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 N = 1e6 + 100;
  const long long mod = 998244353;
  long long t, n, m, k;
  long long a[N];
  string s;
  vector<long long> pos;
  void init() { return; }
  long long quickpow(long long a, long long b) {
    long long res = 1;
    while (b > 0) {
      if (b & 1) res = res * a % mod;
      b >>= 1;
      a = a * a % mod;
    }
    return res;
  }
  long long C(long long n, long long m) {
    if (n < m) return 0;
    long long a = 1, b = 1;
    m = min(m, n - m);
    for (long long i = 1; i <= m; i++) {
      (a *= n - i + 1) %= mod;
      (b *= i) %= mod;
    }
    return a % mod * quickpow(b, mod - 2) % mod;
  }
  void solve() {
    init();
    cin >> s;
    s = "." + s;
    pos.clear();
    pos.push_back(0);
    for (long long i = 1; i <= n; i++) {
      if (s[i] == '1') pos.push_back(i);
    }
    pos.push_back(n + 1);
    if (k == 0 || pos.size() - 2 < k) {
      cout << 1 << '\n';
      return;
    }
    long long ans = 0;
    for (long long l = k - 1; l <= k; l++) {
      long long sum = 0;
      for (long long i = k - l; i + k + 1 < pos.size(); i++) {
        sum = (sum + C(pos[i + l + 1] - 1 - pos[i], l)) % mod;
      }
      if ((k - l) % 2 == 0)
        ans = (ans + sum) % mod;
      else
        ans = (ans - sum + mod) % mod;
    }
    cout << ans << '\n';
    return;
  }
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    while (cin >> n >> k) {
      solve();
    }
    return 0;
  }
  