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 maxN = 1e4;
  long long F[maxN + 1];
  void fact() {
    F[0] = F[1] = 1;
    for (int i = 2; i <= maxN; i++) F[i] = (1LL * F[i - 1] * i) % 998244353;
  }
  long long power(long long a, long long b) {
    long long res = 1;
    while (b) {
      if (b & 1) {
        res = (1LL * res * a) % 998244353, b--;
      } else {
        a = (1LL * a * a) % 998244353, b >>= 1;
      }
    }
    return res;
  }
  long long C(long long n, long long r) {
    if (r > n) return 0;
    long long res = F[n];
    res = (1LL * res * power(F[r], 998244353 - 2)) % 998244353;
    res = (1LL * res * power(F[n - r], 998244353 - 2)) % 998244353;
    return res;
  }
  void solve() {
    long long n, k;
    cin >> n >> k;
    string second;
    cin >> second;
    if (!k) {
      cout << "1\n";
      return;
    }
    long long ans = 1, l = 0, r = 0, cnt = 0;
    while (r < n) {
      if (second[r] == '1') {
        if (cnt == k) {
          (ans += (C(r - l, k) - 1 + 998244353) % 998244353) %= 998244353;
          while (l < n && second[l] != '1') l++;
          l++, cnt--;
          ans = (ans - C(r - l, k - 1) + 1 + 998244353) % 998244353;
        }
        cnt++;
      }
      r++;
    }
    if (cnt == k) (ans += (C(r - l, k) - 1 + 998244353) % 998244353) %= 998244353;
    cout << ans << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0), cout.tie(0), fact();
    solve();
    return 0;
  }
  