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 INF = 0x3f3f3f3f;
  const long long LINF = 0x3f3f3f3f3f3f3f3fLL;
  const int MAX = 5e3 + 10;
  const int MOD = 998244353;
  long long fat[MAX], ifat[MAX];
  long long choose(int a, int b) {
    if (a < b) return 0;
    return fat[a] * ifat[b] % MOD * ifat[a - b] % MOD;
  }
  long long fexp(long long a, long long e) {
    long long ans = 1;
    while (e) {
      if (e & 1) ans = ans * a % MOD;
      a = a * a % MOD;
      e /= 2;
    }
    return ans;
  }
  void compute() {
    fat[0] = 1;
    ifat[0] = 1;
    for (int i = 1; i < MAX; i++) {
      fat[i] = i * fat[i - 1] % MOD;
      ifat[i] = fexp(fat[i], MOD - 2);
    }
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    ;
    compute();
    int n, k;
    cin >> n >> k;
    string s;
    cin >> s;
    if (k == 0 || count(s.begin(), s.end(), '1') < k) {
      cout << 1 << '\n';
      return 0;
    }
    long long ans = 1;
    for (int l = 0; l < n; l++) {
      int cnt = 0, r = l;
      while (r < n && cnt + (s[r] == '1') <= k) cnt += s[r++] == '1';
      long long add = 0;
      if (s[l] == '0' && cnt > 0) add = choose((r - l) - 1, cnt - 1);
      if (s[l] == '1' && ((r - l) - cnt) > 0) add = choose((r - l) - 1, cnt);
      ans = (ans + add) % MOD;
    }
    cout << ans << '\n';
    return 0;
  }
  