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>
  const long long mod = 998244353;
  using namespace std;
  bool isprime(long long k) {
    for (long long i = 2; i <= sqrt(k); i++)
      if (k % i == 0) return false;
    return true;
  }
  long long bm(long long a, long long b, long long mod) {
    if (b == 0) return 1;
    long long t = bm(a, b / 2, mod);
    t = t * t % mod;
    return (b % 2 == 1 ? t * a % mod : t);
  }
  long long inv(long long a, long long mod) { return bm(a, mod - 2, mod); }
  void gay(long long TC) { cout << "Case #" << TC << ": "; }
  long long ncr_pre[5005][5005];
  long long ncr(long long n, long long r) {
    if (n < r) return 0;
    return ncr_pre[n][r];
  }
  void solve(long long TC) {
    long long n, k;
    cin >> n >> k;
    string s;
    cin >> s;
    long long ans = 1;
    long long tc = 0;
    for (auto x : s) tc += x - '0';
    if (tc < k || k == 0) {
      cout << "1\n";
      return;
    }
    for (long long i = 0; i < n; i++) {
      if (i != 0 && s[i - 1] != '1') continue;
      long long j = i - 1, id = -1, cnt = 0;
      while (cnt <= k && !(j == n - 1 || (cnt == k && s[j + 1] == '1'))) {
        j++;
        cnt += s[j] - '0';
        if (s[j] == '1' && id == -1) id = j;
      }
      if (cnt == 0) continue;
      long long len = j - i + 1;
      long long len2 = j - id;
      ans += ncr(len, cnt) - ncr(len2, cnt - 1);
      ans %= mod;
    }
    ans = ans + mod;
    ans %= mod;
    cout << ans << '\n';
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    long long t = 1;
    for (long long i = 0; i <= 5000; i++) {
      for (long long j = 0; j <= i; j++) {
        if (j == 0 || j == i)
          ncr_pre[i][j] = 1;
        else
          ncr_pre[i][j] = ncr_pre[i - 1][j] + ncr_pre[i - 1][j - 1];
        ncr_pre[i][j] %= mod;
      }
    }
    for (long long i = 1; i <= t; i++) solve(i);
    return 0;
  }
  