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 = 5005, mod = 998244353;
  int n, k, sum, ans, tem;
  int fac[N], inv[N], pre[N], suf[N];
  char s[N];
  int C(int i, int j) {
    if (i < j) return 0;
    return 1ll * fac[i] * inv[j] % mod * inv[i - j] % mod;
  }
  int qp(int x, int y) {
    int res = 1;
    while (y) {
      if (y % 2) res = 1ll * res * x % mod;
      x = 1ll * x * x % mod;
      y >>= 1;
    }
    return res;
  }
  void add(int& x, int y) {
    x += y;
    if (x >= mod) x -= mod;
  }
  int main() {
    scanf("%d%d", &n, &k);
    scanf("%s", s + 1);
    ans = 1;
    for (int i = 1; i <= n; ++i) {
      pre[i] = pre[i - 1] + s[i] - '0';
    }
    for (int i = n; i >= 1; --i) suf[i] = suf[i + 1] + s[i] - '0';
    if (pre[n] < k) {
      cout << 1;
      return 0;
    }
    fac[0] = 1;
    for (int i = 1; i <= n; ++i) fac[i] = 1ll * i * fac[i - 1] % mod;
    inv[n] = qp(fac[n], mod - 2);
    for (int i = n - 1; i >= 0; --i) inv[i] = 1ll * inv[i + 1] * (i + 1) % mod;
    for (int i = 1; i <= n; ++i) {
      for (int j = i + 1; j <= n; ++j) {
        tem = pre[j] - pre[i - 1];
        if (tem > k) continue;
        if (s[i] == '0') tem--;
        if (s[j] == '0') tem--;
        add(ans, C(j - i - 1, tem));
      }
    }
    printf("%d\n", ans);
  }
  