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 N = 3e5 + 10, M = 2 * N, mod = 998244353;
  long long inv[N], fac[N], ifac[N];
  void init(int n) {
    inv[1] = fac[0] = fac[1] = ifac[0] = ifac[1] = 1;
    for (int i = 2; i <= n; i++) {
      inv[i] = (mod - mod / i) * inv[mod % i] % mod;
      fac[i] = fac[i - 1] * i % mod;
      ifac[i] = ifac[i - 1] * inv[i] % mod;
    }
  }
  long long C(int m, int n) { return fac[m] * ifac[n] % mod * ifac[m - n] % mod; }
  int pre[N], fix[N];
  string s;
  int pos[N];
  void solve() {
    int n, k;
    cin >> n >> k >> s;
    s = " " + s;
    int last = 1;
    for (int i = 1; i <= n; i++) {
      if (s[i] == '1') pre[i] = last, last = i + 1;
    }
    last = n;
    for (int i = n; i >= 1; i--) {
      if (s[i] == '1') fix[i] = last, last = i - 1;
    }
    int cnt = 0;
    for (int i = 1; i <= n; i++)
      if (s[i] == '1') pos[++cnt] = i;
    if (cnt < k || k == 0) {
      cout << 1 << '\n';
      return;
    }
    long long res = 0;
    for (int i = 1; i <= cnt - k + 1; i++) {
      int L = pre[pos[i]], R = fix[pos[i + k - 1]];
      int L1 = pre[pos[i + 1]], R1 = R;
      res += C(R - L + 1, k);
      if (i != cnt - k + 1) res -= C(R1 - L1 + 1, k - 1);
      res = (res + mod) % mod;
    }
    cout << res << '\n';
  }
  int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    cout.precision(2), cout.setf(ios::fixed);
    init(1e4 + 10);
    solve();
    return 0;
  }
  