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;
  long long int inf = 998244353;
  long long int facto[1000010];
  long long int power(long long int a, long long int pow) {
    long long int ans = 1;
    if (a < 0) a += inf;
    while (pow) {
      if (pow % 2 == 1) {
        ans = ((ans % inf) * (a % inf)) % inf;
      }
      a = ((a % inf) * (a % inf)) % inf;
      if (a < 0) a += inf;
      pow /= 2;
    }
    return ans;
  }
  void fact(int n) {
    long long int fac = 1;
    for (int i = 1; i <= n; i++) {
      fac = ((fac % inf) * (i % inf)) % inf;
      facto[i] = fac;
    }
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    facto[0] = 1;
    fact(6000);
    int n, k;
    cin >> n >> k;
    string s;
    cin >> s;
    vector<int> pos;
    for (int i = 0; i < n; i++) {
      if (s[i] == '1') pos.push_back(i);
    }
    if (pos.size() < k) {
      cout << 1 << "\n";
      return 0;
    }
    if (k == 0) {
      cout << 1 << "\n";
      return 0;
    }
    int i = 0, j = k - 1, sz = pos.size();
    long long int ans = 0;
    while (j < sz) {
      if (j != sz - 1) {
        int szz;
        if (i)
          szz = pos[j + 1] - 1 - pos[i - 1];
        else
          szz = pos[j + 1] - 1 - min(pos[i], 0) + 1;
        long long int tot = facto[szz];
        tot = ((tot % inf) * (power(facto[k], inf - 2) % inf)) % inf;
        tot = ((tot % inf) * (power(facto[szz - k], inf - 2) % inf)) % inf;
        szz = pos[j + 1] - 1 - pos[i];
        long long int tot1 = facto[szz];
        tot1 = ((tot1 % inf) * (power(facto[k - 1], inf - 2) % inf)) % inf;
        tot1 = ((tot1 % inf) * (power(facto[szz - k + 1], inf - 2) % inf)) % inf;
        ans = (ans + tot - tot1) % inf;
      } else {
        int szz;
        if (i)
          szz = max(pos[j], n - 1) - pos[i - 1];
        else
          szz = max(pos[j], n - 1) - min(pos[i], 0) + 1;
        long long int tot = facto[szz];
        tot = ((tot % inf) * (power(facto[k], inf - 2) % inf)) % inf;
        tot = ((tot % inf) * (power(facto[szz - k], inf - 2) % inf)) % inf;
        ans = (ans + tot) % inf;
      }
      j += 1;
      i += 1;
    }
    cout << (ans + inf) % inf << "\n";
    return 0;
  }
  