#include <bits/stdc++.h>
using namespace std;
int n, k;
char s[5005];
long long C[5005][5005];
int main() {
  scanf("%d%d", &n, &k);
  C[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    C[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % 998244353;
    }
  }
  scanf("%s", s + 1);
  int cur = 0, x, y;
  for (int i = 1; i <= n; i++)
    if (s[i] == '1') cur++;
  if (cur < k) {
    printf("1\n");
    return 0;
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    cur = s[i] - '0';
    for (int j = i + 1; j <= n; j++) {
      cur += s[j] - '0';
      if (cur > k) break;
      x = cur;
      y = j - i + 1 - cur;
      if (s[i] == '0')
        x--;
      else
        y--;
      if (s[j] == '0')
        x--;
      else
        y--;
      if (x < 0 || y < 0) continue;
      ans += C[x + y][x];
      ans %= 998244353;
    }
  }
  printf("%lld\n", (ans + 1) % 998244353);
  return 0;
}

### Prompt

Your task is to create a CPP solution to 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;
int n, k;
char s[5005];
long long C[5005][5005];
int main() {
  scanf("%d%d", &n, &k);
  C[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    C[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % 998244353;
    }
  }
  scanf("%s", s + 1);
  int cur = 0, x, y;
  for (int i = 1; i <= n; i++)
    if (s[i] == '1') cur++;
  if (cur < k) {
    printf("1\n");
    return 0;
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    cur = s[i] - '0';
    for (int j = i + 1; j <= n; j++) {
      cur += s[j] - '0';
      if (cur > k) break;
      x = cur;
      y = j - i + 1 - cur;
      if (s[i] == '0')
        x--;
      else
        y--;
      if (s[j] == '0')
        x--;
      else
        y--;
      if (x < 0 || y < 0) continue;
      ans += C[x + y][x];
      ans %= 998244353;
    }
  }
  printf("%lld\n", (ans + 1) % 998244353);
  return 0;
}

```