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;
  const int MOD = 998244353;
  const int N = 5005;
  int n, k;
  int a[N], dem = 0;
  long long C[N][N];
  void pre_calc() {
    C[0][0] = 1;
    for (int i = 1; i <= n; i++) C[0][i] = 1;
    for (int j = 1; j <= n; j++) {
      for (int i = 1; i <= j; i++) {
        C[i][j] = (C[i][j - 1] + C[i - 1][j - 1]) % MOD;
      }
    }
  }
  long long solve(int x, int l, int r) {
    int st = 1, cnt = 0;
    long long res = 0;
    for (int i = x; i < r; i++) {
      if (i - x + 1 < l) continue;
      long long range = (a[i + 1] - 1) - (a[i - x] + 1) + 1;
      res += C[x][range];
      res %= MOD;
    }
    return res;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> k;
    pre_calc();
    for (int i = 1; i <= n; i++) {
      char c;
      cin >> c;
      if (c == '1') a[++dem] = i;
    }
    if ((k > dem) || k == 0) {
      cout << 1;
      return 0;
    }
    a[++dem] = n + 1;
    long long res = 0;
    int t = 1;
    res += solve(k, 1, dem);
    res -= solve(k - 1, 2, dem - 1);
    res = ((res % MOD) + MOD) % MOD;
    cout << res;
    return 0;
  }
  