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;
  const int mod = 998244353;
  void bmod(int &x) { x -= mod, x += x >> 31 & mod; }
  int norm(int x) { return x < 0 ? x + mod : x; }
  void Add(int &x, int y) { x += y - mod, x += x >> 31 & mod; }
  void Sub(int &x, int y) { x -= y, x += x >> 31 & mod; }
  int inv(int x, int k = mod - 2) {
    int r = 1;
    while (k) {
      if (k & 1) r = (long long)x * r % mod;
      x = (long long)x * x % mod;
      k >>= 1;
    }
    return r;
  }
  const int Flen = N;
  int fact[Flen], infact[Flen];
  void preF(int n = Flen - 1) {
    fact[0] = 1;
    for (int i = 1; i <= n; ++i) fact[i] = (long long)fact[i - 1] * i % mod;
    infact[n] = inv(fact[n]);
    for (int i = n; i; --i) infact[i - 1] = (long long)infact[i] * i % mod;
  }
  int C(int x, int y) {
    return y < 0 || x < y
               ? 0
               : (long long)fact[x] * infact[y] % mod * infact[x - y] % mod;
  }
  int n, m;
  char str[N];
  int f[N], s[N];
  int main() {
    preF();
    scanf("%d%d%s", &n, &m, str + 1);
    if (!m) return puts("1"), 0;
    f[0] = 1;
    for (int i = 1; i <= n; ++i) {
      s[i] = s[i - 1] + (str[i] -= 48);
    }
    for (int i = 1, j = 0; i <= n; ++i) {
      f[i] = f[i - 1];
      if (str[i]) {
        while (j < i && s[i] - s[j] > m) ++j;
        if (s[i] - s[j] != m) continue;
        if (s[j])
          Add(f[i], C(i - j - 1, m));
        else
          Add(f[i], C(i - j, m) - 1);
      } else {
        if (s[i] - s[j] != m) continue;
        Add(f[i], C(i - j - 1, m - 1));
      }
    }
    printf("%d\n", f[n]);
    return 0;
  }
  