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 int mod = 998244353;
  const long long int inf = 1e18;
  const long long int N = 5000;
  long long int factorialNumInverse[N + 1];
  long long int naturalNumInverse[N + 1];
  long long int fact[N + 1];
  void InverseofNumber(long long int p) {
    naturalNumInverse[0] = naturalNumInverse[1] = 1;
    for (long long int i = 2; i <= N; i++)
      naturalNumInverse[i] = naturalNumInverse[p % i] * (p - p / i) % p;
  }
  void InverseofFactorial(long long int p) {
    factorialNumInverse[0] = factorialNumInverse[1] = 1;
    for (long long int i = 2; i <= N; i++)
      factorialNumInverse[i] =
          (naturalNumInverse[i] * factorialNumInverse[i - 1]) % p;
  }
  void factorial(long long int p) {
    fact[0] = 1;
    for (long long int i = 1; i <= N; i++) {
      fact[i] = (fact[i - 1] * i) % p;
    }
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long int p = mod;
    InverseofNumber(p);
    InverseofFactorial(p);
    factorial(p);
    long long int n, k;
    string s;
    cin >> n >> k >> s;
    long long int l = 0, r = 0;
    vector<long long int> a;
    for (long long int i = 0; i < n; i++) {
      if (s[i] == '1') a.push_back(i);
    }
    if (a.size() < k || k == 0) {
      cout << 1 << "\n";
      return 0;
    }
    r = (a.size() == k ? n - 1 : a[k] - 1);
    long long int ans =
        (fact[r - l + 1] *
         (factorialNumInverse[k] * factorialNumInverse[r - l + 1 - k] % mod)) %
        mod;
    long long int pl = l, pr = r;
    bool done = 0;
    long long int g = k;
    while (g < a.size()) {
      pr = r;
      pl = l;
      l = a[g - k] + 1;
      r = (g + 1 == a.size() ? n - 1 : a[g + 1] - 1);
      long long int t1 =
          (fact[r - l + 1] *
           (factorialNumInverse[k] * factorialNumInverse[r - l + 1 - k] % mod)) %
          mod;
      long long int x = pr - l + 1;
      long long int t2 = (fact[x] * (factorialNumInverse[k - 1] *
                                     factorialNumInverse[x - k + 1] % mod)) %
                         mod;
      ans = (ans + t1 - t2 + mod) % mod;
      g++;
    }
    cout << ans << "\n";
    return 0;
  }
  