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 maxn = 1e6 + 9;
  const long long MOD = 1e9 + 7;
  const int mod = 998244353;
  inline long long qpow(long long b, long long e, long long m) {
    long long a = 1;
    for (; e; e >>= 1, b = b * b % m)
      if (e & 1) a = a * b % m;
    return a;
  }
  long long exgcd(long long a, long long b, long long &x, long long &y) {
    if (b == 0) {
      x = 1, y = 0;
      return a;
    }
    long long d = exgcd(b, a % b, x, y);
    long long z = x;
    x = y, y = z - y * (a / b);
    return d;
  }
  char s[maxn];
  long long C[5009][5009];
  void init() {
    C[0][0] = 1;
    for (int i = 1; i <= 5000; i++) {
      C[i][0] = 1;
      for (int j = 1; j <= i; j++) {
        C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
        C[i][j] %= mod;
      }
    }
  }
  vector<int> ve;
  int main() {
    init();
    int n, k;
    scanf("%d%d", &n, &k);
    scanf("%s", s + 1);
    if (k == 0) {
      cout << "1" << '\n';
      return 0;
    }
    ve.push_back(0);
    for (int i = 1; i <= n; i++) {
      if (s[i] == '1') ve.push_back(i);
    }
    ve.push_back(n + 1);
    if (ve.size() - 2 < k) {
      cout << 1 << '\n';
      return 0;
    }
    long long ans = 0;
    for (int i = 1;; i++) {
      if (i + k >= ve.size()) break;
      ans += C[ve[i + k] - ve[i - 1] - 1][k];
      if (i + k < (int)ve.size() - 1) ans -= C[ve[i + k] - ve[i] - 1][k - 1];
      ans %= mod;
    }
    ans = (ans % mod + mod) % mod;
    cout << ans << '\n';
  }
  