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 long long mod = 998244353, siz = 1e6 + 5;
  long long t, n, m, k, a[siz], fact[siz], dp[siz];
  long long add(long long x, long long y) {
    x += y;
    x %= mod;
    return x;
  }
  long long sub(long long x, long long y) {
    x -= y;
    x += mod;
    x %= mod;
    return x;
  }
  long long mul(long long x, long long y) {
    x *= y;
    x %= mod;
    return x;
  }
  long long bix(long long a, long long n) {
    long long ret = 1;
    while (n > 0) {
      if (n & 1) (ret *= a) %= mod;
      (a *= a) %= mod;
      n /= 2;
    }
    return ret;
  }
  long long div(long long x) {
    long long ret = bix(x, mod - 2);
    return ret;
  }
  void pre() {
    fact[0] = 1;
    for (long long i = 1; i < siz; i++) fact[i] = (fact[i - 1] * i) % mod;
  }
  long long ncr(long long n, long long r) {
    if (r > n or n < 0) return 0;
    if (n == 0 and r == 0) return 1;
    long long ret = fact[n];
    (ret *= bix(fact[r], mod - 2)) %= mod;
    (ret *= bix(fact[n - r], mod - 2)) %= mod;
    return ret;
  }
  int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    pre();
    cin >> n >> k;
    string s;
    cin >> s;
    long long ones = 0;
    for (long long i = 1; i <= n; i++) {
      a[i] = s[i - 1] - '0';
      ones += a[i];
    }
    if (k < 1 or k > ones) return cout << 1 << endl, 0;
    bool first_chunk = 1;
    ones = 0;
    for (long long i = 1; i <= n; i++) {
      ones += a[i];
      if (ones < k) continue;
      long long cur = 0, stops = i;
      for (long long j = i; j >= 1; j--) {
        cur += a[j];
        if (cur > k) break;
        stops = j;
      }
      long long len = i - stops + 1;
      if (first_chunk) {
        dp[i] = ncr(len, k);
        first_chunk = 0;
      } else {
        if (a[i]) {
          dp[i] += dp[i - 1];
          dp[i] += ncr(len - 1, k);
          dp[i] %= mod;
        } else {
          dp[i] += dp[i - 1];
          dp[i] += ncr(len - 1, k - 1);
          dp[i] %= mod;
        }
      }
    }
    cout << dp[n] << endl;
  }
  