description: 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.
solution: #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;
  }
  