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;
  using pii = pair<int, int>;
  using pll = pair<ll, ll>;
  using vi = vector<int>;
  using vl = vector<ll>;
  using vii = vector<pii>;
  using vll = vector<pll>;
  void debug_out() { cerr << endl; }
  template <typename Head, typename... Tail>
  void debug_out(Head H, Tail... T) {
    cerr << " " << to_string(H);
    debug_out(T...);
  }
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  const long double eps = 1e-9;
  const long long mod = 998244353;
  const int MAXN = 200000;
  ll pow_mod(ll a, ll b, ll P) {
    ll ans = 1;
    while (b) {
      if (b & 1) ans = ans * a % P;
      a = a * a % P;
      b >>= 1;
    }
    return ans;
  }
  ll inv_mod(ll a, ll P) { return pow_mod(a, P - 2ll, P); }
  vector<ll> fact;
  void calc_fact(int n, ll P) {
    if (fact.size() < 1) fact.push_back(1);
    for (int i = fact.size(), nnnn = (n + 1); i < nnnn; ++i)
      fact.push_back(fact[i - 1] * i % P);
  }
  vector<ll> fact_inv;
  void calc_fact_inv(int n, ll P) {
    if (fact_inv.size() < 1) fact_inv.push_back(1);
    for (int i = fact_inv.size(), nnnn = (n + 1); i < nnnn; ++i)
      fact_inv.push_back(fact_inv[i - 1] * inv_mod(i, P) % P);
  }
  ll nCr(int n, int r, ll P) {
    if (r > n) return 0;
    calc_fact(n, P);
    calc_fact_inv(n, P);
    return (fact[n] * fact_inv[r] % P) * fact_inv[n - r] % P;
  }
  ll nCr_inv(int n, int r, ll P) {
    if (r > n) return 0;
    calc_fact(n, P);
    calc_fact_inv(n, P);
    return (fact_inv[n] * fact[r] % P) * fact[n - r] % P;
  }
  void solve() {
    ll n, k;
    cin >> n >> k;
    string s;
    cin >> s;
    s = "1" + s + "1";
    int cur = -1;
    ll ans = 0;
    for (int i = 0, nnnn = (n + 2); i < nnnn; ++i) {
      if (s[i] == '1') {
        int j = i, cnt = 0, last = 0;
        while (j < n + 2) {
          if (s[j] == '1') {
            cnt++;
            if (j >= i + 2 && cnt == k + 2) {
              int num = cnt - 2;
              int spaces = j - 1 - i - 1 + 1;
              int calc;
              if (ans != 0)
                calc =
                    nCr(spaces, num, mod) - nCr(spaces - j + last, num - 1, mod);
              else
                calc = nCr(spaces, num, mod) - 1;
              if (calc < 0) calc += mod;
              ans += calc;
            }
            last = j;
          }
          if (cnt == k + 2) {
            break;
          }
          j++;
        }
      }
    }
    cout << (ans + 1) % mod << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout << setprecision(12) << fixed;
    solve();
  }
  