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;
  template <typename T>
  T gcd(T a, T b) {
    if (a % b) return gcd(b, a % b);
    return b;
  }
  template <typename T>
  T lcm(T a, T b) {
    return (a * (b / gcd(a, b)));
  }
  vector<string> tokenizer(string str, char ch) {
    std::istringstream var((str));
    vector<string> v;
    string t;
    while (getline((var), t, (ch))) {
      v.push_back(t);
    }
    return v;
  }
  void err(istream_iterator<string> it) {}
  template <typename T, typename... Args>
  void err(istream_iterator<string> it, T a, Args... args) {
    cout << *it << " = " << a << "\n";
    err(++it, args...);
  }
  void file_i_o() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  }
  long long solve(string s, long long k) {
    if (!k) return 1;
    long long n = s.length();
    vector<long long> dp(n + 1), lenofsubstr(n);
    dp[0] = 1;
    long long cnt = 0, si = 0;
    for (long long i = 0; i < n; i++) {
      if (s[i] == '1') cnt++;
      if (cnt < k)
        lenofsubstr[i] = -1;
      else {
        while (cnt > k) {
          if (s[si] == '1') cnt--;
          si++;
        }
        lenofsubstr[i] = i - si + 1;
      }
    }
    vector<vector<long long> > comb(5001, vector<long long>(5001, 1));
    for (long long i = 1; i <= n; i++) {
      for (long long j = 1; j <= i; j++) {
        if (i == j) continue;
        comb[i][j] =
            (comb[i - 1][j - 1] % 998244353 + comb[i - 1][j] % 998244353) %
            998244353;
      }
    }
    for (long long i = 0; i < n; i++) {
      if (lenofsubstr[i] == -1)
        dp[i + 1] = 1;
      else if (lenofsubstr[i] == i + 1 && s[i] == '1')
        dp[i + 1] = comb[i + 1][k];
      else {
        dp[i + 1] = dp[i];
        if (s[i] == '1' && lenofsubstr[i] != k)
          dp[i + 1] += comb[lenofsubstr[i] - 1][k];
        else if (s[i] == '0')
          dp[i + 1] += comb[lenofsubstr[i] - 1][k - 1];
        dp[i + 1] %= 998244353;
      }
    }
    return dp[n];
  }
  int32_t main() {
    clock_t begin = clock();
    file_i_o();
    long long n, k;
    cin >> n >> k;
    string s;
    cin >> s;
    cout << solve(s, k) << "\n";
    return 0;
  }
  