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 = 5010;
  const int mod = 998244353;
  long long read() {
    long long x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') x = x * 10ll + ch - '0', ch = getchar();
    return x * f;
  }
  int n, k, pw[N], a[N];
  char s[N];
  void up(int& x, int y) {
    x += y;
    if (x >= mod) x -= mod;
  }
  int Pow(int x, int y) {
    if (!y) return 1;
    int t = Pow(x, y >> 1);
    t = (long long)t * t % mod;
    if (y & 1) t = (long long)t * x % mod;
    return t;
  }
  int fac[N], inv[N], fin[N];
  void init() {
    fac[0] = 1;
    for (int i = 1; i < N; i++) fac[i] = (long long)fac[i - 1] * i % mod;
    inv[0] = inv[1] = 1;
    for (int i = 2; i < N; i++)
      inv[i] = (long long)(mod - mod / i) * inv[mod % i] % mod;
    fin[0] = 1;
    for (int i = 1; i < N; i++) fin[i] = (long long)fin[i - 1] * inv[i] % mod;
  }
  int C(int n, int m) {
    if (n < m) return 0;
    return (long long)fac[n] * fin[m] % mod * fin[n - m] % mod;
  }
  int main() {
    n = read(), k = read();
    init();
    int l = 0;
    scanf("%s", s + 1);
    a[l++] = 0;
    for (int i = 1; i <= n; i++)
      if (s[i] == '1') a[l++] = i;
    a[l++] = n + 1;
    if (!k || k > l - 2) return puts("1"), 0;
    int ans = 0;
    for (int i = 0; i + 1 + k < l; i++) {
      up(ans, C(a[i + 1 + k] - a[i] - 1, k));
      if (i + 1 + k != l - 1)
        up(ans, mod - C(a[i + 1 + k] - a[i + 1] - 1, k - 1));
    }
    cout << ans;
  }
  