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 long long xxx = 5e3;
  const long long maxn = xxx + 5;
  const long long inf = 0x3f3f3f3f3f3f3f3f;
  const long long mod = 998244353;
  const double pi = acos(-1.0);
  inline void updmin(long long& a, long long b) { a > b ? a = b : 0; }
  inline void updmax(long long& a, long long b) { a < b ? a = b : 0; }
  long long fac[maxn], pos[maxn];
  char s[maxn];
  long long inv(long long x) {
    x %= mod;
    if (x == 1) {
      return 1;
    }
    return (-mod / x * inv(mod % x) % mod + mod) % mod;
  }
  long long C(long long n, long long m) {
    return fac[n] * inv(fac[m] * fac[n - m] % mod) % mod;
  }
  int main(int argc, char const* argv[]) {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int cases;
    cases = 1;
    for (int cas = 1; cas <= cases; cas++) {
      long long n, k;
      cin >> n >> k;
      cin >> (s + 1);
      pos[0] = 0;
      long long cnt1 = 0;
      fac[0] = 1;
      for (long long i = 1; i <= n; i++) {
        if (s[i] == '1') {
          pos[++cnt1] = i;
        }
        fac[i] = fac[i - 1] * i % mod;
      }
      pos[cnt1 + 1] = n + 1;
      if (k == 0 || k > cnt1) {
        cout << 1 << "\n";
        continue;
      }
      long long ans = 0;
      for (long long i = k; i <= cnt1; i++) {
        if (i == k) {
          ans += C(pos[i + 1] - pos[i - k] - 1, k);
        } else {
          ans += (C(pos[i + 1] - pos[i - k] - 1, k) -
                  C(pos[i] - pos[i - k] - 1, k - 1) + mod) %
                 mod;
        }
        ans %= mod;
      }
      cout << ans << "\n";
    }
    return 0;
  }
  