#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
long long dp[5000 + 10], cnt[5000 + 10], c[5000 + 10][5000 + 10];
int main() {
  ios_base::sync_with_stdio(false);
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  for (int i = 0; i <= 5000; i++) {
    for (int j = 0; j <= i; j++) {
      if (j == 0 | j == i) {
        c[i][j] = 1;
      } else {
        c[i][j] = c[i - 1][j - 1] % mod + c[i - 1][j] % mod;
        c[i][j] %= mod;
      }
    }
  }
  int ind = 0;
  int f = 0;
  cnt[0] = -1;
  for (int i = 0; i < n; i++) {
    if (s[i] == '1') {
      f++;
      cnt[f] = i;
    }
    if (f == k) {
      break;
    }
    ind++;
    dp[i] = 1;
  }
  dp[ind] = c[ind + 1][k];
  for (int i = ind + 1; i < n; i++) {
    dp[i] = dp[i - 1];
    if (s[i] == '1') {
      f++;
      cnt[f] = i;
      int x = cnt[f - k] + 1;
      dp[i] += c[i - x][k] % mod;
    } else {
      int x = cnt[f - k] + 1;
      dp[i] += c[i - x][k - 1] % mod;
    }
    dp[i] %= mod;
  }
  cout << dp[n - 1] % mod;
}

### 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;
long long dp[5000 + 10], cnt[5000 + 10], c[5000 + 10][5000 + 10];
int main() {
  ios_base::sync_with_stdio(false);
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  for (int i = 0; i <= 5000; i++) {
    for (int j = 0; j <= i; j++) {
      if (j == 0 | j == i) {
        c[i][j] = 1;
      } else {
        c[i][j] = c[i - 1][j - 1] % mod + c[i - 1][j] % mod;
        c[i][j] %= mod;
      }
    }
  }
  int ind = 0;
  int f = 0;
  cnt[0] = -1;
  for (int i = 0; i < n; i++) {
    if (s[i] == '1') {
      f++;
      cnt[f] = i;
    }
    if (f == k) {
      break;
    }
    ind++;
    dp[i] = 1;
  }
  dp[ind] = c[ind + 1][k];
  for (int i = ind + 1; i < n; i++) {
    dp[i] = dp[i - 1];
    if (s[i] == '1') {
      f++;
      cnt[f] = i;
      int x = cnt[f - k] + 1;
      dp[i] += c[i - x][k] % mod;
    } else {
      int x = cnt[f - k] + 1;
      dp[i] += c[i - x][k - 1] % mod;
    }
    dp[i] %= mod;
  }
  cout << dp[n - 1] % mod;
}

```