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 MAXN = 5002;
  const long long mod = 998244353;
  long long fact[MAXN];
  long long rfact[MAXN];
  long long fastpow(long long x, long long p) {
    if (p == 0) return 1;
    if (p == 1) return x;
    long long t = fastpow(x, p / 2);
    t *= t;
    t %= mod;
    if (p % 2 == 1) {
      t *= x;
      t %= mod;
    }
    return t;
  }
  void precompute() {
    fact[0] = 1;
    rfact[0] = fastpow(fact[0], mod - 2);
    for (long long i = 1; i < MAXN; i++) {
      fact[i] = fact[i - 1] * i;
      fact[i] %= mod;
      rfact[i] = fastpow(fact[i], mod - 2);
    }
  }
  long long comb(long long n, long long k) {
    if (k > n) return 0;
    long long ret = fact[n];
    ret *= rfact[k];
    ret %= mod;
    ret *= rfact[n - k];
    ret %= mod;
    return ret;
  }
  vector<pair<long long, long long> > segments;
  long long ans = 1, n, k;
  long long pref[MAXN];
  string s;
  int main() {
    precompute();
    cin >> n >> k;
    long long l, r = -1, cnt1 = 0;
    cin >> s;
    for (int i = 0; i < s.size(); i++) {
      pref[i + 1] = pref[i] + (s[i] == '1');
    }
    for (l = 0; l < s.size(); l++) {
      bool moved_r = false;
      while (r + 1 < s.size() && cnt1 + (s[r + 1] == '1') <= k) {
        r++;
        cnt1 += (s[r] == '1');
        moved_r = 1;
      }
      if (moved_r && cnt1 == k) segments.push_back({l, r});
      if (s[l] == '1') cnt1--;
    }
    for (int i = 0; i < segments.size(); i++) {
      ans += comb(segments[i].second - segments[i].first + 1, k) + mod - 1;
      ans %= mod;
    }
    for (int i = 1; i < segments.size(); i++) {
      int l1 = segments[i].first, r1 = segments[i - 1].second;
      if (l1 > r1) continue;
      ans += mod - comb(r1 - l1 + 1, pref[r1 + 1] - pref[l1]) + 1;
      ans %= mod;
    }
    cout << ans << endl;
    return 0;
  }
  