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 inf = 1000000000000000000;
  const long long mod = 998244353;
  long long fact[1000005], n, k, pr, cnt, na, a[5005], ans;
  string s;
  long long bin_pow(long long x, long long step) {
    if (step == 0) return 1;
    if (step == 1) return x;
    if (step % 2 == 0) {
      long long p = bin_pow(x, step / 2);
      return ((p * p) % mod);
    }
    return ((x * bin_pow(x, step - 1)) % mod);
  }
  long long inv(long long x) { return bin_pow(x, mod - 2); }
  long long c(long long n, long long k) {
    return ((((fact[n] * inv(fact[n - k])) % mod) * inv(fact[k])) % mod);
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    fact[0] = fact[1] = 1;
    for (long long i = 2; i <= 1000000; ++i) fact[i] = (fact[i - 1] * i) % mod;
    cin >> n >> k;
    cin >> s;
    s = '$' + s;
    pr = cnt = 0;
    for (long long i = 1; i <= n; ++i) {
      if (s[i] == '1') a[++na] = i;
    }
    if (k > na || k == 0) {
      cout << 1;
      return 0;
    }
    if (k == na) {
      cout << c(n, k);
      return 0;
    }
    a[0] = 0;
    a[++na] = n + 1;
    for (long long i = k + 1; i <= na; ++i) {
      ans += c(a[i] - a[i - k - 1] - 1, k);
      ans %= mod;
      if (i > k + 1) {
        ans -= c(a[i - 1] - a[i - k - 1] - 1, k - 1);
        ans += mod;
        ans += mod;
        ans %= mod;
      }
    }
    cout << ans;
    return 0;
  }
  