 1000000007, N = 200005, LOG = 25;

ll mul(const ll &a, const ll &b) {
    return (a % MOD + MOD) * (b % MOD + MOD) % MOD;
}

ll add(const ll &a, const ll &b) {
    return (a + b + 2 * MOD) % MOD;
}

ll pw(ll x, ll y) {
    ll ret = 1;
    while (y > 0) {
        if (y % 2 == 0) {
            x = mul(x, x);
            y = y / 2;
        } else {
            ret = mul(ret, x);
            y = y - 1;
        }
    }
    return ret;
}

void solve() {
    int n;
    cin >> n;
    string s;
    cin >> s;
    vector<char>vec = {s[0]};
    for (int i = 1; i < n; ++i) {
        if(s[i] != s[i - 1])
            vec.push_back(s[i]);
    }
    while (vec.size() >= 4){
        vec.pop_back();
        vec.pop_back();
    }
    cout << vec.size() << endl;
}

signed main() {
    Drakon();
    int t = 1;
    cin >> t;
    while (t--) {
        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
8
10101010
4
1010
6
111111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Folding Strip
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have a strip of paper with a binary string $$$s$$$ of length $$$n$$$. You can fold the paper in between any pair of adjacent digits.

A set of folds is considered valid if after the folds, all characters that are on top of or below each other match. Note that all folds are made at the same time, so the characters don't have to match in between folds.

For example, these are valid foldings of $$$s = \mathtt{110110110011}$$$ and $$$s = \mathtt{01110}$$$:

The length of the folded strip is the length seen from above after all folds are made. So for the two above examples, after the folds shown above, the lengths would be $$$7$$$ and $$$3$$$, respectively.

Notice that for the above folding of $$$s = \mathtt{01110}$$$, if we made either of the two folds on their own, that would not be a valid folding. However, because we don't check for validity until all folds are made, this folding is valid.

After performing a set of valid folds, what is the minimum length strip you can form?

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the size of the strip.

The second line of each test case contains a string $$$s$$$ of $$$n$$$ characters '0' and '1' — a description of the digits on the strip.

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

Output: For each test case, output a single integer — the minimum possible length of the strip after a valid folding.

Examples:
input:
6
6
101101
1
0
12
110110110011
5
01110
4
1111
2
01
output:
3
1
3
3
1
2

Note: For the first example case, one optimal folding is to fold the strip in the middle, which produces a strip of length 3.

The third and fourth example cases correspond to the images above. Note that the folding shown above for $$$s = \mathtt{110110110011}$$$ is not of minimal length.

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

public class C {

	static IOHandler sc = new IOHandler();
	static PrintWriter out = new PrintWriter(System.out);

	public static void main(String[] args) throws IOException {

        new Thread(null, new Solution(), "solution", 1 << 26 ).start();
    }

	private static class Solution implements Runnable {

		public void run() {
			// TODO Auto-generated method stub
			int testCases = sc.nextInt();

			for (int i = 1; i <= testCases; ++i) {
				solve(i);
			}

			out.flush();
		}

		private void solve(int t) {
			// Read question and Test cases thoroughly
			// Check for overflow , Avoid Integer ==
			// Handle invalid cases
			// Think before solving
			// Search for concepts
			// expected alg like seg tree, trie, binSearch, sqrt decom ...

			int n = sc.nextInt();

			String s = sc.next();

			int [] arr = new int [n + 2];

			for (int i = 1; i <= n; ++i) {
				arr[i] = s.charAt(i - 1) - '0';
			}

			int lastZero = n + 1;
			int lastOne = n + 1;

			int [] nextDiff = new int [n + 2];

			for (int i = n; i > 0; --i) {
				if (arr[i] == 1) {
					lastOne = i;
					nextDiff[i] = lastZero;
				}else {
					lastZero = i;
					nextDiff[i] = lastOne;
				}
			}

			MinSegmentTree oddSeg = new MinSegmentTree(n + 3);
			MinSegmentTree evenSeg = new MinSegmentTree(n + 3);

			BIT bit = new BIT(n + 3);
			MaxBIT maxBit = new MaxBIT(n + 3);

			RollingHash [] hashes = new RollingHash[3];

			int [] mod = new int [] {1_000_000_007, 989_573_938, 765_938_731};

			Random rand = new Random();

			for (int i = 0; i < 3; ++i) {
				hashes[i] = new RollingHash(arr, 2 + rand.nextInt(10), mod[i]);
			}

			long [] dp = new long [n + 2];

			int maxAllowed;

			MinSegmentTree seg;

			int min, max, mid;

			int maxPos;

			for (int i = n; i > 0; --i) {
				dp[i] = 1 + dp[i + 1];
				dp[i] = Math.min(dp[i], bit.get(i));

				maxPos = (int)maxBit.get(i);

				int diff = maxPos - i;

				if (qualifies(i, diff, hashes, n)) {
					dp[i] = Math.min(dp[i], dp[i + 2 * diff]);
				}

				maxAllowed = nextDiff[i] - 1;
				seg = i % 2 == 0 ? evenSeg : oddSeg;

				dp[i] = Math.min(dp[i], seg.query(i, maxAllowed));
				seg.update(i, (int)dp[i]);

				min = 0;
				max = Math.min(i - 1 - 0, n + 1 - i);

				if (i > 1 && arr[i - 1] == arr[i]) {
					while (max - min > 1) {
						mid = (max + min) / 2;

						if (same(hashes, i - 1 - mid, i - 1, i, i + mid)) {
							min = mid;
						}else {
							max = mid;
						}
					}

					int target = (i - 1) - min;

					bit.update(target, dp[i]);
					maxBit.update(target, i);
				}
			}

			out.println(dp[1]);
		}

		private boolean qualifies(int start, int diff, RollingHash [] hashes, int n) {
			if (diff <= 0)
				return false;

			int fs = start;
			int ss = fs + diff;
			int ts = ss + diff;

			int fe = ss - 1;
			int se = ts - 1;
			int te = ts + diff - 1;

			if (te > n)
				return false;

			return same(hashes, fs, fe, ss, se) && same(hashes, ss, se, ts, te);
		}

		private boolean same(RollingHash [] hashes, int ll, int lm, int rm, int rr) {

			for (RollingHash hash : hashes) {
				if (hash.getHash(lm, ll) != hash.getHash(rm, rr))
					return false;
			}

			return true;
		}
	}

	private static class BIT {

		long [] arr;

		BIT(int n){
			arr = new long [n + 1];
			Arrays.fill(arr, Integer.MAX_VALUE);
		}

		void update (int idx, long val) {
			for (int i = idx; i < arr.length; i += (i & -i)) {
				arr[i] = Math.min(val, arr[i]);
			}
		}

		long get(int idx) {
			long result = Integer.MAX_VALUE;

			for (int i = idx; i > 0; i -= (i & -i)) {
				result = Math.min(arr[i], result);
			}

			return result;
		}

		long query(int a, int b) {
			return get(b) - get(a - 1);
		}

	}

	private static class MaxBIT {

		long [] arr;

		MaxBIT(int n){
			arr = new long [n + 1];
		}

		void update (int idx, long val) {
			for (int i = idx; i < arr.length; i += (i & -i)) {
				arr[i] = Math.max(val, arr[i]);
			}
		}

		long get(int idx) {
			long result = 0;

			for (int i = idx; i > 0; i -= (i & -i)) {
				result = Math.max(arr[i], result);
			}

			return result;
		}

		long query(int a, int b) {
			return get(b) - get(a - 1);
		}

	}

	private static class MinSegmentTree {
	    List<int []> values;

	    MinSegmentTree(int n){
	        int size = 1;

	        while (size < n)
	        	size = size << 1;

	        size += 2;

	        values = new ArrayList<>();

	        int [] useArr;

	        while (size > 1){
	        	useArr = new int [size];
	        	Arrays.fill(useArr, Integer.MAX_VALUE);
	            values.add(useArr);
	            size += 1;
	            size /= 2;
	        }

	        useArr = new int [1];
        	Arrays.fill(useArr, Integer.MAX_VALUE);
	        values.add(useArr);
	    }

	    int query (int l, int h){

	        int result = Integer.MAX_VALUE;

	        l = Math.max(l, 0);

	        for (int [] arr : values){
	            result = Math.min(result, Math.min(arr[l], arr[h]));
	            //System.out.println(Arrays.toString(arr));
	            //System.out.println(l + " " + h + " " + arr[l] + " " + arr[h]);

	            if (h == l) break;

	            if (l % 2 == 1) ++l;
	            if (h % 2 == 0) --h;

	            l /= 2;
	            h /= 2;

	            if (h < l) break;
	        }

	        return result;
	    }

	    void update(int idx, int val){

	    	int next, other;
	    	next = val;

	        for (int [] arr : values){
	            arr[idx] = next;

	            other = (idx % 2) == 0 ? idx + 1 : idx - 1;

	            if (other < arr.length)
	            	next = Math.min(arr[idx], arr[other]);
	            else
	            	next = arr[idx];

	            idx /= 2;
	        }
	    }
	}

	private static class RollingHash {

		long [] mArr;
		long [] forwardArr;
		long [] backArr;

		int [] requestArr;
		int reqMul;

		long mod;

		// lower case alpha
		public static RollingHash getHashFromString(String s, int mul, int mod) {
			int [] arr = new int[s.length()];

			for (int i = 0; i < s.length(); ++i) {
				arr[i] = s.charAt(i) - 'a';
			}

			return new RollingHash(arr, mul, mod);
		}

		// zero indexex
		public RollingHash(int [] arr, int mul, int mod) {

			long cHash = 0;
			long mHash = 1;

			mArr = new long [arr.length + 1];
			forwardArr = new long [arr.length];
			backArr = new long [arr.length];

			this.mod = mod;
			this.requestArr = arr;
			this.reqMul = mul;

			mArr[0] = 1;

			for (int i = 0; i < arr.length; ++i) {
				cHash *= mul;
				cHash += arr[i];
				cHash %= mod;

				forwardArr[i] = cHash;

				mHash *= mul;
				mHash %= mod;

				mArr[i + 1] = mHash;
			}

			cHash = 0;

			for (int i = arr.length - 1; i >= 0; --i) {
				cHash *= mul;
				cHash += arr[i];
				cHash %= mod;

				backArr[i] = cHash;
			}
		}

		// abcdefg
		// def - abcdef - abc * 3

		public long getHash(int from, int to) {
			long result;
			int prev;

			if (to >= from) {
				result = forwardArr[to];
				prev = from - 1;

				long sub = prev == -1 ? 0 : forwardArr[prev];
				sub *= mArr[to - prev];
				sub %= mod;

				result += mod;
				result -= sub;
				result %= mod;
			}else {
				result = backArr[to];
				prev = from + 1;

				long sub = prev == backArr.length ? 0 : backArr[prev];
				sub *= mArr[prev - to];
				sub %= mod;

				result += mod;
				result -= sub;
				result %= mod;
			}

			return result;
		}

		public long getHashBruteForce(int from, int to) {
			long result = 0;

			int diff = from >= to ? -1 : 1;

			result = requestArr[from];
			int idx = from;

			while (idx != to) {
				idx += diff;
				result *= reqMul;
				result += requestArr[idx];
				result %= mod;
			}

			return result;
		}

	}

	private static class IOHandler {
      BufferedReader br;
      StringTokenizer st;

      public IOHandler() {
         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());
      }

      int [] readArray(int n) {
    	  int [] res = new int [n];

    	  for (int i = 0; i < n; ++i)
    		  res[i] = nextInt();

    	  return res;
      }

      int [] readArray2(int n) {
    	  int [] res = new int [n + 1];

    	  for (int i = 1; i <= n; ++i)
    		  res[i] = nextInt();

    	  return res;
      }

      int [][] readGraph(int n, int m, int c) {
    	  int [][] graph = new int [n + 1][];

    	  int [] count = new int [n + 1];
    	  int [][] arr = new int [m][1 + c];

    	  int a, b;

    	  for (int i = 0; i < m; ++i) {
    		  a = sc.nextInt();
    		  b = sc.nextInt();

    		  arr[i][0] = a;
    		  arr[i][1] = b;

    		  ++count[a];
    		  ++count[b];

    		  for (int j = 2; j <= c; ++j) {
    			  arr[i][j] = sc.nextInt();
    		  }
    	  }

    	  for (int i = 0; i <= n; ++i) {
    		  graph[i] = new int [count[i] * c];
    		  count[i] = 0;
    	  }
    	  int swap;

    	  for (int [] tArr : arr) {
    		  for (int j = 1; j < tArr.length; ++j) {
    			  graph[tArr[0]][count[tArr[0]]++] = tArr[j];
    		  }

    		  swap = tArr[0];
    		  tArr[0] = tArr[1];
    		  tArr[1] = swap;

    		  for (int j = 1; j < tArr.length; ++j) {
    			  graph[tArr[0]][count[tArr[0]]++] = tArr[j];
    		  }
    	  }

    	  return graph;
      }

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

1. **Edge Cases**: Small inputs that are at the boundary of the input constraints.
2. **Complex Patterns**: Inputs that have complex patterns of '0's and '1's which might cause the algorithm to fail.
3. **Symmetry**: Inputs that have symmetric properties, which might be incorrectly handled by the algorithm.
4. **Long Sequences**: Inputs that are long enough to stress the algorithm's performance and correctness.

Given the constraints and the need for a small-scale test input, let's focus on a string that is small but has a complex pattern. A good candidate is a string with alternating '0's and '1's, which can be particularly tricky for folding algorithms.

Here is a small-scale test input that should be challenging:

```plaintext
1
8
10101010
```

### Explanation:
- **Test Case**: `1` (Only one test case)
- **Length**: `8` (A small length that is manageable but still has enough complexity)
- **String**: `10101010` (An alternating pattern of '0's and '1's)

This input is designed to test the algorithm's ability to handle alternating patterns and ensure that it correctly identifies valid foldings. The alternating pattern can cause issues if the algorithm does not correctly handle the symmetry and matching of characters.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Folding Strip
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have a strip of paper with a binary string $$$s$$$ of length $$$n$$$. You can fold the paper in between any pair of adjacent digits.

A set of folds is considered valid if after the folds, all characters that are on top of or below each other match. Note that all folds are made at the same time, so the characters don't have to match in between folds.

For example, these are valid foldings of $$$s = \mathtt{110110110011}$$$ and $$$s = \mathtt{01110}$$$:

The length of the folded strip is the length seen from above after all folds are made. So for the two above examples, after the folds shown above, the lengths would be $$$7$$$ and $$$3$$$, respectively.

Notice that for the above folding of $$$s = \mathtt{01110}$$$, if we made either of the two folds on their own, that would not be a valid folding. However, because we don't check for validity until all folds are made, this folding is valid.

After performing a set of valid folds, what is the minimum length strip you can form?

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the size of the strip.

The second line of each test case contains a string $$$s$$$ of $$$n$$$ characters '0' and '1' — a description of the digits on the strip.

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

Output: For each test case, output a single integer — the minimum possible length of the strip after a valid folding.

Examples:
input:
6
6
101101
1
0
12
110110110011
5
01110
4
1111
2
01
output:
3
1
3
3
1
2

Note: For the first example case, one optimal folding is to fold the strip in the middle, which produces a strip of length 3.

The third and fourth example cases correspond to the images above. Note that the folding shown above for $$$s = \mathtt{110110110011}$$$ is not of minimal length.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
const int maxn=1000005;
int n,m,T,ans,flg,U,D;
int a[maxn];
string s,t;
int main(){
	ios::sync_with_stdio(false);
	cin>>T;
	while(T--){
		cin>>n>>s,t="";
		for(int i=0;i<s.size();i++){
			if(i+2<s.size()&&s[i]==s[i+1]&&s[i+1]==s[i+2])
				i++;
			else t=t+s[i];
		}
		m=0;
		for(int i=0;i<t.size();i++){
			int j=i;
			while(j+1<t.size()&&t[j+1]!=t[j])
				j++;
			a[++m]=j-i+1,i=j;
		}
//		for(int i=1;i<=m;i++)
//			printf("%d%c",a[i],i==m? '\n':' ');
		int L=1,R=m;
		while(L<R&&a[L]<=a[L+1])
			L++;
		while(R>L&&a[R]<=a[R-1])
			R--;
		int flg=L<R,s=0;
		for(int i=L;i<=R;i++)
			flg&=(a[L+R-i]==a[i]),s+=a[i];
		if(flg)
			printf("%d\n",s/2);
		else printf("%d\n",s);
	}
	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. Here are some considerations:

1. **Repeating Patterns**: The buggy code seems to handle sequences of three identical characters in a special way, which might not be correct. We can create a string with repeating patterns to test this.
2. **Mixed Patterns**: Strings with mixed patterns of '0' and '1' can also reveal issues in the folding logic.
3. **Edge Cases**: Strings of length 1 or 2, and strings with all characters the same, can also be good test cases.

Let's construct a test input that includes these elements:

```plaintext
4
9
1110111
7
1010101
3
111
2
01
```

### Explanation:
1. **Test Case 1**: `1110111`
   - This string has a repeating pattern of '1' and a '0' in the middle. The buggy code might not handle the middle '0' correctly.
2. **Test Case 2**: `1010101`
   - This string has a mixed pattern of alternating '0' and '1'. The buggy code might not handle the alternating pattern correctly.
3. **Test Case 3**: `111`
   - This string has all characters the same. The buggy code might not handle the repeating '1' correctly.
4. **Test Case 4**: `01`
   - This is a simple edge case with the minimum length of 2. The buggy code should handle this correctly.

This test input should help expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Folding Strip
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have a strip of paper with a binary string $$$s$$$ of length $$$n$$$. You can fold the paper in between any pair of adjacent digits.

A set of folds is considered valid if after the folds, all characters that are on top of or below each other match. Note that all folds are made at the same time, so the characters don't have to match in between folds.

For example, these are valid foldings of $$$s = \mathtt{110110110011}$$$ and $$$s = \mathtt{01110}$$$:

The length of the folded strip is the length seen from above after all folds are made. So for the two above examples, after the folds shown above, the lengths would be $$$7$$$ and $$$3$$$, respectively.

Notice that for the above folding of $$$s = \mathtt{01110}$$$, if we made either of the two folds on their own, that would not be a valid folding. However, because we don't check for validity until all folds are made, this folding is valid.

After performing a set of valid folds, what is the minimum length strip you can form?

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the size of the strip.

The second line of each test case contains a string $$$s$$$ of $$$n$$$ characters '0' and '1' — a description of the digits on the strip.

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

Output: For each test case, output a single integer — the minimum possible length of the strip after a valid folding.

Examples:
input:
6
6
101101
1
0
12
110110110011
5
01110
4
1111
2
01
output:
3
1
3
3
1
2

Note: For the first example case, one optimal folding is to fold the strip in the middle, which produces a strip of length 3.

The third and fourth example cases correspond to the images above. Note that the folding shown above for $$$s = \mathtt{110110110011}$$$ is not of minimal length.

Buggy Code:
#include<bits/stdc++.h>
#define Gc() getchar()
#define Me(x,y) memset(x,y,sizeof(x))
#define Mc(x,y) memcpy(x,y,sizeof(x))
#define d(x,y) ((m)*(x-1)+(y))
#define R(n) (rnd()%(n)+1)
#define Pc(x) putchar(x)
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#define eb emplace_back
#define all(x) x.begin(),x.end()
using namespace std;using ll=long long;using db=double;using lb=long db;using ui=unsigned;using ull=unsigned long long;using pii=pair<int,int>;
const int N=4e5+5,M=(1<<20)+5,K=1000+5,mod=998244353,Mod=mod-1;const db eps=1e-9;const int INF=1e9+7;mt19937 rnd(263082);
#define Tp template<typename T>
#define Ts template<typename T,typename... Ar>
namespace Debug{
	Tp void _debug(char* f,T t){cerr<<f<<'='<<t<<endl;}
	Ts void _debug(char* f,T x,Ar... y){while(*f!=',') cerr<<*f++;cerr<<'='<<x<<",";_debug(f+1,y...);}
	#ifdef LOCAL
	#define gdb(...) _debug((char*)#__VA_ARGS__,__VA_ARGS__)
	#else
	#define gdb(...) void()
	#endif
}using namespace Debug;
int n,m,f[N];char s[N],t[N];
void Solve(){
	int i,j;scanf("%d",&n);
	scanf("%s",s+1);
	m=0;
	for(i=1;i<=n;i++){
		t[++m]=s[i];
		if(m>=3&&t[m-2]==t[m-1]&&t[m-1]==t[m]) m-=2;
	}
	swap(n,m);swap_ranges(s+1,s+m+1,t+1);
	m=0;
	for(i=1;i<=n;i++){
		t[++m]=s[i];
		if(m>=6&&t[m-5]==t[m-2]&&t[m-2]==t[m-1]&&t[m-4]==t[m-3]&&t[m-3]==t[m]) m-=4;
	}
	swap(n,m);swap_ranges(s+1,s+m+1,t+1);
	m=0;
	for(i=1;i<=n;i++){
		t[++m]=s[i];
		if(m>=9&&t[m-8]==t[m-3]&&t[m-3]==t[m-2]&&t[m-7]==t[m-4]&&t[m-4]==t[m-1]&&t[m-6]==t[m-5]&&t[m-5]==t[m]) m-=6;
	}
	swap(n,m);swap_ranges(s+1,s+m+1,t+1);
	m=0;t[++m]='%';t[++m]='#';
	for(i=1;i<=n;i++) t[++m]=s[i],t[++m]='#';t[++m]='$';
	int mid=0,r=0;
	for(i=2;i<m;i++){
		if(i<=r) f[i]=min(r-i,f[2*mid-i]);
		else f[i]=0;
		while(t[i-f[i]-1]==t[i+f[i]+1]) f[i]++;
		if(i+f[i]>r) mid=i,r=i+f[i];
	}
	for(i=n-1;i;i--) if(f[2*i+2]/2==i) {
		rotate(s+1,s+i+1,s+n+1);n-=i;break;
	}
	reverse(s+1,s+n+1);
	m=0;t[++m]='%';t[++m]='#';
	for(i=1;i<=n;i++) t[++m]=s[i],t[++m]='#';t[++m]='$';
	mid=0,r=0;
	for(i=2;i<m;i++){
		if(i<=r) f[i]=min(r-i,f[2*mid-i]);
		else f[i]=0;
		while(t[i-f[i]-1]==t[i+f[i]+1]) f[i]++;
		if(i+f[i]>r) mid=i,r=i+f[i];
	}
	for(i=n-1;i;i--) if(f[2*i+2]/2==i) {
		rotate(s+1,s+i+1,s+n+1);n-=i;break;
	}
	printf("%d\n",n);
}
int main(){
	int t=1;
	scanf("%d",&t);
	while(t--) Solve();
	cerr<<clock()*1.0/CLOCKS_PER_SEC<<'\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
6
6
101101
1
0
12
110110110011
5
01110
4
1111
2
01
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Folding Strip
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have a strip of paper with a binary string $$$s$$$ of length $$$n$$$. You can fold the paper in between any pair of adjacent digits.

A set of folds is considered valid if after the folds, all characters that are on top of or below each other match. Note that all folds are made at the same time, so the characters don't have to match in between folds.

For example, these are valid foldings of $$$s = \mathtt{110110110011}$$$ and $$$s = \mathtt{01110}$$$:

The length of the folded strip is the length seen from above after all folds are made. So for the two above examples, after the folds shown above, the lengths would be $$$7$$$ and $$$3$$$, respectively.

Notice that for the above folding of $$$s = \mathtt{01110}$$$, if we made either of the two folds on their own, that would not be a valid folding. However, because we don't check for validity until all folds are made, this folding is valid.

After performing a set of valid folds, what is the minimum length strip you can form?

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the size of the strip.

The second line of each test case contains a string $$$s$$$ of $$$n$$$ characters '0' and '1' — a description of the digits on the strip.

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

Output: For each test case, output a single integer — the minimum possible length of the strip after a valid folding.

Examples:
input:
6
6
101101
1
0
12
110110110011
5
01110
4
1111
2
01
output:
3
1
3
3
1
2

Note: For the first example case, one optimal folding is to fold the strip in the middle, which produces a strip of length 3.

The third and fourth example cases correspond to the images above. Note that the folding shown above for $$$s = \mathtt{110110110011}$$$ is not of minimal length.

Buggy Code:
import os, sys
from io import BytesIO, IOBase

from collections import defaultdict, deque, Counter
from bisect import bisect_left, bisect_right
from heapq import heappush, heappop
from functools import lru_cache, reduce
from itertools import accumulate
from operator import xor
import random
import math
import sys

# sys.setrecursionlimit(10 ** 6)
# Fast IO Region
BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

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

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

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

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

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

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
sys.setrecursionlimit(4000)

ii = lambda: int(input())
mii = lambda: map(int, input().split())
fii = lambda: map(float, input().split())
lmii = lambda: list(map(int, input().split()))
i2c = lambda n: chr(ord('a') + n)
c2i = lambda c: ord(c) - ord('a')

mod = 998244353

# mod = 10 ** 9 + 7

flush = lambda: sys.stdout.flush()

def solve():
    n = ii()
    s = input()
    l = []
    cur = '*'
    r = 0
    for c in s:
        if c == cur:
            if r > 0:
                l.append(r)
            r = 1
        else:
            r += 1
            cur = c
    l.append(r)
    l1 = []
    for k in l:
        if len(l1) >= 2 and l1[-1] <= l1[-2] and l1[-1] <= k:
            p = l1.pop()
            l1[-1] += k - p
        else:
            l1.append(k)
    print(max(l1))

for _ in range(ii()):
    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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases or scenarios that can cause the algorithm to fail. The buggy code appears to have issues with how it processes sequences of characters and how it merges segments. Here are the steps to create a test input:

1. **Identify the Bug**: The buggy code seems to incorrectly handle the merging of segments when the segments are of different lengths. It also might not correctly handle cases where the same character is repeated multiple times in a row.

2. **Create a Test Case**:
   - **Small Length**: The test case should be small enough to be manageable but large enough to expose the bug.
   - **Repeating Characters**: Include sequences of repeating characters to test the merging logic.
   - **Mixed Characters**: Include sequences with mixed characters to test the folding logic.

3. **Edge Cases**:
   - **Single Character**: Test with a single character.
   - **Two Characters**: Test with two characters that are the same.
   - **Alternating Characters**: Test with a sequence of alternating characters.
   - **Long Repeating Sequence**: Test with a long sequence of the same character.

Based on these considerations, here is a small-scale test input that is complete and valid:

```plaintext
4
3
111
4
1010
6
111111
8
11010111
```

This test input includes:
- A single character repeated three times.
- A sequence of alternating characters.
- A long sequence of the same character.
- A mixed sequence with a combination of repeating and alternating characters.

This should help expose any issues in the merging and folding logic of the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sharky Surfing
time_limit_ms: 3000
memory_limit_mb: 256
Description: Mualani loves surfing on her sharky surfboard!

Mualani's surf path can be modeled by a number line. She starts at position $$$1$$$, and the path ends at position $$$L$$$. When she is at position $$$x$$$ with a jump power of $$$k$$$, she can jump to any integer position in the interval $$$[x, x+k]$$$. Initially, her jump power is $$$1$$$.

However, her surf path isn't completely smooth. There are $$$n$$$ hurdles on her path. Each hurdle is represented by an interval $$$[l, r]$$$, meaning she cannot jump to any position in the interval $$$[l, r]$$$.

There are also $$$m$$$ power-ups at certain positions on the path. Power-up $$$i$$$ is located at position $$$x_i$$$ and has a value of $$$v_i$$$. When Mualani is at position $$$x_i$$$, she has the option to collect the power-up to increase her jump power by $$$v_i$$$. There may be multiple power-ups at the same position. When she is at a position with some power-ups, she may choose to take or ignore each individual power-up. No power-up is in the interval of any hurdle.

What is the minimum number of power-ups she must collect to reach position $$$L$$$ to finish the path? If it is not possible to finish the surf path, output $$$-1$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$L$$$ ($$$1 \leq n, m \leq 2 \cdot 10^5, 3 \leq L \leq 10^9$$$) — the number of hurdles, the number of power-ups, and the position of the end.

The following $$$n$$$ lines contain two integers $$$l_i$$$ and $$$r_i$$$ ($$$2 \leq l_i \leq r_i \leq L-1$$$) — the bounds of the interval for the $$$i$$$'th hurdle. It is guaranteed that $$$r_i + 1 < l_{i+1}$$$ for all $$$1 \leq i < n$$$ (i.e. all hurdles are non-overlapping, sorted by increasing positions, and the end point of a previous hurdle is not consecutive with the start point of the next hurdle).

The following $$$m$$$ lines contain two integers $$$x_i$$$ and $$$v_i$$$ ($$$1 \leq x_i, v_i \leq L$$$) — the position and the value for the $$$i$$$'th power-up. There may be multiple power-ups with the same $$$x$$$. It is guaranteed that $$$x_i \leq x_{i+1}$$$ for all $$$1 \leq i < m$$$ (i.e. the power-ups are sorted by non-decreasing position) and no power-up is in the interval of any hurdle.

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

Output: For each test case, output the minimum number of power-ups she must collect to reach position $$$L$$$. If it is not possible, output $$$-1$$$.

Examples:
input:
4
2 5 50
7 14
30 40
2 2
3 1
3 5
18 2
22 32
4 3 50
4 6
15 18
20 26
34 38
1 2
8 2
10 2
1 4 17
10 14
1 6
1 2
1 2
16 9
1 2 10
5 9
2 3
2 2
output:
4
-1
1
2

Note: In the first test case, she can collect power-ups $$$1$$$, $$$2$$$, $$$3$$$, and $$$5$$$ to clear all hurdles.

In the second test case, she cannot jump over the first hurdle.

In the fourth test case, by collecting both power-ups, she can jump over the hurdle.

Buggy Code:
#include<iostream>
#include<vector>
#include<math.h>
#include<map>
#include<set>

using namespace std;

struct p{
    int begin;
    int end;
};

int main(){
    long long pep;
    cin >> pep;
    while(pep--) {
        int n, m, L;
        cin >> n >> m >> L;
        multiset<int> set;
        vector<p> vec(n);
        int ans = 0;
        for (int i = 0; i < n; i++) {
            cin >> vec[i].begin >> vec[i].end;
        }

        /*for (auto x : vec) {
            cout << x.begin << ' ' << x.end << '\n';
        }*/

        int i = 0;
        int power = 1;
        for (int j = 0; j < m; j++) {
            int ind, u;
            cin >> ind >> u;
            if (ind < vec[i].begin) {
                set.insert(u);
            } else {
                while(ind > vec[i].end) {
                    while (power <= vec[i].end - vec[i].begin + 1 && ans != -1) {

                        /*int l = 3;
                        while (!set.empty() && l > 0) {
                            cout << *set.rbegin() << ' ';
                            set.erase(--set.end());
                            l--;
                        }*/

                        if (!set.empty() && ans > -1) {
                            power += *set.rbegin();
                            set.erase(set.find(*set.rbegin()));
                            ans++;
                        } else {
                            ans = -1;
                        }
                    }
                    i++;
                }
                set.insert(u);
            }
            /*for (auto x : set) {
                cout << x << ' ';
            }
            cout << '\n';*/
        }

        while(i < n) {
            while (power <= vec[i].end - vec[i].begin + 1 && ans != -1) {
                if (!set.empty() && ans > -1) {
                    power += *set.rbegin();
                    set.erase(set.find(*set.rbegin()));
                    ans++;
                } else ans = -1;
            }
            i++;
        }
        cout << ans << '\n';
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sharky Surfing
time_limit_ms: 3000
memory_limit_mb: 256
Description: Mualani loves surfing on her sharky surfboard!

Mualani's surf path can be modeled by a number line. She starts at position $$$1$$$, and the path ends at position $$$L$$$. When she is at position $$$x$$$ with a jump power of $$$k$$$, she can jump to any integer position in the interval $$$[x, x+k]$$$. Initially, her jump power is $$$1$$$.

However, her surf path isn't completely smooth. There are $$$n$$$ hurdles on her path. Each hurdle is represented by an interval $$$[l, r]$$$, meaning she cannot jump to any position in the interval $$$[l, r]$$$.

There are also $$$m$$$ power-ups at certain positions on the path. Power-up $$$i$$$ is located at position $$$x_i$$$ and has a value of $$$v_i$$$. When Mualani is at position $$$x_i$$$, she has the option to collect the power-up to increase her jump power by $$$v_i$$$. There may be multiple power-ups at the same position. When she is at a position with some power-ups, she may choose to take or ignore each individual power-up. No power-up is in the interval of any hurdle.

What is the minimum number of power-ups she must collect to reach position $$$L$$$ to finish the path? If it is not possible to finish the surf path, output $$$-1$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$L$$$ ($$$1 \leq n, m \leq 2 \cdot 10^5, 3 \leq L \leq 10^9$$$) — the number of hurdles, the number of power-ups, and the position of the end.

The following $$$n$$$ lines contain two integers $$$l_i$$$ and $$$r_i$$$ ($$$2 \leq l_i \leq r_i \leq L-1$$$) — the bounds of the interval for the $$$i$$$'th hurdle. It is guaranteed that $$$r_i + 1 < l_{i+1}$$$ for all $$$1 \leq i < n$$$ (i.e. all hurdles are non-overlapping, sorted by increasing positions, and the end point of a previous hurdle is not consecutive with the start point of the next hurdle).

The following $$$m$$$ lines contain two integers $$$x_i$$$ and $$$v_i$$$ ($$$1 \leq x_i, v_i \leq L$$$) — the position and the value for the $$$i$$$'th power-up. There may be multiple power-ups with the same $$$x$$$. It is guaranteed that $$$x_i \leq x_{i+1}$$$ for all $$$1 \leq i < m$$$ (i.e. the power-ups are sorted by non-decreasing position) and no power-up is in the interval of any hurdle.

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

Output: For each test case, output the minimum number of power-ups she must collect to reach position $$$L$$$. If it is not possible, output $$$-1$$$.

Examples:
input:
4
2 5 50
7 14
30 40
2 2
3 1
3 5
18 2
22 32
4 3 50
4 6
15 18
20 26
34 38
1 2
8 2
10 2
1 4 17
10 14
1 6
1 2
1 2
16 9
1 2 10
5 9
2 3
2 2
output:
4
-1
1
2

Note: In the first test case, she can collect power-ups $$$1$$$, $$$2$$$, $$$3$$$, and $$$5$$$ to clear all hurdles.

In the second test case, she cannot jump over the first hurdle.

In the fourth test case, by collecting both power-ups, she can jump over the hurdle.

Buggy Code:
#include <iostream>
#include <cmath>
#include <vector>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long ll;
void solve() {
    priority_queue <ll, vector<ll>, less<ll>> pq;
    int n,m,L;
    cin >> n >> m >> L;
    vector<ll> st(n+1),ed(n+1);
    vector<ll> pt(m+1),v(m+1);

    for (int i=1; i<=n; i++) cin >> st[i] >> ed[i];
    for (int i=1; i<=m; i++) cin >> pt[i] >> v[i];
    ll jp = 1, num = 1;
    ll ans = 0;
    for (int i=1; i<=n; i++) {
        while (pt[num] < st[i]) {
            pq.push(v[num]);
            num++;
        }
        ll dis = ed[i]-st[i]+2;
        // cout << dis << "\n";
        while (jp < dis && pq.empty() != true){
            ans++;
            jp = jp + pq.top();
            pq.pop();
        }
        // cout << jp << '\n';
        if (jp < dis) {
            cout << -1 << "\n";
            return;
        }
    }
    cout << ans << "\n";
}

int main(void) {
    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:
```plaintext
1
2 3 15
5 9
11 13
2 2
4 3
8 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sharky Surfing
time_limit_ms: 3000
memory_limit_mb: 256
Description: Mualani loves surfing on her sharky surfboard!

Mualani's surf path can be modeled by a number line. She starts at position $$$1$$$, and the path ends at position $$$L$$$. When she is at position $$$x$$$ with a jump power of $$$k$$$, she can jump to any integer position in the interval $$$[x, x+k]$$$. Initially, her jump power is $$$1$$$.

However, her surf path isn't completely smooth. There are $$$n$$$ hurdles on her path. Each hurdle is represented by an interval $$$[l, r]$$$, meaning she cannot jump to any position in the interval $$$[l, r]$$$.

There are also $$$m$$$ power-ups at certain positions on the path. Power-up $$$i$$$ is located at position $$$x_i$$$ and has a value of $$$v_i$$$. When Mualani is at position $$$x_i$$$, she has the option to collect the power-up to increase her jump power by $$$v_i$$$. There may be multiple power-ups at the same position. When she is at a position with some power-ups, she may choose to take or ignore each individual power-up. No power-up is in the interval of any hurdle.

What is the minimum number of power-ups she must collect to reach position $$$L$$$ to finish the path? If it is not possible to finish the surf path, output $$$-1$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$L$$$ ($$$1 \leq n, m \leq 2 \cdot 10^5, 3 \leq L \leq 10^9$$$) — the number of hurdles, the number of power-ups, and the position of the end.

The following $$$n$$$ lines contain two integers $$$l_i$$$ and $$$r_i$$$ ($$$2 \leq l_i \leq r_i \leq L-1$$$) — the bounds of the interval for the $$$i$$$'th hurdle. It is guaranteed that $$$r_i + 1 < l_{i+1}$$$ for all $$$1 \leq i < n$$$ (i.e. all hurdles are non-overlapping, sorted by increasing positions, and the end point of a previous hurdle is not consecutive with the start point of the next hurdle).

The following $$$m$$$ lines contain two integers $$$x_i$$$ and $$$v_i$$$ ($$$1 \leq x_i, v_i \leq L$$$) — the position and the value for the $$$i$$$'th power-up. There may be multiple power-ups with the same $$$x$$$. It is guaranteed that $$$x_i \leq x_{i+1}$$$ for all $$$1 \leq i < m$$$ (i.e. the power-ups are sorted by non-decreasing position) and no power-up is in the interval of any hurdle.

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

Output: For each test case, output the minimum number of power-ups she must collect to reach position $$$L$$$. If it is not possible, output $$$-1$$$.

Examples:
input:
4
2 5 50
7 14
30 40
2 2
3 1
3 5
18 2
22 32
4 3 50
4 6
15 18
20 26
34 38
1 2
8 2
10 2
1 4 17
10 14
1 6
1 2
1 2
16 9
1 2 10
5 9
2 3
2 2
output:
4
-1
1
2

Note: In the first test case, she can collect power-ups $$$1$$$, $$$2$$$, $$$3$$$, and $$$5$$$ to clear all hurdles.

In the second test case, she cannot jump over the first hurdle.

In the fourth test case, by collecting both power-ups, she can jump over the hurdle.

Buggy Code:
#include<iostream>
#include<vector>
#include<algorithm>
#include<map>
#include<set>
#include<cstring>
#include<iomanip>
#include<queue>
#include<cmath>

using namespace std;

const int N = 200010;
typedef pair<int, int>PII;

int main()
{
	int t;
	cin >> t;
	while (t--)
	{
		priority_queue<int>q;

		int n, m, L;
		cin >> n >> m >> L;
		vector<PII>p;
		map<PII, int>mp;
		int ma = 0;
		for (int i = 0; i < n; i++)
		{
			int l, r;
			cin >> l >> r;
			if (r - l + 2 > ma)
			{
				p.push_back({ l,r });
				ma = r - l + 2;
			}
		}
		int j = 0;
		int con = 1;
		int count = 0;
		for (int i = 0; i < m; i++)
		{
			int x, v;
			cin >> x >> v;
			if (j<p.size()&&x > p[j].first&&n)
			{
				while (con < p[j].second - p[j].first + 2&&q.size())
				{
					con += q.top();
					q.pop();
					count++;
				}

				if (con < p[j].second - p[j].first + 2)
				{
					n = 0;
				}
				j++;
			}
			q.push(v);

		}
		if(j < p.size())
		{
			while (con < p[p.size()-1].second - p[p.size()-1].first + 2 && q.size())
			{
				con += q.top();
				q.pop();
				count++;
			}

			if (con < p[p.size() - 1].second - p[p.size() - 1].first + 2)
			{
				n = 0;
			}
		}

		if (n == 0) cout << -1 << endl;
		else
		{
			cout << count << endl;
		}

	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
3 4 20
5 7
10 12
15 18
1 1
8 2
13 3
19 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sharky Surfing
time_limit_ms: 3000
memory_limit_mb: 256
Description: Mualani loves surfing on her sharky surfboard!

Mualani's surf path can be modeled by a number line. She starts at position $$$1$$$, and the path ends at position $$$L$$$. When she is at position $$$x$$$ with a jump power of $$$k$$$, she can jump to any integer position in the interval $$$[x, x+k]$$$. Initially, her jump power is $$$1$$$.

However, her surf path isn't completely smooth. There are $$$n$$$ hurdles on her path. Each hurdle is represented by an interval $$$[l, r]$$$, meaning she cannot jump to any position in the interval $$$[l, r]$$$.

There are also $$$m$$$ power-ups at certain positions on the path. Power-up $$$i$$$ is located at position $$$x_i$$$ and has a value of $$$v_i$$$. When Mualani is at position $$$x_i$$$, she has the option to collect the power-up to increase her jump power by $$$v_i$$$. There may be multiple power-ups at the same position. When she is at a position with some power-ups, she may choose to take or ignore each individual power-up. No power-up is in the interval of any hurdle.

What is the minimum number of power-ups she must collect to reach position $$$L$$$ to finish the path? If it is not possible to finish the surf path, output $$$-1$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$L$$$ ($$$1 \leq n, m \leq 2 \cdot 10^5, 3 \leq L \leq 10^9$$$) — the number of hurdles, the number of power-ups, and the position of the end.

The following $$$n$$$ lines contain two integers $$$l_i$$$ and $$$r_i$$$ ($$$2 \leq l_i \leq r_i \leq L-1$$$) — the bounds of the interval for the $$$i$$$'th hurdle. It is guaranteed that $$$r_i + 1 < l_{i+1}$$$ for all $$$1 \leq i < n$$$ (i.e. all hurdles are non-overlapping, sorted by increasing positions, and the end point of a previous hurdle is not consecutive with the start point of the next hurdle).

The following $$$m$$$ lines contain two integers $$$x_i$$$ and $$$v_i$$$ ($$$1 \leq x_i, v_i \leq L$$$) — the position and the value for the $$$i$$$'th power-up. There may be multiple power-ups with the same $$$x$$$. It is guaranteed that $$$x_i \leq x_{i+1}$$$ for all $$$1 \leq i < m$$$ (i.e. the power-ups are sorted by non-decreasing position) and no power-up is in the interval of any hurdle.

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

Output: For each test case, output the minimum number of power-ups she must collect to reach position $$$L$$$. If it is not possible, output $$$-1$$$.

Examples:
input:
4
2 5 50
7 14
30 40
2 2
3 1
3 5
18 2
22 32
4 3 50
4 6
15 18
20 26
34 38
1 2
8 2
10 2
1 4 17
10 14
1 6
1 2
1 2
16 9
1 2 10
5 9
2 3
2 2
output:
4
-1
1
2

Note: In the first test case, she can collect power-ups $$$1$$$, $$$2$$$, $$$3$$$, and $$$5$$$ to clear all hurdles.

In the second test case, she cannot jump over the first hurdle.

In the fourth test case, by collecting both power-ups, she can jump over the hurdle.

Buggy Code:
t = int(input())
for _ in range(t):
    count = 0
    c = 1
    n, m, l = [int(i) for i in input().split()]
    pr = []
    ult = []
    f = True
    for i in range(n):
        pr.append([int(i) for i in input().split()])
    for i in range(m):
        ult.append([int(i) for i in input().split()])
    for i in pr:
        d = i[1] - i[0]
        while c - 2 < d:
            temp = list(filter(lambda x: x[0] < i[1], ult))
            if temp == []:
                count = -1
                f = False
                break
            q = max(temp, key=lambda x: x[1])
            ult.remove(q)
            c += q[1]
            count += 1
        if not f:
            break
        c -= d
    print(count)

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

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

Think step by step.

Response:
```plaintext
3
1 1 10
2 8
1 2
2 2 10
2 8
5 9
1 2
1 2
1 1 10
2 8
1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sharky Surfing
time_limit_ms: 3000
memory_limit_mb: 256
Description: Mualani loves surfing on her sharky surfboard!

Mualani's surf path can be modeled by a number line. She starts at position $$$1$$$, and the path ends at position $$$L$$$. When she is at position $$$x$$$ with a jump power of $$$k$$$, she can jump to any integer position in the interval $$$[x, x+k]$$$. Initially, her jump power is $$$1$$$.

However, her surf path isn't completely smooth. There are $$$n$$$ hurdles on her path. Each hurdle is represented by an interval $$$[l, r]$$$, meaning she cannot jump to any position in the interval $$$[l, r]$$$.

There are also $$$m$$$ power-ups at certain positions on the path. Power-up $$$i$$$ is located at position $$$x_i$$$ and has a value of $$$v_i$$$. When Mualani is at position $$$x_i$$$, she has the option to collect the power-up to increase her jump power by $$$v_i$$$. There may be multiple power-ups at the same position. When she is at a position with some power-ups, she may choose to take or ignore each individual power-up. No power-up is in the interval of any hurdle.

What is the minimum number of power-ups she must collect to reach position $$$L$$$ to finish the path? If it is not possible to finish the surf path, output $$$-1$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$L$$$ ($$$1 \leq n, m \leq 2 \cdot 10^5, 3 \leq L \leq 10^9$$$) — the number of hurdles, the number of power-ups, and the position of the end.

The following $$$n$$$ lines contain two integers $$$l_i$$$ and $$$r_i$$$ ($$$2 \leq l_i \leq r_i \leq L-1$$$) — the bounds of the interval for the $$$i$$$'th hurdle. It is guaranteed that $$$r_i + 1 < l_{i+1}$$$ for all $$$1 \leq i < n$$$ (i.e. all hurdles are non-overlapping, sorted by increasing positions, and the end point of a previous hurdle is not consecutive with the start point of the next hurdle).

The following $$$m$$$ lines contain two integers $$$x_i$$$ and $$$v_i$$$ ($$$1 \leq x_i, v_i \leq L$$$) — the position and the value for the $$$i$$$'th power-up. There may be multiple power-ups with the same $$$x$$$. It is guaranteed that $$$x_i \leq x_{i+1}$$$ for all $$$1 \leq i < m$$$ (i.e. the power-ups are sorted by non-decreasing position) and no power-up is in the interval of any hurdle.

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

Output: For each test case, output the minimum number of power-ups she must collect to reach position $$$L$$$. If it is not possible, output $$$-1$$$.

Examples:
input:
4
2 5 50
7 14
30 40
2 2
3 1
3 5
18 2
22 32
4 3 50
4 6
15 18
20 26
34 38
1 2
8 2
10 2
1 4 17
10 14
1 6
1 2
1 2
16 9
1 2 10
5 9
2 3
2 2
output:
4
-1
1
2

Note: In the first test case, she can collect power-ups $$$1$$$, $$$2$$$, $$$3$$$, and $$$5$$$ to clear all hurdles.

In the second test case, she cannot jump over the first hurdle.

In the fourth test case, by collecting both power-ups, she can jump over the hurdle.

Buggy Code:
def merge(list1,list2):
    merged = []
    i, j = 0, 0
    while i < len(list1) and j < len(list2):
        if list1[i] <= list2[j]:
            merged.append(list1[i])
            i += 1
        else:
            merged.append(list2[j])
            j += 1
    merged.extend(list1[i:])
    merged.extend(list2[j:])
    return merged

t=int(input())
for case in range(t):
    n,m,L=list(map(int,input().split()))
    path=[0 for _ in range(L)]
    power = [[] for _ in range(L)]
    priority=[]
    for i in range(n):
        hur=list(map(int,input().split()))
        l,r=hur[0]-1,hur[1]-1
        path[l]=r-l+1
        path[r+1]=-r+l-1
    for i in range(m):
        power_ups=list(map(int,input().split()))
        power[power_ups[0]-1].append(power_ups[1])
        power[power_ups[0]-1].sort()

    for i in range(1,L):
        path[i]+=path[i-1]

    cnt=0
    k=1
    for i in range(L):
        if path[i]==0 and power[i]!=[]:
            priority=merge(priority,power[i])
        if i+1<L and path[i]!=0:
            while k<=path[i]:
                if priority==[]:
                    cnt=-1
                    break
                k+=priority.pop()
                cnt+=1
    print(cnt)

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

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

Think step by step.

Response:
```plaintext
1
1 1 10
5 8
2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sharky Surfing
time_limit_ms: 3000
memory_limit_mb: 256
Description: Mualani loves surfing on her sharky surfboard!

Mualani's surf path can be modeled by a number line. She starts at position $$$1$$$, and the path ends at position $$$L$$$. When she is at position $$$x$$$ with a jump power of $$$k$$$, she can jump to any integer position in the interval $$$[x, x+k]$$$. Initially, her jump power is $$$1$$$.

However, her surf path isn't completely smooth. There are $$$n$$$ hurdles on her path. Each hurdle is represented by an interval $$$[l, r]$$$, meaning she cannot jump to any position in the interval $$$[l, r]$$$.

There are also $$$m$$$ power-ups at certain positions on the path. Power-up $$$i$$$ is located at position $$$x_i$$$ and has a value of $$$v_i$$$. When Mualani is at position $$$x_i$$$, she has the option to collect the power-up to increase her jump power by $$$v_i$$$. There may be multiple power-ups at the same position. When she is at a position with some power-ups, she may choose to take or ignore each individual power-up. No power-up is in the interval of any hurdle.

What is the minimum number of power-ups she must collect to reach position $$$L$$$ to finish the path? If it is not possible to finish the surf path, output $$$-1$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$L$$$ ($$$1 \leq n, m \leq 2 \cdot 10^5, 3 \leq L \leq 10^9$$$) — the number of hurdles, the number of power-ups, and the position of the end.

The following $$$n$$$ lines contain two integers $$$l_i$$$ and $$$r_i$$$ ($$$2 \leq l_i \leq r_i \leq L-1$$$) — the bounds of the interval for the $$$i$$$'th hurdle. It is guaranteed that $$$r_i + 1 < l_{i+1}$$$ for all $$$1 \leq i < n$$$ (i.e. all hurdles are non-overlapping, sorted by increasing positions, and the end point of a previous hurdle is not consecutive with the start point of the next hurdle).

The following $$$m$$$ lines contain two integers $$$x_i$$$ and $$$v_i$$$ ($$$1 \leq x_i, v_i \leq L$$$) — the position and the value for the $$$i$$$'th power-up. There may be multiple power-ups with the same $$$x$$$. It is guaranteed that $$$x_i \leq x_{i+1}$$$ for all $$$1 \leq i < m$$$ (i.e. the power-ups are sorted by non-decreasing position) and no power-up is in the interval of any hurdle.

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

Output: For each test case, output the minimum number of power-ups she must collect to reach position $$$L$$$. If it is not possible, output $$$-1$$$.

Examples:
input:
4
2 5 50
7 14
30 40
2 2
3 1
3 5
18 2
22 32
4 3 50
4 6
15 18
20 26
34 38
1 2
8 2
10 2
1 4 17
10 14
1 6
1 2
1 2
16 9
1 2 10
5 9
2 3
2 2
output:
4
-1
1
2

Note: In the first test case, she can collect power-ups $$$1$$$, $$$2$$$, $$$3$$$, and $$$5$$$ to clear all hurdles.

In the second test case, she cannot jump over the first hurdle.

In the fourth test case, by collecting both power-ups, she can jump over the hurdle.

Buggy Code:
def isPrime(n):
    for i in range(2, int(n**0.5)+1):
        if n % i == 0:
            return False

    return True

def check():
    global ju,obstacles,boosters
    data = []
    for i in obstacles:
        for j in boosters:
            if j in data:
                continue
            # print(i[0], j[0])
            if i[0] > j[0]:
                ju += i[1]
                data.append(j)
            # print("JUMP", ju, i[-1] - i[0])
        # print(ju,i[-1] - i[0])
        if ju <= i[-1] - i[0]:
            return False
    return True

for _ in range(int(input())):
    n,m,l = map(int, input().split())
    obstacles = []
    boosters = []
    maxi = 0
    for _ in range(n):
        obstacles.append(list(map(int,input().split())))
        maxi = max(maxi,obstacles[-1][1] - obstacles[-1][0])
    for _ in range(m):
        boosters.append(list(map(int, input().split())))
    jump = 1
    count = 0
    for i in range(len(boosters)):
        jump += boosters[i][1]
        count += 1
        if maxi < jump:
            break
    ju = 1
    if check():
        if count == 3:
            print(-1)
        else:
            print(count)
    else:
        print(-1)

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

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

Think step by step.

Response:
```plaintext
1
2 3 15
5 10
12 14
2 2
6 3
11 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sharky Surfing
time_limit_ms: 3000
memory_limit_mb: 256
Description: Mualani loves surfing on her sharky surfboard!

Mualani's surf path can be modeled by a number line. She starts at position $$$1$$$, and the path ends at position $$$L$$$. When she is at position $$$x$$$ with a jump power of $$$k$$$, she can jump to any integer position in the interval $$$[x, x+k]$$$. Initially, her jump power is $$$1$$$.

However, her surf path isn't completely smooth. There are $$$n$$$ hurdles on her path. Each hurdle is represented by an interval $$$[l, r]$$$, meaning she cannot jump to any position in the interval $$$[l, r]$$$.

There are also $$$m$$$ power-ups at certain positions on the path. Power-up $$$i$$$ is located at position $$$x_i$$$ and has a value of $$$v_i$$$. When Mualani is at position $$$x_i$$$, she has the option to collect the power-up to increase her jump power by $$$v_i$$$. There may be multiple power-ups at the same position. When she is at a position with some power-ups, she may choose to take or ignore each individual power-up. No power-up is in the interval of any hurdle.

What is the minimum number of power-ups she must collect to reach position $$$L$$$ to finish the path? If it is not possible to finish the surf path, output $$$-1$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$L$$$ ($$$1 \leq n, m \leq 2 \cdot 10^5, 3 \leq L \leq 10^9$$$) — the number of hurdles, the number of power-ups, and the position of the end.

The following $$$n$$$ lines contain two integers $$$l_i$$$ and $$$r_i$$$ ($$$2 \leq l_i \leq r_i \leq L-1$$$) — the bounds of the interval for the $$$i$$$'th hurdle. It is guaranteed that $$$r_i + 1 < l_{i+1}$$$ for all $$$1 \leq i < n$$$ (i.e. all hurdles are non-overlapping, sorted by increasing positions, and the end point of a previous hurdle is not consecutive with the start point of the next hurdle).

The following $$$m$$$ lines contain two integers $$$x_i$$$ and $$$v_i$$$ ($$$1 \leq x_i, v_i \leq L$$$) — the position and the value for the $$$i$$$'th power-up. There may be multiple power-ups with the same $$$x$$$. It is guaranteed that $$$x_i \leq x_{i+1}$$$ for all $$$1 \leq i < m$$$ (i.e. the power-ups are sorted by non-decreasing position) and no power-up is in the interval of any hurdle.

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

Output: For each test case, output the minimum number of power-ups she must collect to reach position $$$L$$$. If it is not possible, output $$$-1$$$.

Examples:
input:
4
2 5 50
7 14
30 40
2 2
3 1
3 5
18 2
22 32
4 3 50
4 6
15 18
20 26
34 38
1 2
8 2
10 2
1 4 17
10 14
1 6
1 2
1 2
16 9
1 2 10
5 9
2 3
2 2
output:
4
-1
1
2

Note: In the first test case, she can collect power-ups $$$1$$$, $$$2$$$, $$$3$$$, and $$$5$$$ to clear all hurdles.

In the second test case, she cannot jump over the first hurdle.

In the fourth test case, by collecting both power-ups, she can jump over the hurdle.

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

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        StringTokenizer st;
        int t = Integer.parseInt(br.readLine());
        while(t-- >0){
            st = new StringTokenizer(br.readLine());
            int n = Integer.parseInt(st.nextToken());
            int m = Integer.parseInt(st.nextToken());
            long L = Long.parseLong(st.nextToken());
            // Read hurdles
            Hurdle[] hurdles = new Hurdle[n];
            for(int i=0;i<n;i++){
                st = new StringTokenizer(br.readLine());
                long li = Long.parseLong(st.nextToken());
                long ri = Long.parseLong(st.nextToken());
                hurdles[i] = new Hurdle(li, ri);
            }
            Arrays.sort(hurdles, Comparator.comparingLong(h -> h.li));

            PowerUp[] powerUps = new PowerUp[m];
            for(int i=0;i<m;i++){
                st = new StringTokenizer(br.readLine());
                long xi = Long.parseLong(st.nextToken());
                long vi = Long.parseLong(st.nextToken());
                powerUps[i] = new PowerUp(xi, vi);
            }
            Arrays.sort(powerUps, Comparator.comparingLong(p -> p.xi));

            long last_pos =1;
            long k =1;
            int count =0;
            int puIndex =0;
            PriorityQueue<Long> heap = new PriorityQueue<>(Collections.reverseOrder());
            boolean possible = true;
            for(int i=0;i<n;i++){
                Hurdle h = hurdles[i];
                long required_k = h.ri - h.li +2;
                // While k < required_k
                while(k < required_k){
                    while(puIndex <m && powerUps[puIndex].xi <= last_pos +k){
                        heap.offer(powerUps[puIndex].vi);
                        puIndex++;
                    }
                    if(heap.isEmpty()){
                        possible = false;
                        break;
                    }
                    long vi = heap.poll();
                    k += vi;
                    count++;
                }
                if(!possible){
                    break;
                }
                last_pos = h.ri +1;
            }
            if(possible){
                long required_k_final = L - last_pos +1;
                if(required_k_final <=0){
                }
                else{
                    while(k < required_k_final){
                        while(puIndex <m && powerUps[puIndex].xi <= last_pos +k){
                            heap.offer(powerUps[puIndex].vi);
                            puIndex++;
                        }
                        if(heap.isEmpty()){
                            possible = false;
                            break;
                        }
                        long vi = heap.poll();
                        k += vi;
                        count++;
                    }
                }
            }
            if(possible && (last_pos +k >= L)){
                bw.write(String.valueOf(count));
                bw.newLine();
            }
            else{
                bw.write("-1");
                bw.newLine();
            }
        }
        bw.flush();
    }
}

class Hurdle {
    long li;
    long ri;
    Hurdle(long li, long ri){
        this.li = li;
        this.ri = ri;
    }
}

class PowerUp {
    long xi;
    long vi;
    PowerUp(long xi, long vi){
        this.xi = xi;
        this.vi = vi;
    }
}

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

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

Think step by step.

Response:
```plaintext
1
3 4 25
5 10
15 20
22 23
1 2
4 3
14 1
21 5
```