#include <bits/stdc++.h>
using namespace std;
const int MOD = 998244353;
const int N = 5000 + 3;
vector<vector<long long>> C(N, vector<long long>(N, 0));
inline void init() {
  for (int n = 0; n < N; ++n) {
    C[n][0] = 1;
  }
  for (int n = 1; n < N; ++n) {
    for (int k = 1; k <= n; ++k) {
      C[n][k] = (C[n - 1][k] + C[n - 1][k - 1]) % MOD;
    }
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  init();
  int n, k;
  string s;
  while (cin >> n >> k >> s) {
    vector<int> pref(n + 1, 0);
    for (int i = 0; i < n; ++i) {
      s[i] -= '0';
      pref[i + 1] = pref[i] + s[i];
    }
    long long ans = 1;
    if (pref.back() >= k) {
      for (int i = 0; i < n; ++i) {
        for (int j = i + 1; j < n; ++j) {
          vector<int> cnt = {0, 0};
          cnt[1] = pref[j + 1] - pref[i];
          if (cnt[1] > k) {
            continue;
          }
          cnt[0] = j - i + 1 - cnt[1];
          if (cnt[s[i] ^ 1]-- == 0) {
            continue;
          }
          if (cnt[s[j] ^ 1]-- == 0) {
            continue;
          }
          int d = C[j - i - 1][cnt[0]];
          ans = (ans + d) % MOD;
        }
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

### Prompt

Create a solution in Cpp for the following problem:

You are given a binary string (i. e. a string consisting of characters 0 and/or 1) s of length n. You can perform the following operation with the string s at most once: choose a substring (a contiguous subsequence) of s having exactly k characters 1 in it, and shuffle it (reorder the characters in the substring as you wish).

Calculate the number of different strings which can be obtained from s by performing this operation at most once.

Input

The first line contains two integers n and k (2 ≤ n ≤ 5000; 0 ≤ k ≤ n).

The second line contains the string s of length n, consisting of characters 0 and/or 1.

Output

Print one integer — the number of different strings which can be obtained from s by performing the described operation at most once. Since the answer can be large, output it modulo 998244353.

Examples

Input


7 2
1100110


Output


16


Input


5 0
10010


Output


1


Input


8 1
10001000


Output


10


Input


10 8
0010011000


Output


1

Note

Some strings you can obtain in the first example:

  * to obtain 0110110, you can take the substring from the 1-st character to the 4-th character, which is 1100, and reorder its characters to get 0110; 
  * to obtain 1111000, you can take the substring from the 3-rd character to the 7-th character, which is 00110, and reorder its characters to get 11000; 
  * to obtain 1100101, you can take the substring from the 5-th character to the 7-th character, which is 110, and reorder its characters to get 101. 



In the second example, k = 0 so you can only choose the substrings consisting only of 0 characters. Reordering them doesn't change the string at all, so the only string you can obtain is 10010.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MOD = 998244353;
const int N = 5000 + 3;
vector<vector<long long>> C(N, vector<long long>(N, 0));
inline void init() {
  for (int n = 0; n < N; ++n) {
    C[n][0] = 1;
  }
  for (int n = 1; n < N; ++n) {
    for (int k = 1; k <= n; ++k) {
      C[n][k] = (C[n - 1][k] + C[n - 1][k - 1]) % MOD;
    }
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  init();
  int n, k;
  string s;
  while (cin >> n >> k >> s) {
    vector<int> pref(n + 1, 0);
    for (int i = 0; i < n; ++i) {
      s[i] -= '0';
      pref[i + 1] = pref[i] + s[i];
    }
    long long ans = 1;
    if (pref.back() >= k) {
      for (int i = 0; i < n; ++i) {
        for (int j = i + 1; j < n; ++j) {
          vector<int> cnt = {0, 0};
          cnt[1] = pref[j + 1] - pref[i];
          if (cnt[1] > k) {
            continue;
          }
          cnt[0] = j - i + 1 - cnt[1];
          if (cnt[s[i] ^ 1]-- == 0) {
            continue;
          }
          if (cnt[s[j] ^ 1]-- == 0) {
            continue;
          }
          int d = C[j - i - 1][cnt[0]];
          ans = (ans + d) % MOD;
        }
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

```