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 calculate(long long int p, long long int q) {
    long long int modx = 998244353, expo;
    expo = modx - 2;
    while (expo) {
      {
        if (expo & 1) p = (p * q) % modx;
      }
      q = (q * q) % modx;
      expo >>= 1;
    }
    return p % modx;
  }
  long long int power(long long int x, long long int y, long long int p) {
    long long int res = 1;
    x = x % p;
    if (x == 0) return 0;
    while (y > 0) {
      if (y & 1) {
        res = (res * x) % p;
      }
      y = y >> 1;
      x = (x * x) % p;
    }
    return res;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long int t;
    t = 1;
    while (t--) {
      long long int n, k;
      cin >> n >> k;
      string s;
      long long int ans = 1;
      cin >> s;
      if (k == 0) {
        cout << 1 << "\n";
        continue;
      }
      long long int i = 0, j = 0;
      long long int dp[n + 5][k + 5];
      memset(dp, 0, sizeof(dp));
      for (int p = 1; p <= k; p++) {
        dp[0][p] = dp[0][p - 1] + 1;
      }
      for (int p = 1; p <= n; p++) {
        for (int q = 1; q <= k; q++)
          dp[p][q] = (dp[p - 1][q] + dp[p][q - 1]) % 998244353;
      }
      long long int zero = 0, lastzero = 0, one = 0, b = 0;
      while (i < n) {
        if (s[i] == '0') {
          zero++;
          if (one == k) lastzero++;
        } else
          one++;
        if (one > k) {
          for (int p = 0; p < zero; p++) {
            ans = (ans + dp[p][k]) % 998244353;
          }
          long long int temp = 0;
          if (b > 0) {
            for (int p = 0; p < zero - lastzero; p++) {
              temp = (temp + dp[p][k - 1]) % 998244353;
            }
          }
          lastzero = 0;
          b = 1;
          ans = (998244353 + ans - temp) % 998244353;
          while (j < i) {
            if (s[j] == '1') {
              one--;
              j++;
              break;
            }
            j++;
            zero--;
          }
        }
        i++;
      }
      if (one == k) {
        for (int p = 0; p < zero; p++) {
          ans = (ans + dp[p][k]) % 998244353;
        }
        long long int temp = 0;
        if (b > 0) {
          for (int p = 0; p < zero - lastzero; p++) {
            temp = (temp + dp[p][k - 1]) % 998244353;
          }
        }
        ans = (998244353 + ans - temp) % 998244353;
      }
      cout << ans << "\n";
    }
    return 0;
  }
  