#include <bits/stdc++.h>
using namespace std;
int N, K;
string str;
int ps[5005];
long long ans;
long long C[5005][5005];
void ini() {
  for (int i = 0; i <= N; i++) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j <= i - 1; j++) {
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % 998244353;
    }
  }
}
long long cnt(int l, int r) {
  if (ps[r] - ps[l - 1] > K) return 0LL;
  int d = 0;
  if (str[l - 1] == '0')
    d++;
  else
    d--;
  if (str[r - 1] == '0')
    d++;
  else
    d--;
  int I = ps[r - 1] - ps[l] - d;
  int L = r - 1 - l;
  if (I < 0 || I > L) return 0LL;
  return C[L][I];
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> K;
  cin >> str;
  for (int i = 1; i <= N; i++) {
    ps[i] = (str[i - 1] == '1') ? 1 : 0;
    ps[i] += ps[i - 1];
  }
  ini();
  ans = 1;
  if (ps[N] < K) {
    cout << ans << '\n';
    return 0;
  }
  for (int i = 1; i <= N - 1; i++)
    for (int j = i + 1; j <= N; j++) ans = (ans + cnt(i, j)) % 998244353;
  cout << ans << '\n';
  return 0;
}

### Prompt

Generate 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;
string str;
int ps[5005];
long long ans;
long long C[5005][5005];
void ini() {
  for (int i = 0; i <= N; i++) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j <= i - 1; j++) {
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % 998244353;
    }
  }
}
long long cnt(int l, int r) {
  if (ps[r] - ps[l - 1] > K) return 0LL;
  int d = 0;
  if (str[l - 1] == '0')
    d++;
  else
    d--;
  if (str[r - 1] == '0')
    d++;
  else
    d--;
  int I = ps[r - 1] - ps[l] - d;
  int L = r - 1 - l;
  if (I < 0 || I > L) return 0LL;
  return C[L][I];
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> K;
  cin >> str;
  for (int i = 1; i <= N; i++) {
    ps[i] = (str[i - 1] == '1') ? 1 : 0;
    ps[i] += ps[i - 1];
  }
  ini();
  ans = 1;
  if (ps[N] < K) {
    cout << ans << '\n';
    return 0;
  }
  for (int i = 1; i <= N - 1; i++)
    for (int j = i + 1; j <= N; j++) ans = (ans + cnt(i, j)) % 998244353;
  cout << ans << '\n';
  return 0;
}

```