**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring Compression
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Let's define the operation of compressing a string $t$, consisting of at least $2$ digits from $1$ to $9$, as follows:

  * split it into an even number of non-empty substrings — let these substrings be $t_1, t_2, \dots, t_m$ (so, $t = t_1 + t_2 + \dots + t_m$, where $+$ is the concatenation operation);    * write the string $t_2$ $t_1$ times, then the string $t_4$ $t_3$ times, and so on.

For example, for a string "12345", one could do the following: split it into ("1", "23", "4", "5"), and write "235555".

Let the function $f(t)$ for a string $t$ return the minimum length of the string that can be obtained as a result of that process.

You are given a string $s$, consisting of $n$ digits from $1$ to $9$, and an integer $k$. Calculate the value of the function $f$ for all contiguous substrings of $s$ of length exactly $k$.

Input: The first line contains two integers $n$ and $k$ ($2 \le k \le n \le 2 \cdot 10^5$).

The second line contains the string $s$ ($|s| = n$), consisting only of digits from $1$ to $9$.

Output: Output $n - k + 1$ integers — $f(s_{1,k}), f(s_{2,k+1}), \dots, f(s_{n - k + 1, n})$.

Examples:
input:
4 4
5999
output:
14

input:
10 3
1111111111
output:
2 2 2 2 2 2 2 2

input:
11 4
49998641312
output:
12 18 17 15 12 7 7 2

Note:

Buggy Code:
import java.util.*;

public class CompressString
{
    public static void main(String[] args)
    {
        Scanner sc = new Scanner(System.in);

        int m = sc.nextInt();
        int k = sc.nextInt();
        String str = sc.next();

        for(int i=0; i <= m - k; i++)
        {
            String sub = str.substring(i, i+k);
            System.out.println(CalculateLength(sub));
        }
        sc.close();
    }

    private static int CalculateLength(String s)
    {
        int len = s.length();
        int minLength = Integer.MAX_VALUE;

        for(int i=1; i <= len / 2; i++)
        {
            if(len % i == 0)
            {
                int splitCount = len / i;
                StringBuilder sb = new StringBuilder();

                for(int j=0; j < splitCount; j++)
                {
                    sb.append(s.substring(j*i, (j+1) * i));
                }

                int compressLength = 0;
                for(int j=0; j < splitCount; j += 2)
                {
                    String t1 = sb.substring(0, i);
                    String t2 = sb.substring(i, 2 * i);
                    compressLength += t2.length() * Integer.parseInt(t1);
                    sb.delete(0, 2*i);
                }

                minLength = Math.min(minLength, compressLength);
            }
        }

        return minLength;
    }
}

**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 2
112233
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring Compression
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Let's define the operation of compressing a string $t$, consisting of at least $2$ digits from $1$ to $9$, as follows:

  * split it into an even number of non-empty substrings — let these substrings be $t_1, t_2, \dots, t_m$ (so, $t = t_1 + t_2 + \dots + t_m$, where $+$ is the concatenation operation);    * write the string $t_2$ $t_1$ times, then the string $t_4$ $t_3$ times, and so on.

For example, for a string "12345", one could do the following: split it into ("1", "23", "4", "5"), and write "235555".

Let the function $f(t)$ for a string $t$ return the minimum length of the string that can be obtained as a result of that process.

You are given a string $s$, consisting of $n$ digits from $1$ to $9$, and an integer $k$. Calculate the value of the function $f$ for all contiguous substrings of $s$ of length exactly $k$.

Input: The first line contains two integers $n$ and $k$ ($2 \le k \le n \le 2 \cdot 10^5$).

The second line contains the string $s$ ($|s| = n$), consisting only of digits from $1$ to $9$.

Output: Output $n - k + 1$ integers — $f(s_{1,k}), f(s_{2,k+1}), \dots, f(s_{n - k + 1, n})$.

Examples:
input:
4 4
5999
output:
14

input:
10 3
1111111111
output:
2 2 2 2 2 2 2 2

input:
11 4
49998641312
output:
12 18 17 15 12 7 7 2

Note:

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

using ll = long long;
using pii = array<int, 2>;

#define all(x) begin(x), end(x)
#define sz(x) (int) (x).size()

constexpr int INF = 1e7;
using Info = array<ll, 3>;

int main() {
    cin.tie(0) -> sync_with_stdio(0);
    int n, k;
    cin >> n >> k;
    string s;
    cin >> s;

    const auto solve = [&](const string &cur_str) -> ll {
        // {dp[i], dp[i] + multiplier, multiplier}
        vector<int> digits(k + 1);
        for (int i = 1; i <= k; i++) {
            digits[i] = cur_str[i - 1] - '0';
        }

        vector dp(k + 1, Info{INF, INF, INF});
        dp[0] = {0, INF, INF};

        for (int i = 2; i <= k; i++) {
            // case 1: {i - 1, i}
            dp[i] = min(dp[i], Info{
                dp[i - 2][0] + digits[i - 1],
                dp[i - 2][0] + 2 * digits[i - 1],
                digits[i - 1]
            });

            // case 2: we build off previous
            dp[i] = min(dp[i], Info{
                dp[i - 1][1],
                dp[i - 1][1] + dp[i - 1][2],
                dp[i - 1][2]
            });
        }

        return dp[k][0];
    };

    for (int i = 0; i + k <= n; i++) {
        cout << solve(s.substr(i, k)) << " ";
    }
}

**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 2
199999
```

### Explanation:
This test case is designed to expose issues in the buggy code by focusing on a small-scale input that includes a repeated digit sequence and a single different digit. Here’s why this input is challenging:

1. **Small Scale**: The input size is small (n=6, k=2) to ensure it is easily manageable and can be quickly tested.
2. **Repetition and Single Different Digit**: The string "199999" has a single '1' followed by multiple '9's. This setup tests how the algorithm handles a mix of repeated and unique digits.
3. **Edge Case for Compression**: The algorithm needs to correctly handle the compression of substrings like "19" and "99". The substring "19" should be compressed to "999999" (length 6), and "99" should be compressed to "99" (length 2).

This input should help identify any issues in the buggy code, particularly in how it handles the compression logic for different digit sequences.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring Compression
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Let's define the operation of compressing a string $t$, consisting of at least $2$ digits from $1$ to $9$, as follows:

  * split it into an even number of non-empty substrings — let these substrings be $t_1, t_2, \dots, t_m$ (so, $t = t_1 + t_2 + \dots + t_m$, where $+$ is the concatenation operation);    * write the string $t_2$ $t_1$ times, then the string $t_4$ $t_3$ times, and so on.

For example, for a string "12345", one could do the following: split it into ("1", "23", "4", "5"), and write "235555".

Let the function $f(t)$ for a string $t$ return the minimum length of the string that can be obtained as a result of that process.

You are given a string $s$, consisting of $n$ digits from $1$ to $9$, and an integer $k$. Calculate the value of the function $f$ for all contiguous substrings of $s$ of length exactly $k$.

Input: The first line contains two integers $n$ and $k$ ($2 \le k \le n \le 2 \cdot 10^5$).

The second line contains the string $s$ ($|s| = n$), consisting only of digits from $1$ to $9$.

Output: Output $n - k + 1$ integers — $f(s_{1,k}), f(s_{2,k+1}), \dots, f(s_{n - k + 1, n})$.

Examples:
input:
4 4
5999
output:
14

input:
10 3
1111111111
output:
2 2 2 2 2 2 2 2

input:
11 4
49998641312
output:
12 18 17 15 12 7 7 2

Note:

Buggy Code:
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <unordered_map>
using namespace std;

class Calculator
{
public:
    explicit Calculator(const string& s);
    int calculate(int start, int end);

private:
    void divide(int k, int start, int end);
    int operateStack();
    string getkey();

    string s;
    unordered_map<string, int> dict;
    vector<int> stk;
    int min_res;

};

Calculator::Calculator(const string& s0)
    : s(s0),
      min_res(INT_MAX)
{}

string Calculator::getkey()
{
    static char deli = 'a';
    string s = "";
    for(int i = 1; i < stk.size() - 1; ++i) {
        s += string(&s[stk[i - 1]], &s[stk[i]]);
        s += deli;
    }
    return s;
}

void printVec(const vector<int>& vec, const string& s, int res, char delim = ' ')
{
    cout << s << ": (";
    for(int i = 0; i < vec.size() - 1; i++) {
        cout << vec[i] << delim;
    }
    cout << *vec.rbegin() << "), min_res: " << res << endl;
}

int Calculator::operateStack()
{
    string key = getkey();
    if(dict.find(key) != dict.end())
        return dict[key];
    int res = 0;
    for(int i  = 1; i < stk.size() - 1; i += 2) {
        int times = stod(string(&s[stk[i - 1]], &s[stk[i]]));
        int length = stk[i + 1] - stk[i];
        res += times * length;
    }
    dict[key] = res;
    return dict[key];
}

void Calculator::divide(int k, int start, int end)
{
    // printVec(stk, "stk", min_res);

    if(start > end || k < 0) return;
    if(k > end - start + 1) return;

    if(k == 0) {
        stk.push_back(end + 1);
        int tmp_res = operateStack();
        min_res = tmp_res < min_res ? tmp_res : min_res;
        stk.pop_back();
        return;
    }

    for(int i = start; i <= end - k + 1; ++i) {
        stk.push_back(i + 1);
        divide(k - 1, i + 1, end);
        stk.pop_back();
    }
}

int Calculator::calculate(int start, int end)
{
    min_res = INT_MAX;
    for(int k = 1; k <= end - start + 1; k += 2)
    {
        stk.push_back(start);
        divide(k, start, end);
        stk.pop_back();
    }
    return min_res;
}

int main(int argc, char* argv[])
{
    int n, k;
    cin >> n >> k;
    string s;
    cin >> s;
    Calculator cal(s);
    for(int i = k - 1; i < n; ++i) {
        // cout << cal.calculate(i - k + 1, i) << "\n";
        cout << cal.calculate(i - k + 1, i) << " ";
    }
}

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

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

Think step by step.

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

1. **Edge Cases**:
   - Very small strings where the length is just enough to meet the minimum requirements.
   - Strings with repeated digits to test the compression logic.
   - Strings with digits that are close to the boundary values (1 and 9).

2. **Scenarios**:
   - Strings where the compression can be done in multiple valid ways.
   - Strings where the length of the substring is exactly the length of the string.
   - Strings where the length of the substring is close to the length of the string.

3. **Challenging Input**:
   - A string with repeated digits to test the compression logic.
   - A string with a mix of digits to ensure the algorithm handles different values correctly.
   - A string that forces the algorithm to handle multiple splits and combinations.

Given these considerations, a small-scale test input that is complete and valid and likely to expose the bug in the provided code is:

```plaintext
5 2
12345
```

This input has a string of length 5 and a substring length of 2. It will test the compression logic for multiple substrings and ensure that the algorithm handles different digits correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring Compression
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Let's define the operation of compressing a string $t$, consisting of at least $2$ digits from $1$ to $9$, as follows:

  * split it into an even number of non-empty substrings — let these substrings be $t_1, t_2, \dots, t_m$ (so, $t = t_1 + t_2 + \dots + t_m$, where $+$ is the concatenation operation);    * write the string $t_2$ $t_1$ times, then the string $t_4$ $t_3$ times, and so on.

For example, for a string "12345", one could do the following: split it into ("1", "23", "4", "5"), and write "235555".

Let the function $f(t)$ for a string $t$ return the minimum length of the string that can be obtained as a result of that process.

You are given a string $s$, consisting of $n$ digits from $1$ to $9$, and an integer $k$. Calculate the value of the function $f$ for all contiguous substrings of $s$ of length exactly $k$.

Input: The first line contains two integers $n$ and $k$ ($2 \le k \le n \le 2 \cdot 10^5$).

The second line contains the string $s$ ($|s| = n$), consisting only of digits from $1$ to $9$.

Output: Output $n - k + 1$ integers — $f(s_{1,k}), f(s_{2,k+1}), \dots, f(s_{n - k + 1, n})$.

Examples:
input:
4 4
5999
output:
14

input:
10 3
1111111111
output:
2 2 2 2 2 2 2 2

input:
11 4
49998641312
output:
12 18 17 15 12 7 7 2

Note:

Buggy Code:
n, k = map(int, input().split())
s = input()

results = []
for idx in range(n - k + 1):
    t = s[idx:idx + k]
    min_total_length = float('inf')
    length = len(t)
    max_multiplier_length = 3  # Limit the multiplier length to 3 digits

    # Try m = 2 (split into 2 substrings)
    for len_t1 in range(1, min(max_multiplier_length + 1, length)):
        t1 = t[:len_t1]
        len_t2 = length - len_t1
        if len_t2 <= 0:
            continue
        t2 = t[len_t1:]
        int_t1 = int(t1)
        total_length = len_t2 * int_t1
        if total_length < min_total_length:
            min_total_length = total_length

    # Try m = 4 (split into 4 substrings)
    for len_t1 in range(1, min(max_multiplier_length + 1, length - 2)):
        for len_t3 in range(1, min(max_multiplier_length + 1, length - len_t1 - 1)):
            len_t2 = 1
            len_t4 = length - len_t1 - len_t2 - len_t3
            if len_t4 <= 0:
                continue
            t1 = t[:len_t1]
            t2 = t[len_t1:len_t1 + len_t2]
            t3 = t[len_t1 + len_t2:len_t1 + len_t2 + len_t3]
            t4 = t[len_t1 + len_t2 + len_t3:]
            int_t1 = int(t1)
            int_t3 = int(t3)
            total_length = len(t2) * int_t1 + len(t4) * int_t3
            if total_length < min_total_length:
                min_total_length = total_length

    results.append(str(min_total_length))

print(' '.join(results))

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

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

Think step by step.

Response:
```plaintext
5 3
99999
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring Compression
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Let's define the operation of compressing a string $t$, consisting of at least $2$ digits from $1$ to $9$, as follows:

  * split it into an even number of non-empty substrings — let these substrings be $t_1, t_2, \dots, t_m$ (so, $t = t_1 + t_2 + \dots + t_m$, where $+$ is the concatenation operation);    * write the string $t_2$ $t_1$ times, then the string $t_4$ $t_3$ times, and so on.

For example, for a string "12345", one could do the following: split it into ("1", "23", "4", "5"), and write "235555".

Let the function $f(t)$ for a string $t$ return the minimum length of the string that can be obtained as a result of that process.

You are given a string $s$, consisting of $n$ digits from $1$ to $9$, and an integer $k$. Calculate the value of the function $f$ for all contiguous substrings of $s$ of length exactly $k$.

Input: The first line contains two integers $n$ and $k$ ($2 \le k \le n \le 2 \cdot 10^5$).

The second line contains the string $s$ ($|s| = n$), consisting only of digits from $1$ to $9$.

Output: Output $n - k + 1$ integers — $f(s_{1,k}), f(s_{2,k+1}), \dots, f(s_{n - k + 1, n})$.

Examples:
input:
4 4
5999
output:
14

input:
10 3
1111111111
output:
2 2 2 2 2 2 2 2

input:
11 4
49998641312
output:
12 18 17 15 12 7 7 2

Note:

Buggy Code:
dato = input()
n, k = map(int, dato.split())
s = input()
j = 0
res = []

while k <= n:
    s2 = s[j:k]
    res.append(0)
    if k <= 3:
        res[j] = (int(s[j])* (len(s2)-1))
    else:
        res[j] = (int(s[j])* (len(s2)-1))
        for i in range(len(s2)-2,1,-1):
            res2 = (int(s2[0])+int(s2[i]))
            if res[j] > res2:
                res[j] = res2

    j += 1
    k += 1

print(*res)

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

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

Think step by step.

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

1. **Edge Cases**:
   - Strings with repeated digits.
   - Strings with alternating digits.
   - Strings with a mix of small and large digits.
   - Strings where the length of the substring \( k \) is very small or very large.

2. **Behavior of the Buggy Code**:
   - The code seems to have a logical error in the way it calculates the compressed length.
   - It incorrectly handles the compression for substrings of length greater than 3.

3. **Specific Test Input**:
   - A string with a mix of small and large digits to test the compression logic.
   - A substring length \( k \) that is neither too small nor too large to ensure the code's logic is thoroughly tested.

Given these considerations, a challenging test input would be:

```plaintext
10 5
1234567891
```

This input:
- Has a string length \( n = 10 \) and a substring length \( k = 5 \).
- Contains a mix of digits from 1 to 9, which will test the compression logic for various substrings.
- The substrings will have lengths of 5, which is neither too small nor too large, ensuring that the code's logic for compression is thoroughly tested.

This test input should help expose the bug 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: Substring Compression
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Let's define the operation of compressing a string $t$, consisting of at least $2$ digits from $1$ to $9$, as follows:

  * split it into an even number of non-empty substrings — let these substrings be $t_1, t_2, \dots, t_m$ (so, $t = t_1 + t_2 + \dots + t_m$, where $+$ is the concatenation operation);    * write the string $t_2$ $t_1$ times, then the string $t_4$ $t_3$ times, and so on.

For example, for a string "12345", one could do the following: split it into ("1", "23", "4", "5"), and write "235555".

Let the function $f(t)$ for a string $t$ return the minimum length of the string that can be obtained as a result of that process.

You are given a string $s$, consisting of $n$ digits from $1$ to $9$, and an integer $k$. Calculate the value of the function $f$ for all contiguous substrings of $s$ of length exactly $k$.

Input: The first line contains two integers $n$ and $k$ ($2 \le k \le n \le 2 \cdot 10^5$).

The second line contains the string $s$ ($|s| = n$), consisting only of digits from $1$ to $9$.

Output: Output $n - k + 1$ integers — $f(s_{1,k}), f(s_{2,k+1}), \dots, f(s_{n - k + 1, n})$.

Examples:
input:
4 4
5999
output:
14

input:
10 3
1111111111
output:
2 2 2 2 2 2 2 2

input:
11 4
49998641312
output:
12 18 17 15 12 7 7 2

Note:

Buggy Code:
//#pragma GCC optimize("Ofast,unroll-loops")
//#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt,tune=native")

#include <bits/stdc++.h>

using namespace std;

static void solve() {
    int n, k;
    cin >> n >> k;

//    string s;
//    cin >> s;
//
//    vector<int> cnt(k);
//
//    for (int i = 0; i < n - k + 1; ++i) {
//        cnt.assign(k, 1);
//        cnt[k - 1] = 0;
//
//        for (int j = 0; j < k; ++j) {
//            for (int t = j + 2; t < k && s[t] >= s[j]; ++t)
//                ++cnt[j];
//        }
//
//        int ans = 0, j = i;
//
//        while (j < i + k) {
//            int t = j + 2;
//            for (; t < i + k && s[t] >= s[j]; ++t) {}
//
//            if (t == i + k) {
//                ans += (t - j - 1) * (s[j] - '0');
//                break;
//            }
//
//            assert(k > 2 && s[t] < s[j]);
//
//            if (t == i + k - 1) {
//                ans += (t - j - 1) * (s[j] - '0');
//                break;
//            }
//
//            assert(k > 2 && s[t] < s[j] && t < i + k - 1);
//
//            int c0 = s[j] - '0';
//            int c1 = s[t] - '0';
//            int c2 = s[t] - '0';
//
//            if (c1 * cnt[t] >= c0 + c2 * cnt[t + 1]) {
//                ans += (t - j - 1) * (s[j] - '0');
//                j = t;
//            } else {
//                ans += (t + 1 - j - 1) * (s[j] - '0');
//                j = t + 1;
//            }
//        }
//
//        cout << ans << '\n';
//    }

//    string s;
//    cin >> s;
//
//    vector<int> a(n);
//
//    for (int i = 0; i < n; ++i)
//        a[i] = s[i] - '0';
//
//    for (int i = 0; i < n - k + 1; ++i) {
//        int t = i;
//        int ans = a[t];
//
//        for (int j = i + 2; j < i + k; ++j) {
//            int c1 = (i + k - t - 1) * a[t];
//            int c2 = j >= i + k - 1 ? INT_MAX : (j - t - 1) * a[t] + (i + k - j - 1) * a[j];
//            int c3 = j >= i + k - 2 ? INT_MAX : (j - t) * a[t] + (i + k - j - 2) * a[j + 1];
//
//            if (c1 <= c2 && c1 <= c3)
//                ans += a[t];
//            else if (c2 <= c1 && c2 <= c3) {
//                ans += a[j];
//                j += 1;
//            }
//            else {
//                ans += a[t] + a[j + 1];
//                j += 2;
//            }
//        }
//
//        cout << ans << ' ';
//    }

    string s;
    cin >> s;

    vector<int> a(n);

    for (int i = 0; i < n; ++i)
        a[i] = s[i] - '0';

    for (int i = 0; i < n - k + 1; ++i) {
        int t = i;
        int ans = a[t];

        for (int j = i + 2; j < i + k; ++j) {
            int c1 = (i + k - t - 1) * a[t];
            int c2 = j >= i + k - 1 ? INT_MAX : (j - t - 1) * a[t] + (i + k - j - 1) * a[j];
            int c3 = j >= i + k - 2 ? INT_MAX : (j - t) * a[t] + (i + k - j - 2) * a[j + 1];

            if (c1 <= c2 && c1 <= c3)
                ans += a[t];
            else if (c2 <= c1 && c2 <= c3) {
                ans += a[j];
                j += 1;
            }
            else {
                ans += a[t] + a[j + 1];
                j += 2;
            }
        }

        cout << ans << ' ';
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

//    int t;
//    cin >> t;

//    while (t--) solve();

    solve();

    return 0;
}

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

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

Think step by step.

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

1. **Understanding the Buggy Code**:
   - The code attempts to compress substrings of length `k` by comparing and combining digits in a specific way.
   - The main logic involves iterating through the string and comparing different segments to determine the minimum length of the compressed string.

2. **Identifying Potential Issues**:
   - The code uses a combination of nested loops and conditions to determine the compression.
   - It might fail when the input string has repeating digits or when the lengths and positions of the substrings are not handled correctly.

3. **Edge Cases to Consider**:
   - **Repeating Digits**: Strings with many repeated digits.
   - **Boundary Conditions**: Substrings that are at the boundaries of the input string.
   - **Special Patterns**: Strings with specific patterns that might confuse the compression logic.

4. **Generating the Test Input**:
   - A small-scale test input that can expose these issues is a string with a mix of repeating and non-repeating digits, and a specific length `k` that can cause the code to fail.

Based on these considerations, a challenging test input could be:

```plaintext
10 5
1111122222
```

This input has:
- A string length of 10.
- A substring length of 5.
- A pattern of repeating digits followed by a different set of repeating digits.

This input should challenge the code to correctly handle the compression of substrings with repeating digits and ensure that the boundary conditions are handled correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring Compression
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Let's define the operation of compressing a string $t$, consisting of at least $2$ digits from $1$ to $9$, as follows:

  * split it into an even number of non-empty substrings — let these substrings be $t_1, t_2, \dots, t_m$ (so, $t = t_1 + t_2 + \dots + t_m$, where $+$ is the concatenation operation);    * write the string $t_2$ $t_1$ times, then the string $t_4$ $t_3$ times, and so on.

For example, for a string "12345", one could do the following: split it into ("1", "23", "4", "5"), and write "235555".

Let the function $f(t)$ for a string $t$ return the minimum length of the string that can be obtained as a result of that process.

You are given a string $s$, consisting of $n$ digits from $1$ to $9$, and an integer $k$. Calculate the value of the function $f$ for all contiguous substrings of $s$ of length exactly $k$.

Input: The first line contains two integers $n$ and $k$ ($2 \le k \le n \le 2 \cdot 10^5$).

The second line contains the string $s$ ($|s| = n$), consisting only of digits from $1$ to $9$.

Output: Output $n - k + 1$ integers — $f(s_{1,k}), f(s_{2,k+1}), \dots, f(s_{n - k + 1, n})$.

Examples:
input:
4 4
5999
output:
14

input:
10 3
1111111111
output:
2 2 2 2 2 2 2 2

input:
11 4
49998641312
output:
12 18 17 15 12 7 7 2

Note:

Buggy Code:
#include "bits/stdc++.h"
using namespace std;
#define all(x) x.begin(),x.end()
template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { return os << p.first << " " << p.second; }
template<typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type> ostream& operator<<(ostream &os, const T_container &v) { string sep; for (const T &x : v) os << sep << x, sep = " "; return os; }
#ifdef LOCAL
#include "debug.h"
#else
#define debug(...) 42
#define ASSERT(...) 42
#endif
typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> pi;
const int oo = 1e9;
void cmin(int& a, int b) {a=min(a,b);}
struct Mat {
    array<int,10> dp[10];

    Mat() {
        for(auto& i : dp) i.fill(oo);
        for(int i=0;i<10;++i) {
            dp[i][i]=0;
        }
    }
    Mat(char c) {
        for(auto& i : dp) i.fill(oo);
        for(int i=1;i<10;++i) {
            dp[i][i]=i;
            dp[i][0]=i;
        }
        dp[0][c-'0']=0;
    }
    Mat operator*(const Mat& o) const {
        Mat res;
        for(auto& i : res.dp) i.fill(oo);

        for(int j=0;j<10;++j) for(int k=0;k<10;++k) {
            cmin(res.dp[0][k],dp[0][j]+o.dp[j][k]);
            cmin(res.dp[j][0],dp[j][k]+o.dp[k][0]);
            cmin(res.dp[j][k], dp[j][0]+o.dp[0][k]);
        }
        for(int i=1;i<10;++i) for(int j=i;j>=1;--j) for(int k=j;k>=1;--k) {
            cmin(res.dp[i][k],dp[i][j]+o.dp[j][k]);
        }
        return res;
    }

};

typedef Mat pmi;
struct st1 {
    stack<pmi> st;

    void add(pmi p) {
        // fill p in previous function
        if(st.empty()) {
            st.push(p);
            return;
        }
        auto q = st.top();
        st.push(p*q);
    }
    void pop() {
        st.pop();
    }
    bool empty() {
        return st.empty();
    }
    Mat f() {
        if(st.empty()) return {};
        else return st.top();
    }
};
struct st2 {
    stack<pair<pmi,pmi>> st;

    void add(pmi p) {
        // fill p in previous function
        if(st.empty()) {
            st.push({p,p});
            return;
        }
        auto q = st.top().second;
        st.push({p, q*p});
    }
    void pop() {
        st.pop();
    }
    bool empty() {
        return st.empty();
    }
    Mat f() {
        if(st.empty()) return {};
        return st.top().second;
    }
};
struct Q {
    st1 s1;
    st2 s2;
    void add(pmi p) {
        s2.add(p);
    }
    void pop() {
        if(s1.empty()) {
            while(!s2.empty()) {
                s1.add(s2.st.top().first);
                s2.pop();
            }
        }
        s1.pop();
    }
    Mat f() {
        return s1.f() * s2.f();
    }
};
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int n,k; cin >> n >> k;
    string s; cin >> s;
    Q q;
    for(int i=0;i<k;++i) {
        q.add(Mat(s[i]));
    }
    for(int i=k;i<=n;++i) {
        auto res = q.f();
        cout << res.dp[0][0] << ' ';
        if(i==n) break;
        q.pop();
        q.add(s[i]);
    }
}

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

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

Think step by step.

Response:
```plaintext
6 3
123456
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring Compression
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Let's define the operation of compressing a string $t$, consisting of at least $2$ digits from $1$ to $9$, as follows:

  * split it into an even number of non-empty substrings — let these substrings be $t_1, t_2, \dots, t_m$ (so, $t = t_1 + t_2 + \dots + t_m$, where $+$ is the concatenation operation);    * write the string $t_2$ $t_1$ times, then the string $t_4$ $t_3$ times, and so on.

For example, for a string "12345", one could do the following: split it into ("1", "23", "4", "5"), and write "235555".

Let the function $f(t)$ for a string $t$ return the minimum length of the string that can be obtained as a result of that process.

You are given a string $s$, consisting of $n$ digits from $1$ to $9$, and an integer $k$. Calculate the value of the function $f$ for all contiguous substrings of $s$ of length exactly $k$.

Input: The first line contains two integers $n$ and $k$ ($2 \le k \le n \le 2 \cdot 10^5$).

The second line contains the string $s$ ($|s| = n$), consisting only of digits from $1$ to $9$.

Output: Output $n - k + 1$ integers — $f(s_{1,k}), f(s_{2,k+1}), \dots, f(s_{n - k + 1, n})$.

Examples:
input:
4 4
5999
output:
14

input:
10 3
1111111111
output:
2 2 2 2 2 2 2 2

input:
11 4
49998641312
output:
12 18 17 15 12 7 7 2

Note:

Buggy Code:
#include "bits/stdc++.h"
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx,avx2,sse,sse2")
#define fast ios_base::sync_with_stdio(0) , cin.tie(0) , cout.tie(0)
#define endl '\n'
#define int long long
#define f first
#define mp make_pair
#define s second
using namespace std;

int n, k;
string str;

int get(string a){
    int m = a.length();
    a = ' ' + a;
    int inf = 1e18;
    int dp[m + 1][10], idx[10];
    for(int i = 0; i <= m; i++){
        for(int j =0 ; j <= 9; j++) dp[i][j] = inf;
    }
    memset(idx, -1, sizeof(idx));
    dp[2][a[1] - '0'] = a[1] - '0';
    // dp[2][0] = a[1] - '0';
    for(int i = 2; i < m; i++){
        int x = a[i] - '0';
        for(int j = 1; j <= 9; j++){
            // cout << i << ' ' << j << ": " << dp[i][j] << endl;
            // if(dp[i][j] == inf) continue;
            dp[i + 1][j] = min(dp[i + 1][j], dp[i][j] + j);
            dp[i + 1][0] = min(dp[i + 1][0], dp[i][j]);
        }
        dp[i + 1][x] = min(dp[i + 1][x], dp[i][0] + x);
    }
    int ret = LONG_LONG_MAX;
    for(int i = 1; i <= 9; i++){
        // cout << m << " " << i << ": " << dp[m][i] << endl;
        ret = min(ret, dp[m][i]);
    }
    return ret;
}

signed main()
{
    fast;
    cin >> n >> k;
    cin >> str;
    for(int i = 0; i < n; i++){
        if(i + k > n) break;
        string x = str.substr(i, k);
        int ans = get(x);
        cout << ans << ' ';
    }
    cout << endl;
}

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

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

Think step by step.

Response:
```plaintext
6 2
111111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring Compression
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Let's define the operation of compressing a string $t$, consisting of at least $2$ digits from $1$ to $9$, as follows:

  * split it into an even number of non-empty substrings — let these substrings be $t_1, t_2, \dots, t_m$ (so, $t = t_1 + t_2 + \dots + t_m$, where $+$ is the concatenation operation);    * write the string $t_2$ $t_1$ times, then the string $t_4$ $t_3$ times, and so on.

For example, for a string "12345", one could do the following: split it into ("1", "23", "4", "5"), and write "235555".

Let the function $f(t)$ for a string $t$ return the minimum length of the string that can be obtained as a result of that process.

You are given a string $s$, consisting of $n$ digits from $1$ to $9$, and an integer $k$. Calculate the value of the function $f$ for all contiguous substrings of $s$ of length exactly $k$.

Input: The first line contains two integers $n$ and $k$ ($2 \le k \le n \le 2 \cdot 10^5$).

The second line contains the string $s$ ($|s| = n$), consisting only of digits from $1$ to $9$.

Output: Output $n - k + 1$ integers — $f(s_{1,k}), f(s_{2,k+1}), \dots, f(s_{n - k + 1, n})$.

Examples:
input:
4 4
5999
output:
14

input:
10 3
1111111111
output:
2 2 2 2 2 2 2 2

input:
11 4
49998641312
output:
12 18 17 15 12 7 7 2

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
// #define int long long
#define ll long long
#define lb(x) ((x) & (-x))
#define double long double
#define ff first
#define ss second
#define mr make_pair
#define l(p) t[p].l
#define r(p) t[p].r
#define lz(p) t[p].lz
#define ans(p) t[p].ans
#define res(p) t[p].res
const int mod = 998244353 ;
const int P = 1e9 + 7 ;
// const int modu[10] = {1000000007,1000000009,1000000021,
// 1000000033,1000000087,1000000093,1000000097} ;
void cmin(auto &x, auto y) {
    x = min(x, y) ;
}
void cmax(int &x, int y) {
    x = max(x, y) ;
}
int read() {
    char ch ;int s = 0 ;int w = 1;
    while((ch = getchar()) >'9' || ch < '0' )if(ch == '-')w = -1 ;
    while(ch >= '0' && ch <= '9')s = (s << 1) + (s << 3) + (ch ^ 48) , ch = getchar() ;
    return s * w ;
}
void print(int x) {
    if(x < 0) putchar('-'), x = -x ;
    if(x > 9)print(x / 10) ;
    putchar(x % 10 + '0') ;
}
void prn(int x) {
    print(x) ;
    putchar('\n') ;
}
int pows(int x, int k) {
    long long bace = x, ans = 1 ;
    while(k) {
        if(k & 1) ans *= bace, ans %= mod ;
        bace *= bace, bace %= mod ;
        k >>= 1 ;
    }
    return ans ;
}
mt19937 rnd(time(0)^clock()) ;
int TT ;
int n, m ;
int k, Q ;
int a[200005] ;
int f[200005][11] ;
int g[200005][11][11] ;
int A[11][11] ;
int B[11][11] ;
int tmpa[11][11] ;
int tmpb[11][11] ;
int tmp[11] ;
string s ;
void cf(int (&a)[11][11], const int (&b)[11][11], const int (&c)[11][11]) {
	memcpy(tmpa, b, sizeof(b)) ;
	memcpy(tmpb, c, sizeof(c)) ;
	memset(a, 0x3f, sizeof(a)) ;
	for(int i = 0 ; i < 11 ; i++)
		for(int j = 0 ; j < 11 ; j++) {
			for(int k = 0 ; k < 11 ; k++) {
				cmin(a[i][k], tmpa[i][j] + tmpb[j][k]) ;
			}
		}
}
void ccf(int (&a)[11], const int (&b)[11], const int (&c)[11][11]) {
	memcpy(tmp, b, sizeof(tmp)) ;
	memcpy(tmpa, c, sizeof(c)) ;
	memset(a, 0x3f, sizeof(a)) ;
	for(int i = 0 ; i < 11 ; i++)
		for(int j = 0 ; j < 11 ; j++)
			cmin(a[i], tmp[j] + tmpa[j][i]) ;
}
void work() {
	n = read(), k = read() ;
	cin >> s ;
	for(int i = 0 ; i < s.size() ; i++) a[i+1]=s[i]-'0' ;
	for(int i = 1 ; i <= n ; i++) {
		memset(A, 0x3f, sizeof(A)) ;
		for(int j = 0 ; j < 9 ; j++) A[j][j] = j+1 ;
		if(i>=2) A[9][a[i-1]-1]=a[i-1];
		memset(B, 0x3f, sizeof(B)) ;
		for(int j = 0 ; j < 9 ; j++) A[j][9] = 0 ;
		A[9][10] = 0 ;
		memcpy(g[i], A, sizeof(g[i])) ;
		// cout << i << endl ;
		// for(int j = 0 ; j < 11 ; j++, cout << endl)
		// 	for(int k =0 ; k < 11 ; k++, cout << ' ')
		// 		cout << g[i][j][k] ;
	}
	for(int i = 1 ; i + k - 1 <= n ; i++) {
		memset(f[i], 0x3f, sizeof(f[i])) ;
		f[i][a[i]-1]=a[i] ;
		// cout << "?" << i << ' ' << a[i] << endl ;
		for(int j = i+2 ; j - i + 1 <= k ; j++) {
			ccf(f[i], f[i], g[j]) ;
			// for(int q = 0 ; q <= 10 ; q++)cout<<f[i][q]<<' ';cout<<endl;
		}
		int ans = 1e9 ;
		for(int j = 0 ; j < 9 ; j++) cmin(ans, f[i][j]) ;
		cout << ans << ' ' ;
	}

}
signed main(){
    // freopen("data.in", "r", stdin) ;
    // freopen("ans.out", "w", stdout) ;
	TT = 1 ;
    // TT = read() ;
    while(TT--) {

        work() ;
    }
	return 0;
}
/*

*/

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on edge cases or scenarios that can highlight the issues in the algorithm. Here are some considerations:

1. **Edge Cases for String Length and Substring Length**:
   - When \( k \) is close to \( n \), the number of substrings to process is minimized, which can help in identifying issues with the compression logic.
   - When \( k \) is exactly 2, the algorithm should handle the simplest case of compression.

2. **Special Characters and Patterns**:
   - Strings with repeated characters can test the compression logic.
   - Strings with alternating patterns can test the split and concatenate logic.

3. **Boundary Conditions**:
   - Strings with lengths that are just above the minimum and maximum limits can test the edge cases.

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

```plaintext
5 2
11223
```

This test input has:
- \( n = 5 \) (string length)
- \( k = 2 \) (substring length)
- The string "11223" contains repeated characters and a mix of different digits, which can help in testing the compression logic.

This input should be sufficient to expose any bugs in the algorithm, especially in how it handles the compression of substrings of length 2.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring Compression
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Let's define the operation of compressing a string $t$, consisting of at least $2$ digits from $1$ to $9$, as follows:

  * split it into an even number of non-empty substrings — let these substrings be $t_1, t_2, \dots, t_m$ (so, $t = t_1 + t_2 + \dots + t_m$, where $+$ is the concatenation operation);    * write the string $t_2$ $t_1$ times, then the string $t_4$ $t_3$ times, and so on.

For example, for a string "12345", one could do the following: split it into ("1", "23", "4", "5"), and write "235555".

Let the function $f(t)$ for a string $t$ return the minimum length of the string that can be obtained as a result of that process.

You are given a string $s$, consisting of $n$ digits from $1$ to $9$, and an integer $k$. Calculate the value of the function $f$ for all contiguous substrings of $s$ of length exactly $k$.

Input: The first line contains two integers $n$ and $k$ ($2 \le k \le n \le 2 \cdot 10^5$).

The second line contains the string $s$ ($|s| = n$), consisting only of digits from $1$ to $9$.

Output: Output $n - k + 1$ integers — $f(s_{1,k}), f(s_{2,k+1}), \dots, f(s_{n - k + 1, n})$.

Examples:
input:
4 4
5999
output:
14

input:
10 3
1111111111
output:
2 2 2 2 2 2 2 2

input:
11 4
49998641312
output:
12 18 17 15 12 7 7 2

Note:

Buggy Code:
import sys
def input():return sys.stdin.readline().rstrip()
n,k = map(int,input().split(' '))
a = list(map(int,input()))
ans = [int(1e9)] * (n-k+1)
for l in range(0,n-k+1,k+1):
    mid = l+k-1
    r = min(n-1,mid+k)
    for boss in range(0,10):
        dp1 = [[int(1e9)]*(r-mid+1) for i in range(10)]
        dp1[boss][0] = 0
        for i in range(mid+1,r+1):
            new0 = int(1e9)
            for j in range(1,10):
                new0 = min(new0,dp1[j][i-mid-1] + j)
                dp1[j][i-mid] = dp1[j][i-mid-1] + j
            dp1[a[i]][i-mid] = min(dp1[int(a[i])][i-mid], dp1[0][i-mid-1])
            dp1[0][i-mid] = new0
        dp2 = [[int(1e9)]*(mid-l+2) for i in range(10)]
        dp2[boss][0] = 0
        for i in range(mid,l-1,-1):
            new0 = int(1e9)
            for j in range(1,10):
                dp2[j][mid-i+1] = min(dp2[j][mid-i], dp2[0][mid-i]) + j
                if a[i] == j:
                    new0 = min(new0, dp2[j][mid-i])
            dp2[0][mid-i+1] = new0
        for i in range(l,l+k+1):
            cur_l = i
            cur_r = i+k-1
            if cur_r >= n:
                break
            ans[i] = min(ans[i],dp2[0][mid-cur_l+1]+dp1[0][cur_r-mid])
out = ' '.join(list(map(str,ans)))
print(out)

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

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

Think step by step.

Response:
```plaintext
5 2
11111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring Compression
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Let's define the operation of compressing a string $t$, consisting of at least $2$ digits from $1$ to $9$, as follows:

  * split it into an even number of non-empty substrings — let these substrings be $t_1, t_2, \dots, t_m$ (so, $t = t_1 + t_2 + \dots + t_m$, where $+$ is the concatenation operation);    * write the string $t_2$ $t_1$ times, then the string $t_4$ $t_3$ times, and so on.

For example, for a string "12345", one could do the following: split it into ("1", "23", "4", "5"), and write "235555".

Let the function $f(t)$ for a string $t$ return the minimum length of the string that can be obtained as a result of that process.

You are given a string $s$, consisting of $n$ digits from $1$ to $9$, and an integer $k$. Calculate the value of the function $f$ for all contiguous substrings of $s$ of length exactly $k$.

Input: The first line contains two integers $n$ and $k$ ($2 \le k \le n \le 2 \cdot 10^5$).

The second line contains the string $s$ ($|s| = n$), consisting only of digits from $1$ to $9$.

Output: Output $n - k + 1$ integers — $f(s_{1,k}), f(s_{2,k+1}), \dots, f(s_{n - k + 1, n})$.

Examples:
input:
4 4
5999
output:
14

input:
10 3
1111111111
output:
2 2 2 2 2 2 2 2

input:
11 4
49998641312
output:
12 18 17 15 12 7 7 2

Note:

Buggy Code:
def matmul(A,B):
    return [[min(A[i][k]+B[k][j] for k in range(10)) for j in range(10)] for i in range(10)]

def M(d):
    m = [[1e9]*10 for _ in range(10)]
    for c in range(1,10): m[c][0] = m[c][c] = c
    m[0][int(d)] = 0
    return m

n,k = map(int,input().split())
s = input() + "0"*k

for l in range(n-k+1):
    if l % k == 0:
        stack1 = [M(s[l+k-1])]
        for d in s[l:l+k-1][::-1]:
            stack1.append(matmul(M(d),stack1[-1]))
        print(stack1[-1][0][0])
        stack2 = [0,M(s[l+k])]
        for d in s[l+k+1:l+2*k]:
            stack2.append(matmul(stack2[-1],M(d)))
    else:
        print(matmul(stack1[~(l%k)],stack2[l%k])[0][0])

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

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

Think step by step.

Response:
```plaintext
5 5
12345
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring Compression
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Let's define the operation of compressing a string $t$, consisting of at least $2$ digits from $1$ to $9$, as follows:

  * split it into an even number of non-empty substrings — let these substrings be $t_1, t_2, \dots, t_m$ (so, $t = t_1 + t_2 + \dots + t_m$, where $+$ is the concatenation operation);    * write the string $t_2$ $t_1$ times, then the string $t_4$ $t_3$ times, and so on.

For example, for a string "12345", one could do the following: split it into ("1", "23", "4", "5"), and write "235555".

Let the function $f(t)$ for a string $t$ return the minimum length of the string that can be obtained as a result of that process.

You are given a string $s$, consisting of $n$ digits from $1$ to $9$, and an integer $k$. Calculate the value of the function $f$ for all contiguous substrings of $s$ of length exactly $k$.

Input: The first line contains two integers $n$ and $k$ ($2 \le k \le n \le 2 \cdot 10^5$).

The second line contains the string $s$ ($|s| = n$), consisting only of digits from $1$ to $9$.

Output: Output $n - k + 1$ integers — $f(s_{1,k}), f(s_{2,k+1}), \dots, f(s_{n - k + 1, n})$.

Examples:
input:
4 4
5999
output:
14

input:
10 3
1111111111
output:
2 2 2 2 2 2 2 2

input:
11 4
49998641312
output:
12 18 17 15 12 7 7 2

Note:

Buggy Code:
def leftmul(A,d):
    B = [0]*10
    B[0] = A[d]
    for c in range(1,10): B[c] = min(A[c],A[0])+c
    return B

def rightmul(A,d):
    B = [0]*10
    B[0] = min(A[c]+c for c in range(1,10))
    for c in range(1,10): B[c] = A[c]+c
    for i in range(10): B[d] = min(B[d],A[0])
    return B

INF = 10**7

n,k = map(int,input().split())
s = *map(int,input()),*[0]*k
ans = [INF]*(n-k+1)

for col in range(10):
    for l in range(n-k+1):
        if l % k == 0:
            s1 = [INF]*(k+1)
            s2 = [INF]*(k+1)
            r1 = [INF]*10
            r2 = [INF]*10
            r1[col] = r2[col] = 0
            s1[0] = s2[0] = r1[0]
            for d in range(1,k+1):
                r1 = leftmul(r1,s[l+k-d])
                r2 = rightmul(r2,s[l+k+d-1])
                s1[d] = r1[0]
                s2[d] = r2[0]
        ans[l] = min(ans[l], s1[~(l%k)]+s2[l%k])

print(*ans)

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

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

Think step by step.

Response:
```plaintext
5 2
12345
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Substring Compression
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Let's define the operation of compressing a string $t$, consisting of at least $2$ digits from $1$ to $9$, as follows:

  * split it into an even number of non-empty substrings — let these substrings be $t_1, t_2, \dots, t_m$ (so, $t = t_1 + t_2 + \dots + t_m$, where $+$ is the concatenation operation);    * write the string $t_2$ $t_1$ times, then the string $t_4$ $t_3$ times, and so on.

For example, for a string "12345", one could do the following: split it into ("1", "23", "4", "5"), and write "235555".

Let the function $f(t)$ for a string $t$ return the minimum length of the string that can be obtained as a result of that process.

You are given a string $s$, consisting of $n$ digits from $1$ to $9$, and an integer $k$. Calculate the value of the function $f$ for all contiguous substrings of $s$ of length exactly $k$.

Input: The first line contains two integers $n$ and $k$ ($2 \le k \le n \le 2 \cdot 10^5$).

The second line contains the string $s$ ($|s| = n$), consisting only of digits from $1$ to $9$.

Output: Output $n - k + 1$ integers — $f(s_{1,k}), f(s_{2,k+1}), \dots, f(s_{n - k + 1, n})$.

Examples:
input:
4 4
5999
output:
14

input:
10 3
1111111111
output:
2 2 2 2 2 2 2 2

input:
11 4
49998641312
output:
12 18 17 15 12 7 7 2

Note:

Buggy Code:
import sys
from functools import lru_cache

# Constants similar to those defined in C++
N = 200005
INF = int(1e9)

# Read input
n, k = map(int, input().split())
s = ' ' + input().strip()  # Add a space to match 1-based indexing

# Initialize the result and arrays
L = [[INF] * 10 for _ in range(N)]
R = [[INF] * 10 for _ in range(N)]
res = [INF] * (N)

def calc(l, r):
    if r - l + 1 < k:
        return

    m = (l + r) // 2

    for d in range(10):
        for i in range(10):
            L[m + 1][i] = INF
            R[m][i] = INF

        L[m + 1][d] = 0
        R[m][d] = 0

        # Calculate left side (L array)
        for i in range(m, max(l - 1, m - k), -1):
            L[i][0] = L[i + 1][int(s[i])]
            for p in range(1, 10):
                L[i][p] = p + min(L[i + 1][0], L[i + 1][p])

        # Calculate right side (R array)
        for i in range(m + 1, min(r + 1, m + k)):
            R[i][0] = INF
            for p in range(1, 10):
                R[i][0] = min(R[i][0], p + R[i - 1][p])
                R[i][p] = R[i - 1][p] + p
                if int(s[i]) == p:
                    R[i][p] = min(R[i][p], R[i - 1][0])

        # Calculate results
        for i in range(max(l, m - k + 1), m + 1):
            if i + k - 1 <= r:
                res[i] = min(res[i], L[i][0] + R[i + k - 1][0])

    calc(l, m - 1)
    calc(m + 1, r)

# Main function call
calc(1, n)

# Print the results
print(' '.join(map(str, res[1:n - k + 2])))

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

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

Think step by step.

Response:
```plaintext
5 2
11111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yasya and the Mysterious Tree
time_limit_ms: 2500
memory_limit_mb: 512
Description: Yasya was walking in the forest and accidentally found a tree with $n$ vertices. A tree is a connected undirected graph with no cycles.

Next to the tree, the girl found an ancient manuscript with $m$ queries written on it. The queries can be of two types.

The first type of query is described by the integer $y$. The weight of each edge in the tree is replaced by the [bitwise exclusive OR](http://tiny.cc/xor_wiki_eng) of the weight of that edge and the integer $y$.

The second type is described by the vertex $v$ and the integer $x$. Yasya chooses a vertex $u$ ($1 \le u \le n$, $u \neq v$) and mentally draws a bidirectional edge of weight $x$ from $v$ to $u$ in the tree.

Then Yasya finds a simple cycle in the resulting graph and calculates the [bitwise exclusive OR](http://tiny.cc/xor_wiki_eng) of all the edges in it. She wants to choose a vertex $u$ such that the calculated value is maximum. This calculated value will be the answer to the query. It can be shown that such a cycle exists and is unique under the given constraints (independent of the choice of $u$). If an edge between $v$ and $u$ already existed, a simple cycle is the path $v \to u \to v$.

Note that the second type of query is performed mentally, meaning the tree does not change in any way after it.

Help Yasya answer all the queries.

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

The descriptions of the test cases follow.

The first line of each test case contains two integers $n$, $m$ ($2 \le n \le 2 \cdot 10^5$, $1 \le m \le 2 \cdot 10^5$) — the number of vertices in the tree and the number of queries.

The next $n - 1$ lines of each test case contain three integers $v$, $u$, $w$ ($1 \le v, u \le n$, $1 \le w \le 10^9$) — the ends of some edge in the tree and its weight.

It is guaranteed that the given set of edges forms a tree.

The next $m$ lines of each test case describe the queries:

  * ^ $y$ ($1 \le y \le 10^9$) — parameter of the first type query;    * ? $v$ $x$ ($1 \le v \le n$, $1 \le x \le 10^9$) — parameters of the second type query.

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

Output: For each test case, output the answers to the queries of the second type.

Examples:
input:
2
3 7
1 2 1
3 1 8
^ 5
? 2 9
^ 1
? 1 10
^ 6
? 3 1
? 2 9
5 6
1 2 777
3 2 2812
4 1 16
5 3 1000000000
^ 4
? 3 123
? 5 1000000000
^ 1000000000
? 1 908070
? 2 1
output:
13 15 11 10
1000000127 2812 999756331 999999756

input:
3
8 4
8 6 3
6 3 4
2 5 4
7 6 2
7 1 10
4 1 4
5 1 2
^ 4
^ 7
? 7 8
? 4 10
5 6
3 1 4
2 3 9
4 3 6
5 2 10
? 5 7
^ 1
^ 8
? 4 10
? 1 9
? 3 6
4 2
2 1 4
4 3 5
2 3 4
^ 13
? 1 10
output:
14 13
13 8 11 11
10

Note:

Buggy Code:
// I know that I'm not that important to you
// But to me, girl, you're so much more than gorgeous (yeah)
// So much more than perfect (yeah)
// Right now, I know that I'm not really worth it
// If you give me time, I can work on it
// Give me some time while I work on it

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
typedef long long ll;
#define pb push_back
#define ff first
#define ss second
const int N=2e3+7;
const int mod=1e9+7;
const double eps=1e-9;
const double pi=    3.14159265358979323846;
using namespace std;
using namespace __gnu_pbds;
typedef tree<int, null_type, less_equal<int>, rb_tree_tag,
             tree_order_statistics_node_update>
    op_set;

struct BIT_TRIE
{
    struct node
    {
        int cnt=0;
        node* next[2];
        node()
        {
            next[1]=next[0]=NULL;
        }
    };

    node root;

    void add(int val)
    {
        node* temp=&root;
        temp->cnt++;
        for(int i=30;i>=0;i--)
        {
            int bit=1;
            if((val&(1<<i))==0)
                    bit=0;

            if(temp->next[bit]==NULL)
                    temp->next[bit]=new node();

            temp=temp->next[bit];
            temp->cnt++;
        }
    }

    int max_xor(int val)
    {
        int ans=0;
        node* temp=&root;
        for(int i=30;i>=0;i--)
        {
            int bit=1;
            if((val&(1<<i))==0)
                    bit=0;

            int req=1-bit;

            if(temp->next[req]==NULL)
                    temp=temp->next[1-req];
            else if(temp->next[req]->cnt==0)
                    temp=temp->next[1-req];
            else
            {
                ans+=(1<<i);
                temp=temp->next[req];
            }
        }

        return ans;
    }

    void del(int val)
    {
        node *temp=&root;
        temp->cnt--;

        for(int i=30;i>=0;i--)
        {
            int bit=1;
            if((val&(1<<i))==0)
                    bit=0;

            temp=temp->next[bit];
            temp->cnt--;
        }
    }

};

void dfs(int node,int parent,int typeOfNode,int xorWt,vector<vector<pair<int,int>>>&g,vector<int>&type,vector<int>&xorred)
{
    type[node]=typeOfNode;
    xorred[node]=xorWt;
    for(auto &[child,wt] : g[node])
    {
        if(child==parent)
            continue;
        dfs(child,node,1-typeOfNode,xorWt^wt,g,type,xorred);
    }
}

int32_t main()
{
   ios_base::sync_with_stdio(false);
   cin.tie(0);

   int t;
   cin >> t;
   while(t--)
   {
        int n,m;
        cin >> n >> m;

        vector<vector<pair<int,int>>> g(n+1);
        for(int i=1;i<n;i++)
        {
            int u,v,wt;
            cin >> u >> v >> wt;
            g[u].pb({v,wt});
            g[v].pb({u,wt});
        }

        vector<int> type(n+1);
        vector<int> xorred(n+1);
        dfs(1,0,0,0,g,type,xorred);

        BIT_TRIE bt[2];
        bt[0].add(0);
        bt[1].add(0);

        for(int i=1;i<=n;i++)
            bt[ type[i] ].add(xorred[i]);

        int currXor=0;
        while(m--)
        {
            char op;
            cin >> op;
            if(op=='^')
            {
                int val;
                cin >> val;
                currXor=(currXor^val);
            }
            else
            {
                int node,x;
                cin >> node >> x;

                int nodeWt=0;
                if(type[node]==0)
                {
                    bt[0].del(xorred[node]);
                    nodeWt=xorred[node];
                }
                else
                {
                    bt[1].del(xorred[node]);
                    nodeWt=xorred[node]^currXor;
                }

                int ans=0;
                ans=std::max(ans,bt[0].max_xor(nodeWt^x));
                ans=std::max(ans,bt[1].max_xor(nodeWt^x^currXor));
                cout << ans << " ";

                if(type[node]==0)
                    bt[0].add(xorred[node]);
                else
                    bt[1].add(xorred[node]);
            }
        }
        cout << "\n";
   }
}

/*
Mar Gaye Ehsaas Meray,
Chubay Wo Ilfaaz Teray,
Tumne Dekha Tak Na Jaty Way,
*/

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

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

Think step by step.

Response:
```plaintext
2
3 3
1 2 1
3 1 8
^ 5
? 2 9
? 3 1
5 4
1 2 1
2 3 2
3 4 3
4 5 4
^ 1
? 2 5
? 5 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yasya and the Mysterious Tree
time_limit_ms: 2500
memory_limit_mb: 512
Description: Yasya was walking in the forest and accidentally found a tree with $n$ vertices. A tree is a connected undirected graph with no cycles.

Next to the tree, the girl found an ancient manuscript with $m$ queries written on it. The queries can be of two types.

The first type of query is described by the integer $y$. The weight of each edge in the tree is replaced by the [bitwise exclusive OR](http://tiny.cc/xor_wiki_eng) of the weight of that edge and the integer $y$.

The second type is described by the vertex $v$ and the integer $x$. Yasya chooses a vertex $u$ ($1 \le u \le n$, $u \neq v$) and mentally draws a bidirectional edge of weight $x$ from $v$ to $u$ in the tree.

Then Yasya finds a simple cycle in the resulting graph and calculates the [bitwise exclusive OR](http://tiny.cc/xor_wiki_eng) of all the edges in it. She wants to choose a vertex $u$ such that the calculated value is maximum. This calculated value will be the answer to the query. It can be shown that such a cycle exists and is unique under the given constraints (independent of the choice of $u$). If an edge between $v$ and $u$ already existed, a simple cycle is the path $v \to u \to v$.

Note that the second type of query is performed mentally, meaning the tree does not change in any way after it.

Help Yasya answer all the queries.

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

The descriptions of the test cases follow.

The first line of each test case contains two integers $n$, $m$ ($2 \le n \le 2 \cdot 10^5$, $1 \le m \le 2 \cdot 10^5$) — the number of vertices in the tree and the number of queries.

The next $n - 1$ lines of each test case contain three integers $v$, $u$, $w$ ($1 \le v, u \le n$, $1 \le w \le 10^9$) — the ends of some edge in the tree and its weight.

It is guaranteed that the given set of edges forms a tree.

The next $m$ lines of each test case describe the queries:

  * ^ $y$ ($1 \le y \le 10^9$) — parameter of the first type query;    * ? $v$ $x$ ($1 \le v \le n$, $1 \le x \le 10^9$) — parameters of the second type query.

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

Output: For each test case, output the answers to the queries of the second type.

Examples:
input:
2
3 7
1 2 1
3 1 8
^ 5
? 2 9
^ 1
? 1 10
^ 6
? 3 1
? 2 9
5 6
1 2 777
3 2 2812
4 1 16
5 3 1000000000
^ 4
? 3 123
? 5 1000000000
^ 1000000000
? 1 908070
? 2 1
output:
13 15 11 10
1000000127 2812 999756331 999999756

input:
3
8 4
8 6 3
6 3 4
2 5 4
7 6 2
7 1 10
4 1 4
5 1 2
^ 4
^ 7
? 7 8
? 4 10
5 6
3 1 4
2 3 9
4 3 6
5 2 10
? 5 7
^ 1
^ 8
? 4 10
? 1 9
? 3 6
4 2
2 1 4
4 3 5
2 3 4
^ 13
? 1 10
output:
14 13
13 8 11 11
10

Note:

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

constexpr int N = 2e5 + 8, D = 30;
struct Trie {
    int nxt[N * D][2];
    int icnt[N * D];
    int pcnt;

    void init(int p) {
        nxt[p][0] = nxt[p][1] = 0;
    }

    void init() {
        pcnt = 1;
        init(1);
    }

    void modify(int x, int d, int p = 1) {
        icnt[p] += d;
        for (int bit = D; ~bit; --bit) {
            bool c = x & (1ll << bit);
            if (!nxt[p][c]) init(nxt[p][c] = ++pcnt);
            p = nxt[p][c];
            icnt[p] += d;
        }
    }

    ll getmax(int x, int p = 1, ll max = 0) {
        for (int bit = D; ~bit; --bit) {
            bool c = x & (1ll << bit);
            if (icnt[nxt[p][!c]]) p = nxt[p][!c], max |= (1ll << bit);
            else if (icnt[nxt[p][c]]) p = nxt[p][c];
            else break;
        }
        return max;
    }

    ll getmin(int x, int p = 1, ll min = 0) {
        for (int bit = D; ~bit; --bit) {
            bool c = x & (1ll << bit);
            if (icnt[nxt[p][c]]) p = nxt[p][c];
            else if (icnt[nxt[p][!c]]) p = nxt[p][!c], min |= (1ll << bit);
            else break;
        }
        return min;
    }
} trie[2];

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    int t;
    cin >> t;

    while (t--) {
        int n, q;
        cin >> n >> q;

        vector<vector<pair<int, int>>> E(n);
        for (int i = 1; i < n; i++) {
            int u, v, w;
            cin >> u >> v >> w;
            u--, v--;
            E[u].push_back({ w, v });
            E[v].push_back({ w, u });
        }

        vector<int> dep(n), dis(n);
        auto dfs = [&](auto&& self, int u, int p) -> void {
            for (auto [w, v] : E[u]) {
                if (v == p) continue;
                dep[v] = dep[u] ^ 1;
                dis[v] = dis[u] ^ w;
                self(self, v, u);
            }
        };
        dfs(dfs, 0, -1);

        trie[0].init(), trie[1].init();
        for (int u = 0; u < n; ++u) {
            trie[dep[u]].modify(dis[u], 1);
        }

        int w = 0;
        while (q--) {
            char op;
            cin >> op;

            if (op == '^') {
                int y;
                cin >> y;
                w ^= y;
            }
            else {
                int u, x;
                cin >> u >> x;
                u--;
                x ^= dis[u] ^ (dep[u] * w);
                trie[dep[u]].modify(dis[u], -1);
                cout << max(trie[0].getmax(x), trie[1].getmax(x ^ w)) << " \n"[q == 0];
                trie[dep[u]].modify(dis[u], 1);
            }
        }
    }

    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
1 2 1
3 1 8
^ 5
? 2 9
^ 1
? 1 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yasya and the Mysterious Tree
time_limit_ms: 2500
memory_limit_mb: 512
Description: Yasya was walking in the forest and accidentally found a tree with $n$ vertices. A tree is a connected undirected graph with no cycles.

Next to the tree, the girl found an ancient manuscript with $m$ queries written on it. The queries can be of two types.

The first type of query is described by the integer $y$. The weight of each edge in the tree is replaced by the [bitwise exclusive OR](http://tiny.cc/xor_wiki_eng) of the weight of that edge and the integer $y$.

The second type is described by the vertex $v$ and the integer $x$. Yasya chooses a vertex $u$ ($1 \le u \le n$, $u \neq v$) and mentally draws a bidirectional edge of weight $x$ from $v$ to $u$ in the tree.

Then Yasya finds a simple cycle in the resulting graph and calculates the [bitwise exclusive OR](http://tiny.cc/xor_wiki_eng) of all the edges in it. She wants to choose a vertex $u$ such that the calculated value is maximum. This calculated value will be the answer to the query. It can be shown that such a cycle exists and is unique under the given constraints (independent of the choice of $u$). If an edge between $v$ and $u$ already existed, a simple cycle is the path $v \to u \to v$.

Note that the second type of query is performed mentally, meaning the tree does not change in any way after it.

Help Yasya answer all the queries.

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

The descriptions of the test cases follow.

The first line of each test case contains two integers $n$, $m$ ($2 \le n \le 2 \cdot 10^5$, $1 \le m \le 2 \cdot 10^5$) — the number of vertices in the tree and the number of queries.

The next $n - 1$ lines of each test case contain three integers $v$, $u$, $w$ ($1 \le v, u \le n$, $1 \le w \le 10^9$) — the ends of some edge in the tree and its weight.

It is guaranteed that the given set of edges forms a tree.

The next $m$ lines of each test case describe the queries:

  * ^ $y$ ($1 \le y \le 10^9$) — parameter of the first type query;    * ? $v$ $x$ ($1 \le v \le n$, $1 \le x \le 10^9$) — parameters of the second type query.

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

Output: For each test case, output the answers to the queries of the second type.

Examples:
input:
2
3 7
1 2 1
3 1 8
^ 5
? 2 9
^ 1
? 1 10
^ 6
? 3 1
? 2 9
5 6
1 2 777
3 2 2812
4 1 16
5 3 1000000000
^ 4
? 3 123
? 5 1000000000
^ 1000000000
? 1 908070
? 2 1
output:
13 15 11 10
1000000127 2812 999756331 999999756

input:
3
8 4
8 6 3
6 3 4
2 5 4
7 6 2
7 1 10
4 1 4
5 1 2
^ 4
^ 7
? 7 8
? 4 10
5 6
3 1 4
2 3 9
4 3 6
5 2 10
? 5 7
^ 1
^ 8
? 4 10
? 1 9
? 3 6
4 2
2 1 4
4 3 5
2 3 4
^ 13
? 1 10
output:
14 13
13 8 11 11
10

Note:

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

const int MAX = 2e5 + 5;
const int max_log = 30;

int n, q;
vector<pair<int, int>> adj[MAX];
int state[MAX], f[MAX];

void dfs(int u, int par = -1) {
	for (pair<int, int> e : adj[u]) {
		int v = e.first, w = e.second;
		if (v == par) continue;

		state[v] = state[u] ^ 1;
		f[v] = f[u] ^ w;

		dfs(v, u);
	}
}

namespace Solution {
	struct Trie {
		struct Node {
			int nxt[2];
			int exist, cnt;
			long long sum;
		} nodes[MAX * max_log];

		int cur;
		Trie() : cur(0) {
			memset(nodes[0].nxt, -1, sizeof(nodes[0].nxt));
			nodes[0].cnt = 0;
		};

		int new_node() {
			++cur;
			memset(nodes[cur].nxt, -1, sizeof(nodes[cur].nxt));
			nodes[cur].cnt = 0;
			return cur;
		}

		void Insert(int x) {
			int pos = 0;
			for (int i = max_log; i >= 0; i--) {
				int c = (x >> i) & 1;
				if (nodes[pos].nxt[c] == -1) nodes[pos].nxt[c] = new_node();

				pos = nodes[pos].nxt[c];
				nodes[pos].cnt++;
			}
		}

		void Remove(int x) {
			int pos = 0;
			for (int i = max_log; i >= 0; i--) {
				int c = (x >> i) & 1;

				pos = nodes[pos].nxt[c];
				nodes[pos].cnt--;
			}
		}

		int findMax(int x) {
			int pos = 0, ans = 0;
			for (int i = max_log; i >= 0; i--) {
				int c = (x >> i) & 1;
				if (nodes[pos].nxt[c ^ 1] != -1 && nodes[nodes[pos].nxt[c ^ 1]].cnt > 0) {
					ans |= (1 << i);
					pos = nodes[pos].nxt[c ^ 1];
				} else
					if (nodes[pos].nxt[c] != -1) {
						pos = nodes[pos].nxt[c];
					}
			}

			return ans;
		}

	} t[2];

	void solve() {
		dfs(1);
		for (int i = 1; i <= n; i++) t[state[i]].Insert(f[i]);

		int sumXor = 0;
		while (q--) {
			char type; cin >> type;
			if (type == '^') {
				int m; cin >> m;
				sumXor ^= m;
			} else {
				int u, w; cin >> u >> w;

				t[state[u]].Remove(f[u]);
				int x = t[state[u]].findMax(f[u] ^ w);
				int y = t[state[u] ^ 1].findMax(f[u] ^ w ^ sumXor);

				t[state[u]].Insert(f[u]);
				cout << max(x, y) << ' ';
			}
		}

		cout << '\n';
	}
}

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

	#define TASK ""
	if (fopen(TASK".inp", "r")) {
		freopen(TASK".inp", "r", stdin);
		freopen(TASK".out", "w", stdout);
	}

	int numTest; cin >> numTest;

	while (numTest--) {
		cin >> n >> q;

		for (int i = 1; i <= n; i++) adj[i].clear();

		for (int i = 1; i < n; i++) {
			int u, v, w; cin >> u >> v >> w;
			adj[u].push_back({v, w}), adj[v].push_back({u, w});
		}

		Solution::solve();
	}

	return (0 ^ 0);
}

//hiuwsss

**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 bugs in the provided code, we need to focus on edge cases and scenarios that can maximize the failure probability. Here are the steps to create such a test input:

1. **Small-scale Test Case**: Ensure the test case is small enough to be manageable and easy to verify.
2. **Edge Cases**: Consider edge cases such as:
   - Trees with very few vertices.
   - Queries that involve edge weights and vertex indices that are at the extreme ends of the allowed range.
   - Queries that change the state of the tree in a way that could reveal bugs in the XOR operations.
3. **Mix of Query Types**: Include both types of queries to ensure the code handles all operations correctly.

Let's construct a small-scale test case with these considerations:

```plaintext
1
4 6
1 2 1
2 3 2
3 4 3
^ 4
? 2 5
^ 1
? 1 6
^ 7
? 4 8
```

### Explanation:
- **Test Case 1**:
  - **Tree**:
    - 4 vertices.
    - Edges: (1, 2, 1), (2, 3, 2), (3, 4, 3).
  - **Queries**:
    - `^ 4`: Apply XOR 4 to all edge weights.
    - `? 2 5`: Find the maximum XOR for vertex 2 with weight 5.
    - `^ 1`: Apply XOR 1 to all edge weights.
    - `? 1 6`: Find the maximum XOR for vertex 1 with weight 6.
    - `^ 7`: Apply XOR 7 to all edge weights.
    - `? 4 8`: Find the maximum XOR for vertex 4 with weight 8.

This test case includes a mix of edge weights, vertex indices, and XOR operations that can help expose bugs in the code. The small scale ensures it is manageable and easy to verify.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yasya and the Mysterious Tree
time_limit_ms: 2500
memory_limit_mb: 512
Description: Yasya was walking in the forest and accidentally found a tree with $n$ vertices. A tree is a connected undirected graph with no cycles.

Next to the tree, the girl found an ancient manuscript with $m$ queries written on it. The queries can be of two types.

The first type of query is described by the integer $y$. The weight of each edge in the tree is replaced by the [bitwise exclusive OR](http://tiny.cc/xor_wiki_eng) of the weight of that edge and the integer $y$.

The second type is described by the vertex $v$ and the integer $x$. Yasya chooses a vertex $u$ ($1 \le u \le n$, $u \neq v$) and mentally draws a bidirectional edge of weight $x$ from $v$ to $u$ in the tree.

Then Yasya finds a simple cycle in the resulting graph and calculates the [bitwise exclusive OR](http://tiny.cc/xor_wiki_eng) of all the edges in it. She wants to choose a vertex $u$ such that the calculated value is maximum. This calculated value will be the answer to the query. It can be shown that such a cycle exists and is unique under the given constraints (independent of the choice of $u$). If an edge between $v$ and $u$ already existed, a simple cycle is the path $v \to u \to v$.

Note that the second type of query is performed mentally, meaning the tree does not change in any way after it.

Help Yasya answer all the queries.

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

The descriptions of the test cases follow.

The first line of each test case contains two integers $n$, $m$ ($2 \le n \le 2 \cdot 10^5$, $1 \le m \le 2 \cdot 10^5$) — the number of vertices in the tree and the number of queries.

The next $n - 1$ lines of each test case contain three integers $v$, $u$, $w$ ($1 \le v, u \le n$, $1 \le w \le 10^9$) — the ends of some edge in the tree and its weight.

It is guaranteed that the given set of edges forms a tree.

The next $m$ lines of each test case describe the queries:

  * ^ $y$ ($1 \le y \le 10^9$) — parameter of the first type query;    * ? $v$ $x$ ($1 \le v \le n$, $1 \le x \le 10^9$) — parameters of the second type query.

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

Output: For each test case, output the answers to the queries of the second type.

Examples:
input:
2
3 7
1 2 1
3 1 8
^ 5
? 2 9
^ 1
? 1 10
^ 6
? 3 1
? 2 9
5 6
1 2 777
3 2 2812
4 1 16
5 3 1000000000
^ 4
? 3 123
? 5 1000000000
^ 1000000000
? 1 908070
? 2 1
output:
13 15 11 10
1000000127 2812 999756331 999999756

input:
3
8 4
8 6 3
6 3 4
2 5 4
7 6 2
7 1 10
4 1 4
5 1 2
^ 4
^ 7
? 7 8
? 4 10
5 6
3 1 4
2 3 9
4 3 6
5 2 10
? 5 7
^ 1
^ 8
? 4 10
? 1 9
? 3 6
4 2
2 1 4
4 3 5
2 3 4
^ 13
? 1 10
output:
14 13
13 8 11 11
10

Note:

Buggy Code:
# ﷽
import sys
import heapq
from heapq import *
from math import log2,ceil,floor,gcd,sqrt,log,perm,comb,factorial
from random import randrange
from bisect import bisect_left
from collections import defaultdict, Counter, deque, OrderedDict
from functools import lru_cache, reduce
from bisect import bisect_left, bisect_right
from types import GeneratorType

input = lambda: sys.stdin.buffer.readline().decode().rstrip()

class Node:
    def __init__(self, data):
        self.data = data
        self.r1 = None
        self.l1 = None
        self.r0 = None
        self.l0 = None

class Trie:
    def __init__(self):
        self.root = Node(0)

    def insert(self, preXor, parity):
        pCrawl = self.root
        if parity:
            for i in range(31, -1, -1):
                x = preXor & (1 << i)
                if x:
                    if pCrawl.r1 is None:
                        pCrawl.r1 = Node(0)
                    pCrawl = pCrawl.r1
                else:
                    if pCrawl.l1 is None:
                        pCrawl.l1 = Node(0)
                    pCrawl = pCrawl.l1
        else:
            for i in range(31, -1, -1):
                x = preXor & (1 << i)
                if x:
                    if pCrawl.r0 is None:
                        pCrawl.r0 = Node(0)
                    pCrawl = pCrawl.r0
                else:
                    if pCrawl.l0 is None:
                        pCrawl.l0 = Node(0)
                    pCrawl = pCrawl.l0

        pCrawl.data = preXor

    def query(self, xor, parity):
        pCrawl = self.root
        if parity:
            for i in range(31, -1, -1):
                x = xor & (1 << i)
                if x:
                    if pCrawl.l1 is not None:
                        pCrawl = pCrawl.l1
                    elif pCrawl.r1 is not None:
                        pCrawl = pCrawl.r1
                else:
                    if pCrawl.r1 is not None:
                        pCrawl = pCrawl.r1
                    elif pCrawl.l1 is not None:
                        pCrawl = pCrawl.l1
            return xor ^ (pCrawl.data)
        else:
            for i in range(31, -1, -1):
                x = xor & (1 << i)
                if x:
                    if pCrawl.l0 is not None:
                        pCrawl = pCrawl.l0
                    elif pCrawl.r0 is not None:
                        pCrawl = pCrawl.r0
                else:
                    if pCrawl.r0 is not None:
                        pCrawl = pCrawl.r0
                    elif pCrawl.l0 is not None:
                        pCrawl = pCrawl.l0
            return xor ^ (pCrawl.data)

    def delete(self, preXor, parity):
        def delete_helper(node, preXor, depth, parity):
            if node is None:
                return None

            if depth == -1:
                if node.data == preXor:
                    node.data = 0
                return node if node.r0 or node.l0 or node.r1 or node.l1 else None

            x = preXor & (1 << depth)
            if parity:
                if x:
                    node.r1 = delete_helper(node.r1, preXor, depth - 1, parity)
                else:
                    node.l1 = delete_helper(node.l1, preXor, depth - 1, parity)
            else:
                if x:
                    node.r0 = delete_helper(node.r0, preXor, depth - 1, parity)
                else:
                    node.l0 = delete_helper(node.l0, preXor, depth - 1, parity)

            if node.r0 or node.l0 or node.r1 or node.l1:
                return node
            else:
                return None

        delete_helper(self.root, preXor, 31, parity)

def solve():
    # n = int(input())
    n,m  = list(map(int,input().split()))
    g = [[] for i in range(n+1)]
    for i in range(n-1) :
        u,v,w = list(map(int,input().split()))
        g[u].append((v,w))
        g[v].append((u,w))
    queries = [[] for i in range(n+1)]
    cur = 0
    j = 0

    for i in range(m) :
        query = list(map(str,input().split()))
        if query[0] == "^" :
            cur ^= int(query[1])
        else:
            v,x = [int(x) for x in query[1:]]
            queries[v].append((j,x,cur))
            j += 1

    ans = [0]*j

    q = [[1,0,-1,0]]

    xors = [0]*(n+1)
    odd = [0]*(n+1)
    while q :
        node,xor,par,num = q.pop()
        xors[node] = ((xor,num))
        odd[node] = num
        for ch ,w in g[node] :
            if ch != par   :
                q.append((ch,xor^w,node,num^1))
    count = Counter(xors)
    trie = Trie()
    for i in range(1,n+1) :
        value,parity = xors[i]
        trie.insert(value,parity)

    for i in range(1,n+1) :
        for j,x,cur in queries[i] :

            current  = 0
            value,parity = xors[i]
            if count[(value,parity)] == 1  :
                trie.delete(value,parity)
            # print(j,x,cur,i,xors,value^cur^x,parity)
            current = max(current,trie.query(value^x,parity) )
            current = max(current,trie.query(value^x^cur,parity^1) )
            if count[(value,parity)] == 1  :
                trie.insert(value,parity)
            ans[j] = current

    print(*ans)

###############################################################################
###############################################################################
###############################################################################

for t in range(int(input())):
    # case(t+1)
    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 and scenarios that are likely to cause issues. Here are the steps to create such a test input:

1. **Small-Scale Test Case**: Ensure the test case is small enough to be manageable and valid.
2. **Edge Cases**:
   - **Single Edge Tree**: A tree with only one edge.
   - **Multiple Queries**: Include both types of queries.
   - **Edge Weight and Query Parameters**: Use values that can highlight the issues with bitwise operations and the Trie structure.
3. **Complex Queries**: Ensure the queries are complex enough to test the logic of the algorithm, especially the part involving the Trie and bitwise operations.

Let's construct a test case that meets these criteria:

```plaintext
1
4 4
1 2 1
2 3 2
3 4 3
^ 1
? 2 4
^ 2
? 4 1
```

### Explanation:
1. **Tree Structure**:
   - The tree has 4 vertices and 3 edges.
   - Edges: (1, 2, 1), (2, 3, 2), (3, 4, 3).

2. **Queries**:
   - **Query 1**: `^ 1` - This changes the weights of all edges using XOR with 1.
   - **Query 2**: `? 2 4` - This is a second type query that asks for the maximum XOR value for vertex 2 and edge weight 4.
   - **Query 3**: `^ 2` - This changes the weights of all edges using XOR with 2.
   - **Query 4**: `? 4 1` - This is a second type query that asks for the maximum XOR value for vertex 4 and edge weight 1.

### Expected Output:
- The expected output for the second type queries should be calculated based on the XOR operations and the structure of the Trie.

This test case is small and manageable, but it includes a mix of both types of queries and should help in exposing any bugs in the Trie and bitwise operations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yasya and the Mysterious Tree
time_limit_ms: 2500
memory_limit_mb: 512
Description: Yasya was walking in the forest and accidentally found a tree with $n$ vertices. A tree is a connected undirected graph with no cycles.

Next to the tree, the girl found an ancient manuscript with $m$ queries written on it. The queries can be of two types.

The first type of query is described by the integer $y$. The weight of each edge in the tree is replaced by the [bitwise exclusive OR](http://tiny.cc/xor_wiki_eng) of the weight of that edge and the integer $y$.

The second type is described by the vertex $v$ and the integer $x$. Yasya chooses a vertex $u$ ($1 \le u \le n$, $u \neq v$) and mentally draws a bidirectional edge of weight $x$ from $v$ to $u$ in the tree.

Then Yasya finds a simple cycle in the resulting graph and calculates the [bitwise exclusive OR](http://tiny.cc/xor_wiki_eng) of all the edges in it. She wants to choose a vertex $u$ such that the calculated value is maximum. This calculated value will be the answer to the query. It can be shown that such a cycle exists and is unique under the given constraints (independent of the choice of $u$). If an edge between $v$ and $u$ already existed, a simple cycle is the path $v \to u \to v$.

Note that the second type of query is performed mentally, meaning the tree does not change in any way after it.

Help Yasya answer all the queries.

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

The descriptions of the test cases follow.

The first line of each test case contains two integers $n$, $m$ ($2 \le n \le 2 \cdot 10^5$, $1 \le m \le 2 \cdot 10^5$) — the number of vertices in the tree and the number of queries.

The next $n - 1$ lines of each test case contain three integers $v$, $u$, $w$ ($1 \le v, u \le n$, $1 \le w \le 10^9$) — the ends of some edge in the tree and its weight.

It is guaranteed that the given set of edges forms a tree.

The next $m$ lines of each test case describe the queries:

  * ^ $y$ ($1 \le y \le 10^9$) — parameter of the first type query;    * ? $v$ $x$ ($1 \le v \le n$, $1 \le x \le 10^9$) — parameters of the second type query.

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

Output: For each test case, output the answers to the queries of the second type.

Examples:
input:
2
3 7
1 2 1
3 1 8
^ 5
? 2 9
^ 1
? 1 10
^ 6
? 3 1
? 2 9
5 6
1 2 777
3 2 2812
4 1 16
5 3 1000000000
^ 4
? 3 123
? 5 1000000000
^ 1000000000
? 1 908070
? 2 1
output:
13 15 11 10
1000000127 2812 999756331 999999756

input:
3
8 4
8 6 3
6 3 4
2 5 4
7 6 2
7 1 10
4 1 4
5 1 2
^ 4
^ 7
? 7 8
? 4 10
5 6
3 1 4
2 3 9
4 3 6
5 2 10
? 5 7
^ 1
^ 8
? 4 10
? 1 9
? 3 6
4 2
2 1 4
4 3 5
2 3 4
^ 13
? 1 10
output:
14 13
13 8 11 11
10

Note:

Buggy Code:
# ﷽
import sys
import heapq
from heapq import *
from math import log2,ceil,floor,gcd,sqrt,log,perm,comb,factorial
from random import randrange
from bisect import bisect_left
from collections import defaultdict, Counter, deque, OrderedDict
from functools import lru_cache, reduce
from bisect import bisect_left, bisect_right
from types import GeneratorType

input = lambda: sys.stdin.buffer.readline().decode().rstrip()

class Node:
    def __init__(self):
        self.children = [None, None, None, None]  # [r1, l1, r0, l0]
        self.isEndOfWord = False
        self.data = 0

class Trie:
    def __init__(self):
        self.root = Node()

    def insert(self, preXor, parity):
        pCrawl = self.root
        for i in range(31, -1, -1):
            x = (preXor >> i) & 1
            index = x + 2 if parity else x
            if not pCrawl.children[index]:
                pCrawl.children[index] = Node()
            pCrawl = pCrawl.children[index]
        pCrawl.isEndOfWord = True
        pCrawl.data = preXor

    def query(self, xor, parity):
        pCrawl = self.root
        for i in range(31, -1, -1):
            x = (xor >> i) & 1
            index = 1 - x + 2 if parity else 1 - x
            if pCrawl.children[index]:
                pCrawl = pCrawl.children[index]
            else:
                index = x + 2 if parity else x
                if pCrawl.children[index]:
                    pCrawl = pCrawl.children[index]
                else:
                    break  # if no path exists, break out of the loop
        return xor ^ pCrawl.data

    def isEmpty(self, root):
        for child in root.children:
            if child:
                return False
        return True

    def remove(self, root, preXor, depth=31, parity=False):
        if not root:
            return None

        if depth == -1:
            if root.isEndOfWord:
                root.isEndOfWord = False
            if self.isEmpty(root):
                del root
                root = None
            return root

        x = (preXor >> depth) & 1
        index = x + 2 if parity else x
        root.children[index] = self.remove(root.children[index], preXor, depth - 1, parity)

        if self.isEmpty(root) and not root.isEndOfWord:
            del root
            root = None
        return root

    def delete(self, preXor, parity):
        self.remove(self.root, preXor, 31, parity)

from random import shuffle

def solve():
    # n = int(input())
    n,m  = list(map(int,input().split()))
    g = [[] for i in range(n+1)]
    for i in range(n-1) :
        u,v,w = list(map(int,input().split()))
        g[u].append((v,w))
        g[v].append((u,w))
    queries = [[] for i in range(n+1)]
    cur = 0
    j = 0

    for i in range(m) :
        query = list(map(str,input().split()))
        if query[0] == "^" :
            cur ^= int(query[1])
        else:
            v,x = [int(x) for x in query[1:]]
            queries[v].append((j,x,cur))
            j += 1

    ans = [0]*j

    q = [[1,0,-1,0]]

    xors = [0]*(n+1)
    odd = [0]*(n+1)
    while q :
        node,xor,par,num = q.pop()
        xors[node] = ((xor,num))
        odd[node] = num
        for ch ,w in g[node] :
            if ch != par   :
                q.append((ch,xor^w,node,num^1))
    count = Counter(xors)
    trie = Trie()
    for i in range(1,n+1) :
        value,parity = xors[i]
        trie.insert(value,parity)

    for i in range(1,n+1) :
        # shuffle(queries[i])
        for j,x,cur in queries[i] :

            current  = 0
            value,parity = xors[i]
            if count[(value,parity)] == 1  :
                trie.delete(value,parity)
            # print(j,x,cur,i,xors,value^cur^x,parity)
            current = max(current,trie.query(value^x,parity) )
            current = max(current,trie.query(value^x^cur,parity^1) )
            if count[(value,parity)] == 1  :
                trie.insert(value,parity)
            ans[j] = current

    print(*ans)

###############################################################################
###############################################################################
###############################################################################

for t in range(int(input())):
    # case(t+1)
    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 code, we need to focus on edge cases and scenarios that can cause issues. Here are the steps to think through:

1. **Tree Structure**: Create a tree with a structure that can cause issues in the traversal or XOR calculations.
2. **Query Types**: Include both types of queries in a way that can highlight bugs.
3. **Edge Cases**: Consider edge cases such as:
   - A tree with only two nodes.
   - A tree with many nodes.
   - Queries that involve the root node.
   - Queries that involve leaf nodes.
   - Queries that involve the same node multiple times.
4. **XOR Operations**: Use XOR operations that can lead to unexpected results, especially with large numbers.

Given these considerations, let's create a small-scale test input:

```plaintext
1
4 5
1 2 1
2 3 2
3 4 3
^ 1
? 1 1
? 2 2
? 3 3
? 4 4
```

### Explanation:

- **Tree Structure**:
  - The tree has 4 nodes and 3 edges.
  - The edges are: (1, 2, 1), (2, 3, 2), (3, 4, 3).

- **Queries**:
  - The first query is a type 1 query: `^ 1`. This will XOR all edge weights with 1.
  - The next four queries are type 2 queries:
    - `? 1 1`: Calculate the maximum XOR for vertex 1 with weight 1.
    - `? 2 2`: Calculate the maximum XOR for vertex 2 with weight 2.
    - `? 3 3`: Calculate the maximum XOR for vertex 3 with weight 3.
    - `? 4 4`: Calculate the maximum XOR for vertex 4 with weight 4.

This test input is designed to:
- Test the XOR operation on all edges.
- Test the queries on different vertices, including the root and leaf nodes.
- Ensure that the tree is small enough to be manageable but complex enough to expose any bugs in the code.

Let's see how the code handles this input.