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;
  long long Power(long long aaa, long long bbb) {
    long long result = 1;
    aaa %= 998244353;
    while (bbb) {
      if (bbb % 2 == 1) result *= aaa;
      aaa *= aaa;
      aaa %= 998244353;
      result %= 998244353;
      bbb /= 2;
    }
    return result;
  }
  long long Mod_Inv(long long aaa) {
    aaa %= 998244353;
    aaa = Power(aaa, 998244353 - 2);
    return aaa;
  }
  long long Factorial[1000001];
  int Make_Factorial() {
    Factorial[0] = 1;
    for (long long i = 1; i < 1000001; i++) {
      Factorial[i] = i * Factorial[i - 1];
      Factorial[i] %= 998244353;
    }
    return 0;
  }
  int mk_fact = Make_Factorial();
  vector<long long> fmoe(1000001);
  int fmodf() {
    fmoe[1000001 - 1] = Mod_Inv(Factorial[1000001 - 1]);
    for (long long i = 1000001 - 2; i >= 0; i--) {
      fmoe[i] = fmoe[i + 1] * (i + 1);
      fmoe[i] %= 998244353;
    }
    return 0;
  }
  int fmoed = fmodf();
  long long nCr(long long nnn, long long rrr) {
    if (nnn < rrr) return 0;
    long long aans = Factorial[nnn];
    aans %= 998244353;
    aans *= fmoe[rrr];
    aans %= 998244353;
    aans *= fmoe[nnn - rrr];
    aans %= 998244353;
    return aans;
  }
  long long nPr(long long nnn, long long rrr) {
    long long aans = Factorial[nnn];
    aans %= 998244353;
    aans *= fmoe[nnn - rrr];
    aans %= 998244353;
    return aans;
  }
  void sol() {
    long long n, k;
    cin >> n >> k;
    string s;
    cin >> s;
    if (k == 0) {
      cout << 1 << endl;
      return;
    }
    long long o = 0, z = 0;
    vector<long long> p;
    p.push_back(-1);
    for (int i = 0; i < n; i++) {
      if (s[i] == '1') p.push_back(i);
    }
    p.push_back(n);
    int u = p.size();
    if (u - 2 < k) {
      cout << 1 << endl;
      return;
    }
    long long ans = 0;
    for (int i = 0; i + k < u - 1; i++) {
      long long n0 = (p[i + k + 1] - p[i] - 1) - k;
      long long tot = p[i + k + 1] - p[i] - 1;
      ans += nCr(tot, n0);
      ans %= 998244353;
    }
    for (int i = 1; i + k - 1 < u - 2; i++) {
      long long n0 = (p[i + k] - p[i] - 1) - (k - 1);
      long long tot = p[i + k] - p[i] - 1;
      ans += 998244353 - nCr(tot, n0);
      ans %= 998244353;
    }
    cout << ans << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    sol();
    return 0;
  }
  