#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n, k;
  int mod = 998244353;
  cin >> n >> k;
  string S;
  cin >> S;
  vector<vector<int>> c(n + 1, vector<int>(n + 1, 1));
  for (int i = 2; i <= n; ++i) {
    for (int j = 1; j < i; ++j) {
      c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
    }
  }
  int f = 0;
  for (auto s : S) {
    if (s == '1') {
      ++f;
    }
  }
  if (f < k) {
    cout << 1 << endl;
    return;
  }
  int r = 1;
  for (int i = 0; i < n; ++i) {
    if (S[i] == '1') {
      int j = i + 1, z = 0;
      while (j < n) {
        z += (S[j] == '1');
        if (z >= k) {
          --z;
          break;
        }
        ++j;
      }
      int x = j - i - 1;
      if (x == z || x == 0) {
        continue;
      }
      r += c[x][z + 1];
      r %= mod;
    } else {
      int j = i + 1, z = 0;
      while (j < n) {
        z += (S[j] == '1');
        if (z > k) {
          --z;
          break;
        }
        ++j;
      }
      int x = j - i - 1;
      if (z == 0) {
        continue;
      }
      r += c[x][z - 1];
      r %= mod;
    }
  }
  cout << r << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int T;
  solve();
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
void solve() {
  int n, k;
  int mod = 998244353;
  cin >> n >> k;
  string S;
  cin >> S;
  vector<vector<int>> c(n + 1, vector<int>(n + 1, 1));
  for (int i = 2; i <= n; ++i) {
    for (int j = 1; j < i; ++j) {
      c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
    }
  }
  int f = 0;
  for (auto s : S) {
    if (s == '1') {
      ++f;
    }
  }
  if (f < k) {
    cout << 1 << endl;
    return;
  }
  int r = 1;
  for (int i = 0; i < n; ++i) {
    if (S[i] == '1') {
      int j = i + 1, z = 0;
      while (j < n) {
        z += (S[j] == '1');
        if (z >= k) {
          --z;
          break;
        }
        ++j;
      }
      int x = j - i - 1;
      if (x == z || x == 0) {
        continue;
      }
      r += c[x][z + 1];
      r %= mod;
    } else {
      int j = i + 1, z = 0;
      while (j < n) {
        z += (S[j] == '1');
        if (z > k) {
          --z;
          break;
        }
        ++j;
      }
      int x = j - i - 1;
      if (z == 0) {
        continue;
      }
      r += c[x][z - 1];
      r %= mod;
    }
  }
  cout << r << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int T;
  solve();
  return 0;
}

```