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 double PI = acos(-1);
  const long long nax = 5000 + 10, mod = 998244353, inf = 1e18;
  long long first[nax], invf[nax];
  long long mul(long long x, long long y) { return x * y % mod; }
  long long expo(long long a, long long b) {
    long long res = 1;
    while (b) {
      if (b & 1) res = mul(res, a);
      a = mul(a, a);
      b >>= 1;
    }
    return res;
  }
  long long inv(long long x) { return expo(x, mod - 2); }
  long long nck(long long n, long long r) {
    if (r < 0 or r > n) return 0;
    return mul(first[n], mul(invf[n - r], invf[r]));
  }
  long long sub(long long x, long long y) {
    x -= y;
    if (x < 0) x += mod;
    return x;
  }
  long long add(long long x, long long y) {
    x += y;
    if (x >= mod) x -= mod;
    return x;
  }
  void solve() {
    long long n, res = 1, k;
    cin >> n >> k;
    string second;
    cin >> second;
    vector<long long> a(n), pref(n + 1, 0);
    for (long long i = 0; i < n; i++) a[i] = (second[i] - '0');
    for (long long i = 0; i < n; i++) pref[i + 1] = pref[i] + a[i];
    if (pref[n] >= k) {
      for (long long i = 0; i < n; i++)
        for (long long j = i + 1; j < n; j++) {
          long long ok = pref[j + 1] - pref[i];
          if (ok <= k) {
            ok -= (a[i] ^ 1) + (a[j] ^ 1);
            long long val = j - i - 1;
            res += nck(val, ok);
            res %= mod;
          }
        }
    }
    cout << res;
    cout << '\n';
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    first[0] = 1;
    for (long long i = 1; i < nax; i++) first[i] = mul(first[i - 1], i);
    invf[nax - 1] = inv(first[nax - 1]);
    for (long long i = nax - 1; i; i--) invf[i - 1] = mul(invf[i], i);
    long long tt = 1;
    for (long long i = 1; i <= tt; i++) {
      solve();
    }
    return 0;
  }
  