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 int N = 5e3 + 10;
  const int Mod = 998244353;
  long long n, t, k;
  long long dp[N];
  string s;
  long long ans = 1;
  long long fact[N], inv[N];
  void Entekhab();
  long long Tavan(long long, long long);
  int Check(string);
  long long Ent(long long, long long);
  int main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> n >> k >> s;
    Entekhab();
    if (k == 0 or Check(s) < k) {
      cout << 1 << endl;
      exit(0);
    }
    for (int i = 0; i < n; i++) {
      int l = i, r = i, cnt1 = 0, cnt0 = 0;
      bool seen0 = false, seen1 = false;
      for (int j = i; j < n; j++) {
        if (s[j] == '1')
          cnt1++, seen1 = true;
        else
          seen0 = true, cnt0++;
        if (j == n - 1 or (cnt1 == k and s[j + 1] == '1')) {
          r = j;
          break;
        }
      }
      if (s[i] == '1' and seen0) {
        ans += Ent(cnt1, r - l);
        ans %= Mod;
      } else if (s[i] == '0' and seen1) {
        ans += Ent(cnt0, r - l);
        ans %= Mod;
      }
    }
    cout << ans << endl;
  }
  int Check(string s) {
    int res = 0;
    for (int i = 0; i < s.size(); i++)
      if (s[i] == '1') res++;
    return res;
  }
  void Entekhab() {
    fact[0] = 1, inv[0] = 1;
    for (long long i = 1; i < N; i++)
      fact[i] = (fact[i - 1] * i) % Mod, inv[i] = Tavan(fact[i], (Mod - 2));
  }
  long long Ent(long long x, long long y) {
    if (x > y) return 0;
    if (y < 0 or x < 0) return 0;
    return (fact[y] * ((inv[x] * inv[y - x]) % Mod)) % Mod;
  }
  long long Tavan(long long x, long long y) {
    if (y == 0) return 1;
    long long z = Tavan(x, y / 2);
    z = (z * z) % Mod;
    if (y & 1) return (z * x) % Mod;
    return z;
  }
  