#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 9, inf = 0x3f3f3f3f, mod = 998244353;
long long C[5010][5010];
void solve() {
  for (int i = 0; i <= 5000; i++) {
    C[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
    }
  }
  long long n, k;
  scanf("%lld%lld", &n, &k);
  string s;
  cin >> s;
  int lo = 0, ri = 0, cnt = 0, plo = -inf, pri = -inf;
  long long ans = 0;
  if (k == 0) {
    printf("1\n");
    return;
  }
  while (ri < n) {
    if (s[ri] == '1') cnt++;
    while (ri + 1 < n && s[ri + 1] == '0') ri++;
    while (cnt > k) {
      if (s[lo] == '1') cnt--;
      lo++;
    }
    if (cnt == k) {
      ans = (ans + C[ri - lo + 1][k]) % mod;
      if (plo != -inf) {
        ans = (ans - C[pri - lo + 1][k - 1] + mod) % mod;
      }
      plo = lo;
      pri = ri;
    }
    ri++;
  }
  if (pri == -inf) {
    printf("1\n");
    return;
  }
  printf("%lld\n", (ans + mod) % mod);
}
int main() {
  solve();
  return 0;
}

### Prompt

Please formulate 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;
const int N = 2e5 + 9, inf = 0x3f3f3f3f, mod = 998244353;
long long C[5010][5010];
void solve() {
  for (int i = 0; i <= 5000; i++) {
    C[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
    }
  }
  long long n, k;
  scanf("%lld%lld", &n, &k);
  string s;
  cin >> s;
  int lo = 0, ri = 0, cnt = 0, plo = -inf, pri = -inf;
  long long ans = 0;
  if (k == 0) {
    printf("1\n");
    return;
  }
  while (ri < n) {
    if (s[ri] == '1') cnt++;
    while (ri + 1 < n && s[ri + 1] == '0') ri++;
    while (cnt > k) {
      if (s[lo] == '1') cnt--;
      lo++;
    }
    if (cnt == k) {
      ans = (ans + C[ri - lo + 1][k]) % mod;
      if (plo != -inf) {
        ans = (ans - C[pri - lo + 1][k - 1] + mod) % mod;
      }
      plo = lo;
      pri = ri;
    }
    ri++;
  }
  if (pri == -inf) {
    printf("1\n");
    return;
  }
  printf("%lld\n", (ans + mod) % mod);
}
int main() {
  solve();
  return 0;
}

```