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;
  inline long long read() {
    long long x = 0, f = 1;
    char c = getchar();
    for (; !isdigit(c); c = getchar())
      if (c == '-') f = -1;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
    return x * f;
  }
  inline void out(int x) {
    if (x > 9) out(x / 10);
    putchar(x % 10 ^ 48);
  }
  inline void print(int x) {
    if (x < 0) putchar('-'), x = -x;
    out(x), puts("");
  };
  const int N = 5005, p = 998244353;
  char s[N];
  long long jc[N], inv[N];
  inline int spow(long long x, int y, long long res = 1) {
    while (y) {
      if (y & 1) (res *= x) %= p;
      (x *= x) %= p, y >>= 1;
    }
    return res % p;
  }
  inline int C(int x, int y) { return jc[x] * inv[y] % p * inv[x - y] % p; }
  int sum[N];
  void solve() {
    int n = read(), k = read(), ans = 1;
    scanf("%s", s + 1), jc[0] = 1;
    for (int i = 1; i <= n; i++)
      sum[i] = sum[i - 1] + s[i] - 48, jc[i] = jc[i - 1] * i % p;
    inv[n] = spow(jc[n], p - 2);
    for (int i = n - 1; ~i; i--) inv[i] = inv[i + 1] * (i + 1) % p;
    if (sum[n] < k) return (void)puts("1");
    for (int i = 1; i <= n; i++)
      for (int j = i + 1; j <= n; j++) {
        int cnt = sum[j] - sum[i - 1];
        if (cnt <= k) {
          if (s[i] == '0') cnt--;
          if (s[j] == '0') cnt--;
          if (cnt >= 0 && cnt <= j - i - 1) (ans += C(j - i - 1, cnt)) %= p;
        }
      }
    cout << ans;
  }
  int main() {
    int T = 1;
    while (T--) solve();
  }
  