**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cases
time_limit_ms: 2000
memory_limit_mb: 256
Description: You're a linguist studying a mysterious ancient language. You know that

  1. Its words consist only of the first $c$ letters of the Latin alphabet.    2. Each word has a case which can be unambiguously determined by its last letter (different letters correspond to different cases). For example, words "ABACABA" and "ABA" (if they exist) have the same case in this language because they both have the same ending 'A', whereas "ALICE" and "BOB" have different cases. If the language does not have a case corresponding to some letter, it means that the word cannot end with this letter.    3. The length of each word is $k$ or less.

You have a single text written in this language. Unfortunately, as the language is really ancient, spaces between words are missing and all letters are uppercase. You wonder what is the minimum number of cases the language can have. Can you find this out?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 10\,000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains three integers $n$, $c$, $k$ ($1 \le k \le n \le 2^{18}$, $1 \le c \le 18$) — the length of the text, the number of letters in the language, and the maximum length of the word.

The second line contains a string of $n$ characters — the text itself. Each character is one of the first $c$ uppercase letters of the Latin alphabet.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2^{18}$ and the sum of $2^c$ over all test cases does not exceed $2^{18}$.

Output: For each test case, output a single line consisting of a single integer — the minimum number of cases in the language.

Examples:
input:
7
5 5 1
ABCDE
3 1 2
AAA
3 2 2
AAB
10 2 2
ABABABABAB
4 4 4
DCBA
1 17 1
Q
9 3 2
ABCABCABC
output:
5
1
2
1
1
1
2

Note: In the first test case, there must be five cases in the language (for each of the letters 'A', 'B', 'C', 'D', and 'E' there must be a case that has a corresponding ending).

In the fourth test case, one case with ending 'B' is sufficient.

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

using namespace std;

const int N=1e6+10;

int n,m,k;
bool f[N];
char s[N];

int main() {
	int T;
	cin >> T;
	while(T--) {
		cin >> n >> m >> k;
		for(int i=1; i<=n; i++) cin >> s[i];
		for(int i=0; i<1<<m; i++) f[i]=0;
		map<char,int> M;
		int num=-1;
		for(int i=1; i<=n; i++)
			if(!M.count(s[i])) M[s[i]]=++num;
		for(int i=1; i+k-1<=n; i++) {
			int cnt=0;
			for(int j=i; j<=i+k-1; j++) cnt|=1<<M[s[j]];
			f[(1<<m)-cnt-1]=1;
		}
		for(int i=0; i<m; i++)
			for(int j=0; j<1<<m; j++)
				if(!(j>>i&1)) f[j]|=f[j+(1<<i)];
		f[(1<<m)-(1<<M[s[n]])-1]=1;
		int ans=1e9;
		for(int i=0; i<1<<m; i++) {
			int cnt=0;
			for(int j=0; j<m; j++) if(i>>j&1) cnt++;
			if(!f[i]) ans=min(ans,cnt);
		}
		cout << ans << endl;
	}
	return 0;
}

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided buggy code, we need to focus on edge cases that might cause the algorithm to fail. Here are some considerations:

1. **Single Character Test Case**: A test case with a single character can help check if the algorithm correctly handles the minimum possible input.
2. **Repeating Characters**: A test case with a string of repeating characters can help check if the algorithm correctly identifies that only one case is needed.
3. **Multiple Unique Characters**: A test case with multiple unique characters can help check if the algorithm correctly identifies the number of cases needed.
4. **Maximum Word Length**: A test case where the word length is exactly the maximum allowed can help check if the algorithm correctly processes the input.
5. **Edge Case for `m` and `k`**: A test case where `m` and `k` are at their maximum values can help check if the algorithm handles large inputs correctly.

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

```plaintext
5
1 1 1
A
3 1 2
AAA
3 2 2
AAB
10 2 2
ABABABABAB
10 3 3
ABCABCABCABC
```

This test input includes:
1. A single character test case.
2. A test case with repeating characters.
3. A test case with multiple unique characters.
4. A test case where the word length is exactly the maximum allowed.
5. A test case with a larger `m` and `k` to check for edge cases.

This should help expose any issues in the buggy code.

5
1 1 1
A
3 1 2
AAA
3 2 2
AAB
10 2 2
ABABABABAB
10 3 3
ABCABCABCABC


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cases
time_limit_ms: 2000
memory_limit_mb: 256
Description: You're a linguist studying a mysterious ancient language. You know that

  1. Its words consist only of the first $c$ letters of the Latin alphabet.    2. Each word has a case which can be unambiguously determined by its last letter (different letters correspond to different cases). For example, words "ABACABA" and "ABA" (if they exist) have the same case in this language because they both have the same ending 'A', whereas "ALICE" and "BOB" have different cases. If the language does not have a case corresponding to some letter, it means that the word cannot end with this letter.    3. The length of each word is $k$ or less.

You have a single text written in this language. Unfortunately, as the language is really ancient, spaces between words are missing and all letters are uppercase. You wonder what is the minimum number of cases the language can have. Can you find this out?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 10\,000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains three integers $n$, $c$, $k$ ($1 \le k \le n \le 2^{18}$, $1 \le c \le 18$) — the length of the text, the number of letters in the language, and the maximum length of the word.

The second line contains a string of $n$ characters — the text itself. Each character is one of the first $c$ uppercase letters of the Latin alphabet.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2^{18}$ and the sum of $2^c$ over all test cases does not exceed $2^{18}$.

Output: For each test case, output a single line consisting of a single integer — the minimum number of cases in the language.

Examples:
input:
7
5 5 1
ABCDE
3 1 2
AAA
3 2 2
AAB
10 2 2
ABABABABAB
4 4 4
DCBA
1 17 1
Q
9 3 2
ABCABCABC
output:
5
1
2
1
1
1
2

Note: In the first test case, there must be five cases in the language (for each of the letters 'A', 'B', 'C', 'D', and 'E' there must be a case that has a corresponding ending).

In the fourth test case, one case with ending 'B' is sufficient.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'

//debug
void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}
template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifndef ONLINE_JUDGE
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif
#ifndef ONLINE_JUDGE
#define debarr(arr, n) cerr << "arr: ["; forn(i, n) cerr << arr[i] << (i == n-1 ? "" : ", "); cerr << "]" << endl;
#else
#define debarr(x...)
#endif

// -------------------- = -----------------------

typedef long long ll;
using ld = long double;

#define bg(x) begin(x)
#define all(x) bg(x), end(x)
#define rall(x) x.rbegin(), x.rend()
#define sor(x) sort(all(x))
#define rev(x) reverse(all(x))

#define forn(i, n) for (int i = 0; i < n; i++)
#define fora(i, a, b) for (int i = a; i <= b; i++)
#define rofn(i, n) for (int i = n-1; i >= 0; i--)
#define rofa(i, a, b) for (int i = a; i >= b; i--)

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
        // http://xorshift.di.unimi.it/splitmix64.c
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }

    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};

// ====================== end of template ===========================

void solve() {
    int n, c, k; cin >> n >> c >> k;
    string s; cin >> s;

    vector<vector<int>> pref(c, vector<int>(n + 1));
    for (int j = 0; j < c; j++) {
        for (int i = 0; i < n; i++) {
            pref[j][i+1] += pref[j][i] + (s[i]-'A' == j);
        }
    }

    vector<int> dp(1<<18);
    for (int i = 1; i <= n; i++) {
        if (i - k >= 0) {
            int m = 0;
            for (int j = 0; j < c; j++) if (pref[j][i] - pref[j][i - k] > 0) m |= 1<<j;
            dp[(~m) & ((1<<c) - 1)] = 1;
        }
    }

    int ans = INT_MAX;
    for (int m = (1<<c) - 1; m >= 0; m--) {
        for (int j = 0; j < c; j++) {
            dp[m] |= dp[m | (1<<j)];
        }
        int b = 1<<(s.back()-'A');
        dp[m] |= ((m&b) ^ b) > 0;
        if (!dp[m]) ans = min(ans, __builtin_popcount(m));
    }

    cout << ans << endl;
}

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

    int tt; cin >> tt; while (tt--) 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 create a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that might cause the code to fail. Here are some key points to consider:

1. **Small Test Cases**: Since we need a small-scale test input, we can start with a single test case or a few test cases with small values of \( n \), \( c \), and \( k \).

2. **Edge Cases**:
   - **Single Character Text**: A text with a single character.
   - **All Characters the Same**: A text where all characters are the same.
   - **All Characters Different**: A text where all characters are different.
   - **Maximum Word Length**: A text where the maximum word length \( k \) is equal to the length of the text \( n \).

3. **Complex Cases**:
   - **Repeating Patterns**: A text with repeating patterns that might cause the prefix sum array to behave unexpectedly.
   - **Boundary Conditions**: Test cases where \( n \), \( c \), and \( k \) are at their boundary values (e.g., \( n = 1 \), \( c = 1 \), \( k = 1 \)).

Given these considerations, let's create a test input that includes a mix of these cases:

```plaintext
5
1 1 1
A
3 1 3
AAA
3 3 3
ABC
4 2 2
ABBA
6 3 3
ABCCBA
```

Explanation:
1. **Test Case 1**: \( n = 1 \), \( c = 1 \), \( k = 1 \), Text = "A" - Single character text.
2. **Test Case 2**: \( n = 3 \), \( c = 1 \), \( k = 3 \), Text = "AAA" - All characters the same.
3. **Test Case 3**: \( n = 3 \), \( c = 3 \), \( k = 3 \), Text = "ABC" - All characters different.
4. **Test Case 4**: \( n = 4 \), \( c = 2 \), \( k = 2 \), Text = "ABBA" - Repeating pattern.
5. **Test Case 5**: \( n = 6 \), \( c = 3 \), \( k = 3 \), Text = "ABCCBA" - More complex repeating pattern.

This test input should help expose any bugs in the provided code by covering a range of edge cases and complex scenarios.

5
1 1 1
A
3 1 3
AAA
3 3 3
ABC
4 2 2
ABBA
6 3 3
ABCCBA


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cases
time_limit_ms: 2000
memory_limit_mb: 256
Description: You're a linguist studying a mysterious ancient language. You know that

  1. Its words consist only of the first $c$ letters of the Latin alphabet.    2. Each word has a case which can be unambiguously determined by its last letter (different letters correspond to different cases). For example, words "ABACABA" and "ABA" (if they exist) have the same case in this language because they both have the same ending 'A', whereas "ALICE" and "BOB" have different cases. If the language does not have a case corresponding to some letter, it means that the word cannot end with this letter.    3. The length of each word is $k$ or less.

You have a single text written in this language. Unfortunately, as the language is really ancient, spaces between words are missing and all letters are uppercase. You wonder what is the minimum number of cases the language can have. Can you find this out?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 10\,000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains three integers $n$, $c$, $k$ ($1 \le k \le n \le 2^{18}$, $1 \le c \le 18$) — the length of the text, the number of letters in the language, and the maximum length of the word.

The second line contains a string of $n$ characters — the text itself. Each character is one of the first $c$ uppercase letters of the Latin alphabet.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2^{18}$ and the sum of $2^c$ over all test cases does not exceed $2^{18}$.

Output: For each test case, output a single line consisting of a single integer — the minimum number of cases in the language.

Examples:
input:
7
5 5 1
ABCDE
3 1 2
AAA
3 2 2
AAB
10 2 2
ABABABABAB
4 4 4
DCBA
1 17 1
Q
9 3 2
ABCABCABC
output:
5
1
2
1
1
1
2

Note: In the first test case, there must be five cases in the language (for each of the letters 'A', 'B', 'C', 'D', and 'E' there must be a case that has a corresponding ending).

In the fourth test case, one case with ending 'B' is sufficient.

Buggy Code:
def binary_search(sorted_list, target):
    left, right = 0, len(sorted_list) - 1

    while left <= right:
        mid = (left + right) // 2
        mid_letter = sorted_list[mid]

        if mid_letter == target:
            return True
        elif mid_letter < target:
            left = mid + 1
        else:
            right = mid - 1

    return False

def possible(X,text,k):
    if(text[-1] not in X):
        return False
    Y = sorted(X)
    dist = 1
    for i in text:
        if binary_search(Y,i):
            dist = 1
        else:
            dist += 1
        if dist > k:
            return False
    return True

def solve():
    s = list(map(int, input().split()))
    n = s[0]
    c = s[1]
    k = s[2]
    text = list(map(lambda x: ord(x)-ord('A'), list(input())))

    if k == 1:
        print(len(set(text)))
        return

    ans = [text[-1]]
    r = int(n/k)
    counter = 0
    found = False
    for i in range(n):
        if text[i] in ans:
            counter = 0
        else:
            counter += 1
        if counter == k:
            ans.append(text[i])
    print(len(ans))

t = int(input())
while t:
    solve()
    t -= 1

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

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

Think step by step.

Response:
```plaintext
1
6 3 2
AABABA
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cases
time_limit_ms: 2000
memory_limit_mb: 256
Description: You're a linguist studying a mysterious ancient language. You know that

  1. Its words consist only of the first $c$ letters of the Latin alphabet.    2. Each word has a case which can be unambiguously determined by its last letter (different letters correspond to different cases). For example, words "ABACABA" and "ABA" (if they exist) have the same case in this language because they both have the same ending 'A', whereas "ALICE" and "BOB" have different cases. If the language does not have a case corresponding to some letter, it means that the word cannot end with this letter.    3. The length of each word is $k$ or less.

You have a single text written in this language. Unfortunately, as the language is really ancient, spaces between words are missing and all letters are uppercase. You wonder what is the minimum number of cases the language can have. Can you find this out?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 10\,000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains three integers $n$, $c$, $k$ ($1 \le k \le n \le 2^{18}$, $1 \le c \le 18$) — the length of the text, the number of letters in the language, and the maximum length of the word.

The second line contains a string of $n$ characters — the text itself. Each character is one of the first $c$ uppercase letters of the Latin alphabet.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2^{18}$ and the sum of $2^c$ over all test cases does not exceed $2^{18}$.

Output: For each test case, output a single line consisting of a single integer — the minimum number of cases in the language.

Examples:
input:
7
5 5 1
ABCDE
3 1 2
AAA
3 2 2
AAB
10 2 2
ABABABABAB
4 4 4
DCBA
1 17 1
Q
9 3 2
ABCABCABC
output:
5
1
2
1
1
1
2

Note: In the first test case, there must be five cases in the language (for each of the letters 'A', 'B', 'C', 'D', and 'E' there must be a case that has a corresponding ending).

In the fourth test case, one case with ending 'B' is sufficient.

Buggy Code:
import math as m
w = int(input())

for _ in range(w):
    # 9 3 2
    # ABCABCABC

    # 8 3 2
    # BBBBBABC

    n, c, k = list(map(int, input().split()))
    WORD = input()
    DP = [(20, []) for i in range(n + 1)]
    DP[n] = (0, [set()])

    for i in range(n-1, -1, -1):
        # print(DP)
        for gap in range(0,k):
            if i + gap < n:
                pfx_end = i + gap
                sfx_start = i + gap + 1

                for wordset in DP[sfx_start][1]:
                    if WORD[pfx_end] in wordset:
                        if DP[i][0] > DP[sfx_start][0]:
                            DP[i] = (DP[sfx_start][0], [wordset.union(set(WORD[pfx_end])), ])
                    else:
                        # print("triggered")
                        if DP[i][0] > DP[sfx_start][0] + 1:
                            DP[i] = (DP[sfx_start][0] + 1, [wordset.union(set(WORD[pfx_end])), ])
                        elif DP[i][0] == DP[sfx_start][0] + 1:
                            DP[i][1].append(wordset.union(set(WORD[pfx_end])))
    print(DP[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
3
5 5 1
ABCDE
10 2 2
ABABABABAB
9 3 2
ABCABCABC
```

3
5 5 1
ABCDE
10 2 2
ABABABABAB
9 3 2
ABCABCABC


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cases
time_limit_ms: 2000
memory_limit_mb: 256
Description: You're a linguist studying a mysterious ancient language. You know that

  1. Its words consist only of the first $c$ letters of the Latin alphabet.    2. Each word has a case which can be unambiguously determined by its last letter (different letters correspond to different cases). For example, words "ABACABA" and "ABA" (if they exist) have the same case in this language because they both have the same ending 'A', whereas "ALICE" and "BOB" have different cases. If the language does not have a case corresponding to some letter, it means that the word cannot end with this letter.    3. The length of each word is $k$ or less.

You have a single text written in this language. Unfortunately, as the language is really ancient, spaces between words are missing and all letters are uppercase. You wonder what is the minimum number of cases the language can have. Can you find this out?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 10\,000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains three integers $n$, $c$, $k$ ($1 \le k \le n \le 2^{18}$, $1 \le c \le 18$) — the length of the text, the number of letters in the language, and the maximum length of the word.

The second line contains a string of $n$ characters — the text itself. Each character is one of the first $c$ uppercase letters of the Latin alphabet.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2^{18}$ and the sum of $2^c$ over all test cases does not exceed $2^{18}$.

Output: For each test case, output a single line consisting of a single integer — the minimum number of cases in the language.

Examples:
input:
7
5 5 1
ABCDE
3 1 2
AAA
3 2 2
AAB
10 2 2
ABABABABAB
4 4 4
DCBA
1 17 1
Q
9 3 2
ABCABCABC
output:
5
1
2
1
1
1
2

Note: In the first test case, there must be five cases in the language (for each of the letters 'A', 'B', 'C', 'D', and 'E' there must be a case that has a corresponding ending).

In the fourth test case, one case with ending 'B' is sufficient.

Buggy Code:
T = int(input())

for t in range(T):
    N, C, K = map(int, input().split())
    s = input()

    """
    observations:
    last letter must be a case

    idea:
    divide into chunks split by last string
    find min types of characters in a string of length K-1

    """

    freq = [0 for x in range(C)]
    appear = [False for x in range(C)]
    last = ord(s[-1]) - ord('A')

    for i in range(K-2):
        ch = ord(s[i]) - ord('A')
        freq[ch] += 1
        appear[ch] = True

    if K == 1:
        for i in range(N):
            ch = ord(s[i]) - ord('A')
            appear[ch] = True
        ans = 0
        for i in range(C):
            if appear[i]:
                ans += 1
        print(ans)
        continue

    best = 1
    for i in range(K-2, N):
        ch = ord(s[i]) - ord('A')
        appear[ch] = True
        freq[ch] += 1
        if freq[last] == 0:
            num = 0
            if i-K+1 >= 0 and freq[ord(s[i-K+1])-ord('A')] > 0:
                num -= 1 # if head is in the substring
            if i+1 < N-1 and freq[ord(s[i+1])-ord('A')] > 0:
                num -= 1 # if tail is in the substring
                if i-K+1 >= 0 and s[i-K+1] == s[i+1]:
                    num += 1
            for j in range(C):
                if freq[j] > 0:
                    num += 1
            best = max(best, C-num)
        # print(i-K+2)
        ch = ord(s[i-K+2])-ord('A')
        freq[ch] -= 1

    total = 0
    for i in range(C):
        if appear[i]:
            total += 1

    best = max(1, best - (C-total))
    print(best)

**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
4 2 2
ABAB
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cases
time_limit_ms: 2000
memory_limit_mb: 256
Description: You're a linguist studying a mysterious ancient language. You know that

  1. Its words consist only of the first $c$ letters of the Latin alphabet.    2. Each word has a case which can be unambiguously determined by its last letter (different letters correspond to different cases). For example, words "ABACABA" and "ABA" (if they exist) have the same case in this language because they both have the same ending 'A', whereas "ALICE" and "BOB" have different cases. If the language does not have a case corresponding to some letter, it means that the word cannot end with this letter.    3. The length of each word is $k$ or less.

You have a single text written in this language. Unfortunately, as the language is really ancient, spaces between words are missing and all letters are uppercase. You wonder what is the minimum number of cases the language can have. Can you find this out?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 10\,000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains three integers $n$, $c$, $k$ ($1 \le k \le n \le 2^{18}$, $1 \le c \le 18$) — the length of the text, the number of letters in the language, and the maximum length of the word.

The second line contains a string of $n$ characters — the text itself. Each character is one of the first $c$ uppercase letters of the Latin alphabet.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2^{18}$ and the sum of $2^c$ over all test cases does not exceed $2^{18}$.

Output: For each test case, output a single line consisting of a single integer — the minimum number of cases in the language.

Examples:
input:
7
5 5 1
ABCDE
3 1 2
AAA
3 2 2
AAB
10 2 2
ABABABABAB
4 4 4
DCBA
1 17 1
Q
9 3 2
ABCABCABC
output:
5
1
2
1
1
1
2

Note: In the first test case, there must be five cases in the language (for each of the letters 'A', 'B', 'C', 'D', and 'E' there must be a case that has a corresponding ending).

In the fourth test case, one case with ending 'B' is sufficient.

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

public class Main {
boolean MULTI_CASE = true;
boolean ALWAYS_FLUSH = false;

void go() {
    int n = nextInt();
    int c = nextInt();
    int len = nextInt();
    int[] a = next().chars().map(x -> x - 'A').toArray();
    int[] g = new int[c];
    int[] b = new int[1 << c];
    for (int i = 0; i < n; i++) {
        g[a[i]]++;
        if (i - len >= 0) {
            g[a[i - len]]--;
        }
        if (i >= len - 1) {
            int x = 0;
            for (int j = 0; j < c; j++) {
                if (g[j] > 0) x |= 1 << j;
            }
            b[x] = 1;
        }
    }
    int[] v = new int[1 << c];
    int vn = 0;
    for (int i = 0; i < (1 << c); i++) {
        if (b[i] > 0) v[vn++] = i;
    }
    v = Arrays.copyOf(v, vn);
    Integer[] w = new Integer[vn];
    for (int i = 0; i < vn; i++) w[i] = v[i];
    Arrays.sort(w, (x, y) -> Integer.compare(Integer.bitCount(x), Integer.bitCount(y)));
    int[] pos = new int[1 << c];
//    ruffleSort(v);
    int ret = c;
    for (int i = 1; i < (1 << c); i++) {
        if ((i & (1 << a[n - 1])) == 0) continue;
        boolean ok = true;
        int pp = 0;
        for (int j = 0; j < c; j++) {
            if ((i & (1 << j)) > 0) {
                pp = Math.max(pp, pos[i ^ (1 << j)]);
            }
        }
        for (int j = pp; j < vn; j++) {
            if ((i & w[j]) == 0) {
                pos[i] = j;
                ok = false;
                break;
            }
        }
        if (ok) {
            ret = Math.min(ret, Integer.bitCount(i));
        }
    }
    sl(ret);
}

void ruffleSort(int[] a) {
    Random random = new Random();
    int n = a.length;
    for (int i = 0; i < n; i++) {
        int oi = random.nextInt(n);
        int temp = a[oi];
        a[oi] = a[i];
        a[i] = temp;
    }
    Arrays.sort(a);
}

/**
 * Input.
 */
InputStream inStream;
byte[] inBuff = new byte[1024];
int inBuffCursor = 0, inBuffLen = 0;

boolean isVisibleChar(int c) {
    return 33 <= c && c <= 126;
}

boolean hasNextByte() {
    if (inBuffCursor < inBuffLen) return true;
    inBuffCursor = 0;
    try {
        inBuffLen = inStream.read(inBuff);
    } catch (Exception e) {
        e.printStackTrace(System.out);
    }
    return inBuffLen > 0;
}

boolean hasNext() {
    while (hasNextByte() && !isVisibleChar(inBuff[inBuffCursor])) inBuffCursor++;
    return hasNextByte();
}

int nextByte() {
    return hasNextByte() ? inBuff[inBuffCursor++] : -1;
}

String next() {
    if (!hasNext()) throw new RuntimeException("no next.");
    StringBuilder sb = new StringBuilder();
    int b = nextByte();
    while (isVisibleChar(b)) {
        sb.appendCodePoint(b);
        b = nextByte();
    }
    return sb.toString();
}

long nextLong() {
    if (!hasNext()) throw new RuntimeException("no next.");
    long result = 0;
    boolean negative = false;
    int b = nextByte();
    if (b < '0') {
        if (b == '-') negative = true;
        else if (b != '+') throw new RuntimeException("long number must start with +/-.");
        b = nextByte();
    }
    while (isVisibleChar(b)) {
        if (b < '0' || b > '9') throw new RuntimeException("wrong digit in long:" + (char) b);
        // NOTE: may overflow here, even if it is a valid Long, eg.-(1<<63).
        result = result * 10 + (b - '0');
        b = nextByte();
    }
    return negative ? -result : result;
}

int nextInt() {
    long x = nextLong();
    if (x < Integer.MIN_VALUE || x > Integer.MAX_VALUE)
        throw new RuntimeException("int overflow:" + x);
    return (int) x;
}

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

/**
 * Output.
 */
PrintWriter printOut = new PrintWriter(System.out);

void so(Object obj, boolean newLine) {
    if (newLine) printOut.println(obj);
    else printOut.print(obj);
    if (ALWAYS_FLUSH) printOut.flush();
}

void so(Object obj) {
    so(obj, false);
}

void sl(Object obj) {
    so(obj, true);
}

void sl() {
    sl("");
}

/**
 * Main.
 */
void mainGo() {
    try {
        inStream = new FileInputStream("src/main.in");
        ALWAYS_FLUSH = true;
    } catch (Exception e) {
        inStream = System.in;
    }
    while (hasNext()) {
        if (MULTI_CASE) {
            int T = nextInt();
            if (T == 0) break;
            for (int i = 0; i < T; ++i) {
                go();
            }
        } else {
            go();
        }
    }
    printOut.flush();
}

public static void main(String[] args) throws Exception {
    new Main().mainGo();
}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, let's consider the following steps:

1. **Understand the Bug**: The buggy code has issues with handling the cases and the bit manipulation. Specifically, it may not correctly handle the cases where the same letter can appear multiple times in a word or where the word length is greater than the maximum allowed length.

2. **Edge Cases**:
   - **Single Letter Words**: Test cases where the text consists of a single letter repeated.
   - **Maximum Length Words**: Test cases where the words are exactly the maximum length.
   - **Mixed Length Words**: Test cases where the words have varying lengths.
   - **Multiple Cases**: Test cases where the text contains multiple different letters.

3. **Input Constraints**:
   - The number of test cases should be within the limit of 10,000.
   - The length of the text should be within the limit of \(2^{18}\).
   - The number of letters in the language should be within the limit of 18.
   - The maximum length of the word should be within the limit of the length of the text.

4. **Generate a Challenging Test Input**:
   - We will create a test case where the text consists of a mix of letters and the words have varying lengths, including the maximum length.
   - We will ensure that the text contains all possible letters to maximize the number of cases.

Here is a small-scale test input that should challenge the buggy code:

```plaintext
3
5 5 1
ABCDE
6 3 3
ABCABC
8 4 4
ABCDABCD
```

Explanation:
- **Test Case 1**:
  - \(n = 5\), \(c = 5\), \(k = 1\)
  - Text: "ABCDE"
  - Each letter is a different case, so the minimum number of cases is 5.
- **Test Case 2**:
  - \(n = 6\), \(c = 3\), \(k = 3\)
  - Text: "ABCABC"
  - The text contains all three letters, and each letter can be the end of a word. The minimum number of cases is 3.
- **Test Case 3**:
  - \(n = 8\), \(c = 4\), \(k = 4\)
  - Text: "ABCDABCD"
  - The text contains all four letters, and each letter can be the end of a word. The minimum number of cases is 4.

This test input should expose the bug in the code by ensuring that the algorithm correctly handles different lengths and multiple cases.

3
5 5 1
ABCDE
6 3 3
ABCABC
8 4 4
ABCDABCD


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cases
time_limit_ms: 2000
memory_limit_mb: 256
Description: You're a linguist studying a mysterious ancient language. You know that

  1. Its words consist only of the first $c$ letters of the Latin alphabet.    2. Each word has a case which can be unambiguously determined by its last letter (different letters correspond to different cases). For example, words "ABACABA" and "ABA" (if they exist) have the same case in this language because they both have the same ending 'A', whereas "ALICE" and "BOB" have different cases. If the language does not have a case corresponding to some letter, it means that the word cannot end with this letter.    3. The length of each word is $k$ or less.

You have a single text written in this language. Unfortunately, as the language is really ancient, spaces between words are missing and all letters are uppercase. You wonder what is the minimum number of cases the language can have. Can you find this out?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 10\,000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains three integers $n$, $c$, $k$ ($1 \le k \le n \le 2^{18}$, $1 \le c \le 18$) — the length of the text, the number of letters in the language, and the maximum length of the word.

The second line contains a string of $n$ characters — the text itself. Each character is one of the first $c$ uppercase letters of the Latin alphabet.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2^{18}$ and the sum of $2^c$ over all test cases does not exceed $2^{18}$.

Output: For each test case, output a single line consisting of a single integer — the minimum number of cases in the language.

Examples:
input:
7
5 5 1
ABCDE
3 1 2
AAA
3 2 2
AAB
10 2 2
ABABABABAB
4 4 4
DCBA
1 17 1
Q
9 3 2
ABCABCABC
output:
5
1
2
1
1
1
2

Note: In the first test case, there must be five cases in the language (for each of the letters 'A', 'B', 'C', 'D', and 'E' there must be a case that has a corresponding ending).

In the fourth test case, one case with ending 'B' is sufficient.

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

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
//	#pragma GCC optimize("O3,unroll-loops")
//	#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
template <typename T> using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

#define vt vector
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef vector<int> vi;

template<class T> bool chmin(T& a, const T& b) {
	return b<a?a=b, 1:0;
}
template<class T> bool chmax(T& a, const T& b) {
	return a<b?a=b, 1:0;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng_64(chrono::steady_clock::now().time_since_epoch().count());

const long long B = 11;

void solve() {
	int n, k, c;
	cin >> n >> c >> k;
	string s;
	cin >> s;
	for(int i = 0;i<n;i++)s[i]-='A';
	if(k < B) {
		vt<vt<int>> pref(n+1, vt<int>(c, 0));
		for(int i = 1;i<=n;i++) {
			pref[i][s[i-1]]++;
		}
		for(int i = 1;i<=n;i++) {
			for(int j = 0;j<c;j++) {
				pref[i][j] = pref[i-1][j] + pref[i][j];
			}
		}
		set<int> needed_set;
		for(int i = 0;i+k<=n;i++) {
			int mask=0;
			for(int j = 0;j<c;j++) {
				if(pref[i+k][j]-pref[i][j])mask|=(1<<j);
			}
			needed_set.insert(mask);
		}
		vector<int> needed(all(needed_set));
		int res=c;
		needed.push_back((1<<s.back()));
		reverse(all(needed));
		for(int mask=0;mask<(1<<c);mask++) {
			bool check=true;
			for(int m:needed) {
				if(!(mask&m)) {
					check=false;
					break;
				}
			}
			if(check) {
				res=min(res, __builtin_popcount(mask));
			}
		}
		cout << res << "\n";
	}
	else {

		vt<vt<int>> pref(n+1, vt<int>(c, -1));
		for(int i = 1;i<=n;i++) {
			pref[i][s[i-1]]=i;
		}
		for(int i = 1;i<=n;i++) {
			for(int j = 0;j<c;j++) {
				pref[i][j] = max(pref[i-1][j],pref[i][j]);
			}
		}
		int res=n;
		for(int mask = 0;mask<(1<<c);mask++) {
			int cur = 0;
			bool check=1;
			if(!(mask&(1<<s.back())))continue;
			int cnt=0;
			while(cur < n) {
				int mx=cur;
				for(int j = 0;j<c;j++) {
					if(!(mask&(1<<j)))continue;
					cur=max(cur, pref[min(n, cur+k)][j]);
				}
				for(int j = 0;j<c;j++) {
					if(!(mask&(1<<j)))continue;
					cur=max(cur, pref[min(n, cur+k)][j]);
				}
				cnt++;
				if(cur==mx){
					check=false;
					break;
				}
				if(cur!=n and cur < k*(cnt)){
					check=false;
					break;
				}
			}
			if(check) {
				res=min(res, __builtin_popcount(mask));
			}
		}
		cout << res << "\n";
	}

}
int main() {
	cin.tie(0)->sync_with_stdio(0);
	cin.exceptions(cin.failbit);
  int t=1;
	cin >> t;
  while(t--)
    solve();
}

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

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

Think step by step.

Response:
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 can cause the code to fail. Here are some potential edge cases to consider:

1. **Single Character Texts**: Test cases where the text consists of a single character.
2. **Maximum Length Texts**: Test cases where the text length is at the maximum allowed value.
3. **All Characters Same**: Test cases where all characters in the text are the same.
4. **All Characters Different**: Test cases where all characters in the text are different.
5. **Mixed Cases**: Test cases where the text has a mix of different characters and lengths.

Let's create a small-scale test input that covers these scenarios:

```plaintext
5
1 1 1
A
2 2 1
AB
3 3 2
ABC
10 2 2
ABABABABAB
9 3 2
ABCABCABC
```

This test input includes:
1. A single character text with one letter.
2. A two-character text with two different letters.
3. A three-character text with three different letters.
4. A ten-character text with two different letters, repeated.
5. A nine-character text with three different letters, repeated.

This should help expose any issues in the buggy code, especially in how it handles different lengths and character sets.

5
1 1 1
A
2 2 1
AB
3 3 2
ABC
10 2 2
ABABABABAB
9 3 2
ABCABCABC


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cases
time_limit_ms: 2000
memory_limit_mb: 256
Description: You're a linguist studying a mysterious ancient language. You know that

  1. Its words consist only of the first $c$ letters of the Latin alphabet.    2. Each word has a case which can be unambiguously determined by its last letter (different letters correspond to different cases). For example, words "ABACABA" and "ABA" (if they exist) have the same case in this language because they both have the same ending 'A', whereas "ALICE" and "BOB" have different cases. If the language does not have a case corresponding to some letter, it means that the word cannot end with this letter.    3. The length of each word is $k$ or less.

You have a single text written in this language. Unfortunately, as the language is really ancient, spaces between words are missing and all letters are uppercase. You wonder what is the minimum number of cases the language can have. Can you find this out?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 10\,000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains three integers $n$, $c$, $k$ ($1 \le k \le n \le 2^{18}$, $1 \le c \le 18$) — the length of the text, the number of letters in the language, and the maximum length of the word.

The second line contains a string of $n$ characters — the text itself. Each character is one of the first $c$ uppercase letters of the Latin alphabet.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2^{18}$ and the sum of $2^c$ over all test cases does not exceed $2^{18}$.

Output: For each test case, output a single line consisting of a single integer — the minimum number of cases in the language.

Examples:
input:
7
5 5 1
ABCDE
3 1 2
AAA
3 2 2
AAB
10 2 2
ABABABABAB
4 4 4
DCBA
1 17 1
Q
9 3 2
ABCABCABC
output:
5
1
2
1
1
1
2

Note: In the first test case, there must be five cases in the language (for each of the letters 'A', 'B', 'C', 'D', and 'E' there must be a case that has a corresponding ending).

In the fourth test case, one case with ending 'B' is sufficient.

Buggy Code:
#include <iostream>
#include <vector>
#include <unordered_set>
#include <string>

using namespace std;

bool canFormWords(const string &text, const unordered_set<char> &cases, int k) {
    for (int i = 0; i < text.length();) {
        bool found = false;
        for (int j = 1; j <= k && (i + j) <= text.length(); ++j) {
            if (cases.find(text[i + j - 1]) != cases.end()) {
                i += j;
                found = true;
                break;
            }
        }
        if (!found) return false;
    }
    return true;
}

int main() {
    int t;
    cin >> t;

    while (t--) {
        int n, c, k;
        string text;
        cin >> n >> c >> k >> text;

        // Try to use as few cases as possible
        for (int num_cases = 1; num_cases <= c; ++num_cases) {
            bool possible = false;
            // Check all combinations of 'num_cases' end letters
            for (int mask = 0; mask < (1 << c); ++mask) {
                if (__builtin_popcount(mask) == num_cases) {
                    unordered_set<char> cases;
                    for (int i = 0; i < c; ++i) {
                        if (mask & (1 << i)) {
                            cases.insert('A' + i);
                        }
                    }
                    if (canFormWords(text, cases, k)) {
                        possible = true;
                        break;
                    }
                }
            }
            if (possible) {
                cout << num_cases << endl;
                break;
            }
        }
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
5
10 2 2
ABABABABAB
10 2 1
ABABABABAB
10 3 2
ABCABCABC
10 3 1
ABCABCABC
10 4 4
ABCDABCDAB
```

5
10 2 2
ABABABABAB
10 2 1
ABABABABAB
10 3 2
ABCABCABC
10 3 1
ABCABCABC
10 4 4
ABCDABCDAB


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cases
time_limit_ms: 2000
memory_limit_mb: 256
Description: You're a linguist studying a mysterious ancient language. You know that

  1. Its words consist only of the first $c$ letters of the Latin alphabet.    2. Each word has a case which can be unambiguously determined by its last letter (different letters correspond to different cases). For example, words "ABACABA" and "ABA" (if they exist) have the same case in this language because they both have the same ending 'A', whereas "ALICE" and "BOB" have different cases. If the language does not have a case corresponding to some letter, it means that the word cannot end with this letter.    3. The length of each word is $k$ or less.

You have a single text written in this language. Unfortunately, as the language is really ancient, spaces between words are missing and all letters are uppercase. You wonder what is the minimum number of cases the language can have. Can you find this out?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 10\,000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains three integers $n$, $c$, $k$ ($1 \le k \le n \le 2^{18}$, $1 \le c \le 18$) — the length of the text, the number of letters in the language, and the maximum length of the word.

The second line contains a string of $n$ characters — the text itself. Each character is one of the first $c$ uppercase letters of the Latin alphabet.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2^{18}$ and the sum of $2^c$ over all test cases does not exceed $2^{18}$.

Output: For each test case, output a single line consisting of a single integer — the minimum number of cases in the language.

Examples:
input:
7
5 5 1
ABCDE
3 1 2
AAA
3 2 2
AAB
10 2 2
ABABABABAB
4 4 4
DCBA
1 17 1
Q
9 3 2
ABCABCABC
output:
5
1
2
1
1
1
2

Note: In the first test case, there must be five cases in the language (for each of the letters 'A', 'B', 'C', 'D', and 'E' there must be a case that has a corresponding ending).

In the fourth test case, one case with ending 'B' is sufficient.

Buggy Code:
import math
import re
import functools
import random
import sys
import os
import typing
#from math import gcd, comb,sqrt
from collections import Counter, defaultdict, deque
from functools import lru_cache, reduce
from itertools import accumulate, combinations, permutations
from heapq import nsmallest, nlargest, heappushpop, heapify, heappop, heappush
from io import BytesIO, IOBase
from copy import deepcopy
import threading
from typing import *
from bisect import bisect_left, bisect_right
from types import GeneratorType

# from sortedcontainers import  SortedList

from operator import add

BUFSIZE = 8192

def I():
    return input()

def II():
    return int(input())

def MII():
    return map(int, input().split())

def LI():
    return list(input().split())

def LII():
    return list(map(int, input().split()))

def GMI():
    return map(lambda x: int(x) - 1, input().split())

def LGMI():
    return list(map(lambda x: int(x) - 1, input().split()))

input = lambda: sys.stdin.readline().strip()
# 1000000000000002493

mod = 998244353
inf = 10 ** 15

def solve():
    n,c,k=MII()
    A=I()
    if k>=n:
        print(1)
        return
    cnt=[0]*c
    for i in range(k):
        a=ord(A[i])-ord('A')
        cnt[a]+=1
    dd=ord(A[-1])-ord('A')
    B=[0]*c
    idx=0
    if not cnt[dd]:
        for i in range(c):
            if cnt[i]:
                B[i]=1
        idx+=1

    for i in range(k,n):
        a=ord(A[i])-ord('A')
        b=ord(A[i-k])-ord('A')
        cnt[a]+=1
        cnt[b]-=1
        if not cnt[dd]:
            for j in range(c):
                if cnt[j]:
                    B[j]|=1<<idx
            idx+=1

    ans=c
    full=(1<<(idx))-1

    for i in range(1<<c):
        if i.bit_count()>=ans:
            continue
        cur=0
        for j in range(c):
            if i>>j&1 and j!=dd:
                cur|=B[j]
        if cur==full:
            ans=i.bit_count()
    print(ans+1)

for _ in range(II()):
    solve()

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to create a test case that specifically targets the logic where the code might fail. The key points to consider are:

1. **Edge cases**: Test the boundaries of the input constraints, such as the minimum and maximum values for `n`, `c`, and `k`.
2. **Special cases**: Test cases where the text is very uniform or very diverse.
3. **Logical errors**: Test cases that might expose logical errors in the code, such as incorrect handling of the last letter's case.

Given the constraints:
- `1 <= t <= 10,000`
- `1 <= k <= n <= 2^18`
- `1 <= c <= 18`

Let's create a test case that could potentially expose a logical error in the code. Specifically, we will create a test case where the text is long, and the maximum word length is much smaller than the text length, but the text contains a pattern that might cause the code to miscount the number of cases.

### Test Case
- **t**: 1 (one test case)
- **n**: 16 (a small but non-trivial length)
- **c**: 3 (a small number of letters)
- **k**: 3 (a small maximum word length)
- **Text**: "ABCABCABCABCABCABC" (a repeating pattern)

This test case is designed to ensure that the code correctly handles the repeating pattern and correctly counts the number of cases.

### Output
```plaintext
1
16 3 3
ABCABCABCABCABCABC
```

This test case will help in verifying whether the code correctly handles the repeating pattern and correctly counts the number of cases. If the code fails, it will likely be due to an incorrect handling of the last letter's case or an incorrect counting mechanism.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cases
time_limit_ms: 2000
memory_limit_mb: 256
Description: You're a linguist studying a mysterious ancient language. You know that

  1. Its words consist only of the first $c$ letters of the Latin alphabet.    2. Each word has a case which can be unambiguously determined by its last letter (different letters correspond to different cases). For example, words "ABACABA" and "ABA" (if they exist) have the same case in this language because they both have the same ending 'A', whereas "ALICE" and "BOB" have different cases. If the language does not have a case corresponding to some letter, it means that the word cannot end with this letter.    3. The length of each word is $k$ or less.

You have a single text written in this language. Unfortunately, as the language is really ancient, spaces between words are missing and all letters are uppercase. You wonder what is the minimum number of cases the language can have. Can you find this out?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 10\,000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains three integers $n$, $c$, $k$ ($1 \le k \le n \le 2^{18}$, $1 \le c \le 18$) — the length of the text, the number of letters in the language, and the maximum length of the word.

The second line contains a string of $n$ characters — the text itself. Each character is one of the first $c$ uppercase letters of the Latin alphabet.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2^{18}$ and the sum of $2^c$ over all test cases does not exceed $2^{18}$.

Output: For each test case, output a single line consisting of a single integer — the minimum number of cases in the language.

Examples:
input:
7
5 5 1
ABCDE
3 1 2
AAA
3 2 2
AAB
10 2 2
ABABABABAB
4 4 4
DCBA
1 17 1
Q
9 3 2
ABCABCABC
output:
5
1
2
1
1
1
2

Note: In the first test case, there must be five cases in the language (for each of the letters 'A', 'B', 'C', 'D', and 'E' there must be a case that has a corresponding ending).

In the fourth test case, one case with ending 'B' is sufficient.

Buggy Code:
import sys

def solve():
    n, c, k = map(int, sys.stdin.readline().split())
    s = sys.stdin.readline().strip()

    cumsums = [[0] * (n + 1) for _ in range(c)]
    for i in range(c):
        for j in range(n):
            cumsums[i][j + 1] = cumsums[i][j] + (s[j] == chr(i + ord('A')))

    bms = [False] * (1 << c)
    for lo in range(n - k + 1):
        ms = 0
        for i in range(c):
            if cumsums[i][lo + k] - cumsums[i][lo] != 0:
                ms |= (1 << i)
        bms[ms] = True

    bms[1 << (ord(s[-1]) - ord('A'))] = True

    bad = [False] * (1 << c)
    for i in range(1 << c):
        bad[i] = bms[((1 << c) - 1) ^ i]
    for bm in range((1 << c) - 1, -1, -1):
        for b in range(c):
            bad[bm] = bad[bm] or bad[bm | (1 << b)]

    res = 10**9
    for i in range(1 << c):
        if not bad[i]:
            res = min(res, bin(i).count('1'))
    print(res)

def main():
    t = int(sys.stdin.readline())
    for _ in range(t):
        solve()

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
```plaintext
1
10 3 3
ABCABCABC
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cases
time_limit_ms: 2000
memory_limit_mb: 256
Description: You're a linguist studying a mysterious ancient language. You know that

  1. Its words consist only of the first $c$ letters of the Latin alphabet.    2. Each word has a case which can be unambiguously determined by its last letter (different letters correspond to different cases). For example, words "ABACABA" and "ABA" (if they exist) have the same case in this language because they both have the same ending 'A', whereas "ALICE" and "BOB" have different cases. If the language does not have a case corresponding to some letter, it means that the word cannot end with this letter.    3. The length of each word is $k$ or less.

You have a single text written in this language. Unfortunately, as the language is really ancient, spaces between words are missing and all letters are uppercase. You wonder what is the minimum number of cases the language can have. Can you find this out?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 10\,000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains three integers $n$, $c$, $k$ ($1 \le k \le n \le 2^{18}$, $1 \le c \le 18$) — the length of the text, the number of letters in the language, and the maximum length of the word.

The second line contains a string of $n$ characters — the text itself. Each character is one of the first $c$ uppercase letters of the Latin alphabet.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2^{18}$ and the sum of $2^c$ over all test cases does not exceed $2^{18}$.

Output: For each test case, output a single line consisting of a single integer — the minimum number of cases in the language.

Examples:
input:
7
5 5 1
ABCDE
3 1 2
AAA
3 2 2
AAB
10 2 2
ABABABABAB
4 4 4
DCBA
1 17 1
Q
9 3 2
ABCABCABC
output:
5
1
2
1
1
1
2

Note: In the first test case, there must be five cases in the language (for each of the letters 'A', 'B', 'C', 'D', and 'E' there must be a case that has a corresponding ending).

In the fourth test case, one case with ending 'B' is sufficient.

Buggy Code:
// Source: https://usaco.guide/general/io

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

public class Main {
	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		int T = sc.nextInt();
		for(int i = 1; i <= T; i++){
			int N = sc.nextInt();
			int C = sc.nextInt();
			int K = sc.nextInt();
			sc.nextLine();
			String text = sc.nextLine();
			// Map of letters to the letters that they have never been in an K-interval with
			Map<String,HashSet<Integer>> map = new HashMap<String,HashSet<Integer>>();
			Set<String> ls = new HashSet<String>();
			for(int j = 0; j < text.length(); j++){
				ls.add(text.substring(j,j+1));
			}
			for(char start = 'A'; start <= 'A' + C - 1; start++){
				String st = start + "";
				map.put(st, new HashSet<Integer>());
				for(int interval = 0; interval <= text.length() - K; interval++){
					map.get(st).add(interval);
				}
				/*
				for(char other = 'A'; other <= 'A' + C - 1; other++){
					if(other == start) continue;
					map.get(st).add(other + "");
				} */
			}
			for(int start = 0; start <= text.length() - K; start++){
				// start thru start + K - 1
				HashSet<String> set = new HashSet<String>();
				for(int next = start; next < start + K; next++){
					set.add(text.substring(next,next + 1));
				}
				//System.out.println("set is currently " + set);
				for(String str : set){
					/*
					for(char tang = 'A'; tang <= 'A' + C - 1; tang++){
						if(!set.contains(tang + "")){
							System.out.println("current str " + str);
							System.out.println("tang is " + tang);
							map.get(str).add(tang + "");
							map.get(tang + "").add(str);
						}
					} */
					map.get(str).remove(start);
				}
			}

			//System.out.println(map);

			// Top-down bitmask dp
			int min = C;
			boolean[] dp = new boolean[1<<C];
			List<Set<Integer>> dp2 = new ArrayList<Set<Integer>>();
			for(int num = 0; num < 1 << C; num++){
				dp2.add(new HashSet<Integer>());
			}
			Set<Integer> surj = new HashSet<Integer>();
			for(int n = 0; n < N - K + 1; n++){
				surj.add(n);
			}
			dp2.set(0, surj);
			//System.out.println(ls);
			int last = text.charAt(text.length() - 1) - 'A';
outer:      for(int num = 0; num <= (1 << C) - 1; num++){

				List<String> letters = new ArrayList<String>();
				//Set<Integer> unbits = new HashSet<Integer>();
				for(int l = 0; l < C; l++){
					if((num & (1 << l)) == 0){
						continue;
					}
					if((num & (1 << l)) != 0 && !ls.contains((char)('A' + l) + "")){
						continue outer;
					}
					else letters.add((char)('A' + l) + "");
				}
				//System.out.println("For bitmask " + Integer.toBinaryString(num) + " letters is " + letters);
				if(num == (1 << C) - 1) dp[num] = true;
				else{
					for(int bit = 0; bit < C; bit++){
						if(!ls.contains((char)('A' + bit) + "")){
							continue;
						}
						//if((num & (1 << bit)) != 0){
							int modifiedNum = num ^ (1 << bit);
							if(dp[modifiedNum] == true){

								dp[num] = true;
								break;
								// This is also guaranteed to be false because removing any letters from that won't improve the outcome
							}
							else{
								if((num & (1 << bit)) == 0) continue;
								// Check if removing the letter represented by this extra bit actually has an impact
								/* If all of the other letters currently in subset are in the arraylist
								representing in the value in map of the key of this letter then it is not possible */
								String letter = (char)('A' + bit) + "";
								//System.out.println("Letter being added is " + letter);
								/*
								if((num & (1 << (C - 1))) == 0){
									dp[num] = false;
									break;
								} */
								Set<Integer> stuff = dp2.get(modifiedNum);
								//System.out.println(stuff + " is where " + letters + " except " + letter + " doesn't occur");
								// now intersect with wherever letter doesn't occur
								Set<Integer> nonOccurrences = map.get(letter);
								Set<Integer> codeforces = new HashSet<Integer>();
								for(Integer s : stuff){
									if(nonOccurrences.contains(s)){
										codeforces.add(s);
									}
								}
								dp2.set(num,codeforces);
								/*
								if(codeforces.size() == 0){
									dp[num] = true;
									break;
								} */

								//System.out.println("Nonocurrences for letters " + letters + " are " + codeforces + " for " + letters + " with additional " + letter);
								if(codeforces.size() == 0 && (num & (1 << (last))) != 0){
									dp[num] = true;
									break;
								}

							}
						//}
						//int modifiedNum = num ^ bit;
					}
				}
				if(dp[num]) min = Math.min(min, letters.size());

				//System.out.println("For " + num + " it's " + dp[num]);
			}
			System.out.println(min);
		}
	}
}

**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
10 3 3
ABCABCABC
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cases
time_limit_ms: 2000
memory_limit_mb: 256
Description: You're a linguist studying a mysterious ancient language. You know that

  1. Its words consist only of the first $c$ letters of the Latin alphabet.    2. Each word has a case which can be unambiguously determined by its last letter (different letters correspond to different cases). For example, words "ABACABA" and "ABA" (if they exist) have the same case in this language because they both have the same ending 'A', whereas "ALICE" and "BOB" have different cases. If the language does not have a case corresponding to some letter, it means that the word cannot end with this letter.    3. The length of each word is $k$ or less.

You have a single text written in this language. Unfortunately, as the language is really ancient, spaces between words are missing and all letters are uppercase. You wonder what is the minimum number of cases the language can have. Can you find this out?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 10\,000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains three integers $n$, $c$, $k$ ($1 \le k \le n \le 2^{18}$, $1 \le c \le 18$) — the length of the text, the number of letters in the language, and the maximum length of the word.

The second line contains a string of $n$ characters — the text itself. Each character is one of the first $c$ uppercase letters of the Latin alphabet.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2^{18}$ and the sum of $2^c$ over all test cases does not exceed $2^{18}$.

Output: For each test case, output a single line consisting of a single integer — the minimum number of cases in the language.

Examples:
input:
7
5 5 1
ABCDE
3 1 2
AAA
3 2 2
AAB
10 2 2
ABABABABAB
4 4 4
DCBA
1 17 1
Q
9 3 2
ABCABCABC
output:
5
1
2
1
1
1
2

Note: In the first test case, there must be five cases in the language (for each of the letters 'A', 'B', 'C', 'D', and 'E' there must be a case that has a corresponding ending).

In the fourth test case, one case with ending 'B' is sufficient.

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

public class Main {
boolean MULTI_CASE = true;
boolean ALWAYS_FLUSH = false;

void go() {
    int n = nextInt();
    int c = nextInt();
    int len = nextInt();
    int[] a = next().chars().map(x -> x - 'A').toArray();
    int[] g = new int[c];
    int[] b = new int[1 << c];
    for (int i = 0; i < n; i++) {
        g[a[i]]++;
        if (i - len >= 0) {
            g[a[i - len]]--;
        }
        if (i >= len - 1) {
            int x = 0;
            for (int j = 0; j < c; j++) {
                if (g[j] > 0) x |= 1 << j;
            }
            b[x] = 1;
        }
    }
    int[] v = new int[1 << c];
    int vn = 0;
    for (int i = 0; i < (1 << c); i++) {
        if (b[i] > 0) v[vn++] = i;
    }
    v = Arrays.copyOf(v, vn);
    Integer[] w = new Integer[vn];
    for (int i = 0; i < vn; i++) w[i] = v[i];
    Arrays.sort(w, (x, y) -> Integer.compare(Integer.bitCount(x), Integer.bitCount(y)));
//    ruffleSort(v);
    int ret = c;
    for (int i = 1; i < (1 << c); i++) {
        if ((i & (1 << a[n - 1])) == 0) continue;
        boolean ok = true;
        for (int x : w) {
            if ((i & x) == 0) {
                ok = false;
                break;
            }
        }
        if (ok) {
            ret = Math.min(ret, Integer.bitCount(i));
        }
    }
    sl(ret);
}

void ruffleSort(int[] a) {
    Random random = new Random();
    int n = a.length;
    for (int i = 0; i < n; i++) {
        int oi = random.nextInt(n);
        int temp = a[oi];
        a[oi] = a[i];
        a[i] = temp;
    }
    Arrays.sort(a);
}

/**
 * Input.
 */
InputStream inStream;
byte[] inBuff = new byte[1024];
int inBuffCursor = 0, inBuffLen = 0;

boolean isVisibleChar(int c) {
    return 33 <= c && c <= 126;
}

boolean hasNextByte() {
    if (inBuffCursor < inBuffLen) return true;
    inBuffCursor = 0;
    try {
        inBuffLen = inStream.read(inBuff);
    } catch (Exception e) {
        e.printStackTrace(System.out);
    }
    return inBuffLen > 0;
}

boolean hasNext() {
    while (hasNextByte() && !isVisibleChar(inBuff[inBuffCursor])) inBuffCursor++;
    return hasNextByte();
}

int nextByte() {
    return hasNextByte() ? inBuff[inBuffCursor++] : -1;
}

String next() {
    if (!hasNext()) throw new RuntimeException("no next.");
    StringBuilder sb = new StringBuilder();
    int b = nextByte();
    while (isVisibleChar(b)) {
        sb.appendCodePoint(b);
        b = nextByte();
    }
    return sb.toString();
}

long nextLong() {
    if (!hasNext()) throw new RuntimeException("no next.");
    long result = 0;
    boolean negative = false;
    int b = nextByte();
    if (b < '0') {
        if (b == '-') negative = true;
        else if (b != '+') throw new RuntimeException("long number must start with +/-.");
        b = nextByte();
    }
    while (isVisibleChar(b)) {
        if (b < '0' || b > '9') throw new RuntimeException("wrong digit in long:" + (char) b);
        // NOTE: may overflow here, even if it is a valid Long, eg.-(1<<63).
        result = result * 10 + (b - '0');
        b = nextByte();
    }
    return negative ? -result : result;
}

int nextInt() {
    long x = nextLong();
    if (x < Integer.MIN_VALUE || x > Integer.MAX_VALUE)
        throw new RuntimeException("int overflow:" + x);
    return (int) x;
}

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

/**
 * Output.
 */
PrintWriter printOut = new PrintWriter(System.out);

void so(Object obj, boolean newLine) {
    if (newLine) printOut.println(obj);
    else printOut.print(obj);
    if (ALWAYS_FLUSH) printOut.flush();
}

void so(Object obj) {
    so(obj, false);
}

void sl(Object obj) {
    so(obj, true);
}

void sl() {
    sl("");
}

/**
 * Main.
 */
void mainGo() {
    try {
        inStream = new FileInputStream("src/main.in");
        ALWAYS_FLUSH = true;
    } catch (Exception e) {
        inStream = System.in;
    }
    while (hasNext()) {
        if (MULTI_CASE) {
            int T = nextInt();
            if (T == 0) break;
            for (int i = 0; i < T; ++i) {
                go();
            }
        } else {
            go();
        }
    }
    printOut.flush();
}

public static void main(String[] args) throws Exception {
    new Main().mainGo();
}
}

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

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

Think step by step.

Response:
```plaintext
3
5 5 1
ABCDE
10 2 2
ABABABABAB
9 3 2
ABCABCABC
```


**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 {
        vector<int> digits(k + 1);
        for (int i = 1; i <= k; i++) {
            digits[i] = cur_str[i - 1] - '0';
        }

        array<int, 10> prv_loc;
        prv_loc.fill(-1);

        vector<ll> dp(k + 1, INF);
        dp[0] = 0;
        dp[2] = digits[1];
        prv_loc[digits[1]] = 1;
        prv_loc[digits[2]] = 2;
        for (int i = 3; i <= k; i++) {
            for (int j = 1; j <= 9; j++) {
                int loc = prv_loc[j];
                if (loc == -1) { continue; }
                dp[i] = min(dp[i], dp[loc - 1] + 1ll * j * (i - loc));
            }

            prv_loc[digits[i]] = i;
        }
        return dp[k];
    };

    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:
To generate a challenging test input that exposes the bug in the given code, we need to focus on edge cases and scenarios that can maximize the failure probability. Let's consider the following aspects:

1. **Edge Cases for Substring Lengths:**
   - Small values of \( k \) (e.g., \( k = 2 \)).
   - Large values of \( k \) (e.g., \( k \) close to \( n \)).

2. **Edge Cases for String Content:**
   - Strings with repeated digits.
   - Strings with alternating digits.
   - Strings with a mix of different digits.

3. **Edge Cases for Substring Compression:**
   - Substrings where the first digit is large.
   - Substrings where the first digit is small.
   - Substrings where the first few digits are the same.

Given these considerations, a small-scale test input that can expose the bug might look like this:

```plaintext
6 3
123321
```

### Explanation:
- **n = 6, k = 3**: This is a small-scale input that is easy to verify.
- **String "123321"**: This string contains a mix of different digits and has a pattern that can challenge the compression logic.

### Expected Output:
- The substrings of length 3 are "123", "233", and "332", "321".
- For "123", the compressed length is 3 (no compression possible).
- For "233", the compressed length is 4 (2 times "3").
- For "332", the compressed length is 4 (3 times "2").
- For "321", the compressed length is 3 (no compression possible).

Thus, the expected output should be:
```plaintext
3 4 4 3
```

This test input should help identify any issues in the compression logic, especially in handling different patterns and digit repetitions.


**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 <vector>
#include <algorithm>
#include <map>
#include <cassert>
#include <cmath>
#include <random>
#include <set>
#include <array>
#include <queue>
#include <bitset>
#include <unordered_map>

using namespace std;
#define ll long long
int k;
vector <int> a;
vector <int> ans;
const int MAXN = 2e5;
const int INF = 1e9;
int dp[MAXN + 10][10][10][2][2];

void clear_dp(int i) {
    for (int x = 0; x < 10; ++x) {
        for (int y = 0; y < 10; ++y) {
            for (int f1 = 0; f1 < 2; ++f1) {
                for (int f2 = 0; f2 < 2; ++f2) {
                    dp[i][x][y][f1][f2] = INF;
                }
            }
        }
    }
}

void calc(int l, int r) {
    if (r - l + 1 < k) return;
    int m = (l + r) / 2;

    for (int i = l; i <= r; ++i) clear_dp(i);

    for (int x = 1; x < 10; ++x) {
        dp[m][x][a[m]][0][1] = 0;
        dp[m][x][x][0][0] = x;

        dp[m][x][x][1][0] = x;
    }

    for (int i = m - 1; i >= l; --i) {
        for (int x = 1; x < 10; ++x) {
            for (int y = 1; y < 10; ++y) {
                for (int f2 = 0; f2 < 2; ++f2) {
                    dp[i][x][y][0][f2] = min(dp[i + 1][x][y][0][f2] + x, dp[i + 1][a[i]][y][1][f2]);
                    dp[i][x][y][1][f2] = dp[i + 1][x][y][0][f2] + x;
                }
            }
        }
    }

    for (int x = 1; x < 10; ++x) {
        dp[m + 1][x][a[m + 1]][0][1] = 0;
        dp[m + 1][x][x][0][0] = x;
        dp[m + 1][x][x][1][0] = x;
    }

    for (int i = m + 2; i <= r; ++i) {
        for (int x = 1; x < 10; ++x) {
            for (int y = 1; y < 10; ++y) {
                for (int f1 = 0; f1 < 2; ++f1) {
                    dp[i][x][y][f1][0] = min(dp[i - 1][x][y][f1][0], dp[i - 1][x][y][f1][1]) + y;
                    dp[i][x][a[i]][f1][1] = min(dp[i][x][a[i]][f1][1], dp[i - 1][x][y][f1][0]);
                }
            }
        }
    }

    for (int i = m; i >= l; --i) {
        if (i + k - 1 <= m) break;
        if (i + k - 1 <= r) {
            for (int x = 1; x < 10; ++x) {
                for (int y = 1; y < 10; ++y) {
                    int me = 0;
                    if (i < m) me = dp[i + 1][a[i]][x][1][0];
                    else if (x != a[i]) me = INF;
                    ans[i] = min(ans[i], me + dp[i + k - 1][x][y][0][0]);
                    me = 0;
                    if (i < m) me = dp[i + 1][a[i]][x][1][1];
                    else if (x != a[i]) me = INF;
                    ans[i] = min(ans[i], me + dp[i + k - 1][x][y][1][0]);
                }
            }
        }
    }

    calc(l, m);
    calc(m + 1, r);
}

inline void solve() {
    int n;
    cin >> n >> k;
    string s;
    cin >> s;
    a.resize(n);
    for (int i = 0; i < n; ++i) {
        a[i] = s[i] - '0';
    }
    ans.resize(n, INF);
    calc(0, n - 1);
    for (int i = 0; i + k - 1 < n; ++i) cout << ans[i] << " ";
}

signed main() {
    if (1) {
        ios_base::sync_with_stdio(false);
        cin.tie(nullptr);
        cout.tie(nullptr);
    }
    if (1) {
        int t = 1;
        // cin >> t;
        while (t--) {
            solve();
        }
    }
}

/*

 */

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy 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**:
   - Strings with repeated digits.
   - Strings with alternating digits.
   - Strings with a mix of digits that could cause issues in the compression logic.
   - Strings that are just long enough to test the limits of the algorithm.

2. **Maximizing Failure Probability**:
   - Test with a string that has a pattern that could cause the compression logic to fail, such as a string with a high frequency of a single digit followed by a different digit.
   - Test with a string that has a complex pattern that might not be handled correctly by the dynamic programming approach.

3. **Small-Scale Test Input**:
   - Since the input size is limited, we need to create a test case that is small but still challenging.

Given these considerations, a suitable test input could be:

```plaintext
6 3
111222
```

This test input has the following characteristics:
- It is small and manageable.
- It contains a repeating pattern of digits.
- It has a mix of digits that could expose issues in the compression logic.
- The length of the string is 6, and the substring length is 3, which allows for multiple contiguous substrings to be evaluated.

This test input should help in identifying any bugs in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 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 <vector>
#include <string>

using namespace std;

int main() {
    int n, k;
    cin >> n >> k;
    string s;
    cin >> s;

    vector<int> result;

    // Iterate over all substrings of length k
    for (int i = 0; i <= n - k; i++) {
        int compressed_length = 0;

        // Process the substring s[i:i+k] in pairs
        for (int j = 0; j < k; j += 2) {
            int t1 = s[i + j] - '0'; // Convert char to int
            if (j + 1 < k) {
                int len_t2 = 1; // t2 is one character long
                compressed_length += t1 * len_t2;
            }
        }

        result.push_back(compressed_length);
    }

    // Output the results
    for (int i = 0; i < result.size(); i++) {
        cout << result[i] << " ";
    }
    cout << endl;

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
6 2
121212
```

### Explanation:
This test case is designed to expose the bug in the given buggy code. Here's how it works:

1. **Input Parameters**:
   - `n = 6` (length of the string)
   - `k = 2` (length of the substrings to consider)

2. **String**:
   - `s = "121212"`

3. **Substrings of Length k**:
   - Substring 1: "12"
   - Substring 2: "21"
   - Substring 3: "12"
   - Substring 4: "21"
   - Substring 5: "12"

4. **Expected Output**:
   - For "12": The compressed length is 2 (since "2" is repeated 1 time).
   - For "21": The compressed length is 2 (since "1" is repeated 2 times).
   - For "12": The compressed length is 2 (same as the first one).
   - For "21": The compressed length is 2 (same as the second one).
   - For "12": The compressed length is 2 (same as the third one).

   Therefore, the expected output is: `2 2 2 2 2`.

5. **Buggy Code Behavior**:
   - The buggy code only considers the first character of each pair and multiplies it by the length of the second character (which is always 1). This means it will always output `2` for any valid input of length 2, but it does not handle the actual compression logic correctly.

This test case is small and complete, and it should clearly show that the buggy code is not performing the correct compression logic.


**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 calculate_f(t):
    m = len(t)
    result = 0
    for i in range(0, m, 2):
        count = int(t[i])
        string_length = len(t[i+1])
        result += count * string_length
    return result

def solve(n, k, s):
    results = []
    for i in range(n - k + 1):
        substring = s[i:i+k]
        result = calculate_f(substring)
        results.append(result)
    print(" ".join(map(str, results)))

# Example usage
n, k = map(int, input().split())
s = input().strip()
solve(n, k, s)

**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. **Even Length Substrings**: The code assumes that the string can always be split into an even number of non-empty substrings. If the length of the substring is odd, the code will fail.

2. **Single-Digit Count**: The code assumes that the count is always a single digit. If the count is a multi-digit number, the code will also fail.

3. **Invalid Splitting**: The code does not handle cases where the string cannot be split into valid pairs of count and string.

Given these points, we can create a test input that includes:

- A substring of odd length to expose the issue with the even length assumption.
- A substring where the count is a multi-digit number to expose the issue with the single-digit assumption.
- A substring where the count is not a valid number to expose the issue with invalid splitting.

Let's construct a small-scale test input that includes these scenarios:

```plaintext
10 5
11234
```

### Explanation:
- **n = 10**: The string length is 10.
- **k = 5**: The length of each substring is 5.
- **s = "11234"**: The string contains a substring of length 5.

### Analysis:
- The substring "11234" will be processed by the `calculate_f` function.
- The function will try to split it into pairs: ("1", "1234").
- The count is "1" (valid), and the string length is 4.
- The result should be 4 (1 * 4).

However, if the string length is odd, the code will fail to handle it correctly. This test input will expose the issue with the even length assumption. Additionally, if the count is a multi-digit number or invalid, it will also expose the issues with the single-digit and invalid splitting assumptions.


**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 min_compressed_length(substr):
    n = len(substr)
    min_len = float('inf')

    # Try all possible even splits
    for i in range(1, n // 2 + 1):
        total_len = 0
        valid = True

        for j in range(i, n, 2 * i):
            if j + i > n:
                break
            repeat_count = int(substr[j - i:j])
            repeated_str = substr[j:j + i]
            total_len += repeat_count * len(repeated_str)

            if repeat_count == 0:
                valid = False
                break

        if valid:
            min_len = min(min_len, total_len)

    return min_len

# Read input
n, k = map(int, input().split())
s = input()

# Initialize result array
results = []

# Calculate minimum compressed length for each substring of length k
for i in range(n - k + 1):
    substr = s[i:i + k]
    results.append(min_compressed_length(substr))

# Output the results
print(" ".join(map(str, 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
6 4
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:
def min_compressed_length(substr):
    k = len(substr)
    min_len = float('inf')

    # Try all possible even splits
    for i in range(1, k // 2 + 1):
        if k % (2 * i) == 0:
            length = 0
            for j in range(0, k, 2 * i):
                repeat_count = int(substr[j:j + i])
                repeated_str = substr[j + i:j + 2 * i]
                length += repeat_count * len(repeated_str)
            min_len = min(min_len, length)

    return min_len

# Read input
import sys
input = sys.stdin.read
data = input().split()

n = int(data[0])
k = int(data[1])
s = data[2]

results = []

for i in range(n - k + 1):
    substr = s[i:i + k]
    results.append(min_compressed_length(substr))

print(" ".join(map(str, 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
6 2
123456
```

### Explanation:
This test case is designed to expose potential issues in the buggy code by focusing on the smallest possible value of \( k \) (which is 2) and a string \( s \) that is long enough to generate multiple substrings of length \( k \).

- The input string \( s = "123456" \) is a simple sequence of digits.
- For \( k = 2 \), the substrings are "12", "23", "34", "45", and "56".
- The function \( f(t) \) for each of these substrings should be the length of the repeated string times the repeat count, which is simply the length of the substring itself since the repeat count is always 1.

The expected output for this test case is:
```
2 2 2 2 2
```

This test case will help verify if the code correctly handles the smallest possible value of \( k \) and whether it correctly calculates the length of the compressed string for each substring.


**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.Scanner;
public class CompressString {
      public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        String s = sc.next();
        for (int i = 0; i <= n - k; i++) {
            String t = s.substring(i, i + k);
            System.out.print(compress(t) + " ");
        }
    }
    public static int compress(String t) {
        int m = t.length();
        StringBuilder result = new StringBuilder();
        for (int i = 0; i < m; i += 2) {
            int times = t.charAt(i) - '0';
            char toRepeat = t.charAt(i + 1);
            for (int j = 0; j < times; j++) {
                result.append(toRepeat);
            }
        }
        return result.length();
    }
}

**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
991199
```