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 = 2 * 1e5 + 10;
  const int mod = 998244353;
  inline int add(int a, int b) { return a + b >= mod ? a + b - mod : a + b; }
  inline int mul(int a, int b) { return a * 1ll * b % mod; }
  long long f[N];
  long long fp(long long b, long long p) {
    if (p == 0) return 1;
    return fp(b * b % mod, p / 2) * (p & 1 ? b : 1) % mod;
  }
  long long inv(long long x) { return fp(x, mod - 2); }
  long long nCk(long long n, long long k) {
    return f[n] * inv(f[n - k]) % mod * inv(f[k]) % mod;
  }
  void Ineedhelp() {
    f[0] = 1;
    for (int i = 1; i < N; ++i) {
      f[i] = f[i - 1] * i % mod;
    }
    int n, k;
    string s;
    cin >> n >> k >> s;
    int i = 0, j = -1, cnt1 = 0, last1 = -1;
    while (j < n - 1 && cnt1 + (s[j + 1] == '1') <= k) {
      ++j;
      cnt1 += s[j] == '1';
      if (s[j] == '1') last1 = j;
    }
    if (cnt1 < k || k == 0) {
      cout << "1\n";
      return;
    }
    long long ways = nCk(j - i + 1, k);
    while (j < n - 1) {
      int jj = j;
      ++j;
      while (j < n - 1 && s[j + 1] == '0') ++j;
      while (i < j && s[i] == '0') ++i;
      ++i;
      ways = ways - nCk(jj - i + 1, k - 1) + mod;
      ways %= mod;
      ways += nCk(j - i + 1, k);
      ways %= mod;
    }
    cout << ways << '\n';
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int t = 1;
    while (t--) Ineedhelp();
  }
  