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 = 1e5 + 1;
  bool notPrime[N];
  vector<int> primes;
  void calc_prime() {
    notPrime[1] = true;
    notPrime[2] = false;
    for (int i = 2; i * i < N; ++i) {
      if (!notPrime[i]) {
        primes.push_back(i);
        int j = i * i;
        while (j < N) {
          notPrime[j] = true;
          j += i;
        }
      }
    }
  }
  long long mod = 998244353;
  long long cnk[5001][5001];
  void prep(int k) {
    cnk[0][0] = 1;
    for (int j = 1; j <= k; ++j) {
      cnk[j][0] = 1;
      for (int i = 1; i <= j; ++i) {
        cnk[j][i] = (cnk[j - 1][i - 1] + cnk[j - 1][i]) % mod;
      }
    }
  }
  int main() {
    int n, k;
    cin >> n >> k;
    if (k == 0) {
      cout << 1 << endl;
      return 0;
    }
    prep(n);
    string s;
    cin >> s;
    long long ans = 0;
    vector<long long> p(n);
    long long pref = 0;
    for (int i = 0; i < n; ++i) {
      p[i] = pref;
      if (s[i] == '1') {
        ++p[i];
      }
      pref = p[i];
    }
    int l = -1, r = -1;
    for (int(i) = 0; (i) < (n); ++(i)) {
      if (p[i] == k) {
        r = i;
        l = 0;
      }
    }
    if (l == -1) {
      cout << 1 << endl;
      return 0;
    }
    ans += cnk[r - l + 1][k];
    int pr = r, pl = l;
    vector<long long> a;
    a.push_back(ans);
    while (r < n - 1) {
      ++r;
      while (r < n - 1 && s[r + 1] == '0') {
        ++r;
      }
      while (s[l] != '1') {
        ++l;
      }
      ++l;
      if (l > pr) {
        ans = (ans + cnk[r - l + 1][k]) % mod;
        ans = (ans + mod - 1) % mod;
        a.push_back(ans);
        pr = r;
        pl = l;
        continue;
      }
      long long cur = cnk[pr - l + 1][k - 1];
      long long nw = cnk[r - l + 1][k];
      ans += (nw - cur + mod) % mod;
      ans %= mod;
      pr = r;
      pl = l;
    }
    cout << ans << endl;
  }
  