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 mod = 998244353;
  const long double pi = acos(-1);
  int xc[] = {0, -1, 0, 1, -1, -1, 1, 1};
  int yc[] = {1, 0, -1, 0, -1, 1, -1, 1};
  string Q;
  long long bpow(long long n, long long x) {
    return x <= 0 ? 1 : bpow(n * n % mod, x / 2) % mod * (x % 2 ? n : 1) % mod;
  }
  long long modinverse(long long b) { return bpow(b, mod - 2); }
  long long factN = 1e4 + 1;
  vector<long long> fact(factN, 1);
  void factinit() {
    for (int i = 1; i < factN; ++i) {
      fact[i] = i * fact[i - 1];
      fact[i] %= mod;
    }
  }
  void solve(long long tc) {
    factinit();
    long long n, k;
    cin >> n >> k;
    string s;
    cin >> s;
    vector<long long> v(n + 1);
    for (int i = 1; i <= n; ++i) {
      if (s[i - 1] == '1') v[i] = 1;
      v[i] += v[i - 1];
    }
    if (!k || k > v[n]) {
      cout << "1\n";
      return;
    }
    long long sum = 0, oldj = 0;
    for (int i = 0, j = 1; i <= n;) {
      while (j <= n && v[j] - v[i] <= k) j++;
      j--;
      if (v[j] - v[i] < k) break;
      sum += fact[j - i] * modinverse(fact[k]) % mod *
             modinverse(fact[j - i - k]) % mod;
      if (oldj != 0)
        sum -= fact[oldj - i] * modinverse(fact[k - 1]) % mod *
               modinverse(fact[oldj - i - (k - 1)]) % mod;
      oldj = j;
      while (i <= n && v[j] - v[i] == k) i++;
    }
    cout << (sum % mod + mod) % mod << "\n";
  }
  int main() {
    cin.tie(0);
    cin.sync_with_stdio(0);
    cout.tie(0);
    ;
    long long t = 1, tc = 1;
    while (t--) solve(tc++);
    return 0;
  }
  