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;
  using ll = long long;
  const ll mod = 998244353;
  void __print(int x) { cerr << x; }
  void __print(long x) { cerr << x; }
  void __print(long long x) { cerr << x; }
  void __print(unsigned x) { cerr << x; }
  void __print(unsigned long x) { cerr << x; }
  void __print(unsigned long long x) { cerr << x; }
  void __print(float x) { cerr << x; }
  void __print(double x) { cerr << x; }
  void __print(long double x) { cerr << x; }
  void __print(char x) { cerr << '\'' << x << '\''; }
  void __print(const char *x) { cerr << '\"' << x << '\"'; }
  void __print(const string &x) { cerr << '\"' << x << '\"'; }
  void __print(bool x) { cerr << (x ? "true" : "false"); }
  template <typename T, typename V>
  void __print(const pair<T, V> &x) {
    cerr << '{';
    __print(x.first);
    cerr << ',';
    __print(x.second);
    cerr << '}';
  }
  template <typename T>
  void __print(const T &x) {
    int f = 0;
    cerr << '{';
    for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
    cerr << "}";
  }
  void _print() { cerr << "]\n"; }
  template <typename T, typename... V>
  void _print(T t, V... v) {
    __print(t);
    if (sizeof...(v)) cerr << ", ";
    _print(v...);
  }
  ll add(ll x, ll y) {
    x += y;
    while (x >= mod) {
      x -= mod;
    }
    while (x < 0) {
      x += mod;
    }
    return x;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    t = 1;
    while (t--) {
      ll n, k;
      cin >> n >> k;
      string s;
      cin >> s;
      vector<vector<ll>> c(n + 1);
      vector<ll> p(n + 1);
      for (ll i = 0; i < n; i++) {
        p[i + 1] = p[i] + (s[i] - '0');
      }
      for (ll i = 0; i < n; i++) {
        c[i].resize(i + 1);
        c[i][0] = c[i][i] = 1;
        for (ll j = 1; j < i; j++) {
          c[i][j] = add(c[i - 1][j], c[i - 1][j - 1]);
        }
      }
      int ans = 1;
      for (ll i = 0; i < n; i++) {
        for (ll j = i + 1; j < n; j++) {
          ll cnt = j + 1 - i;
          ll cnt1 = p[j + 1] - p[i];
          if (cnt1 > k or p[n] < k) {
            continue;
          }
          cnt -= 2;
          if (s[i] == '0') {
            cnt1--;
          }
          if (s[j] == '0') {
            cnt1--;
          }
          if (cnt1 <= cnt and cnt1 >= 0 and cnt >= 0) {
            ans = add(ans, c[cnt][cnt1]);
          }
        }
      }
      cout << ans << "\n";
    }
  }
  