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>
  template <typename Sep, typename Head, typename... Tail>
  void fold_print(Sep sep, Head&& h, Tail&&... args) {
    ((std::cerr << h) << ... << (std::cerr << sep, args));
  }
  std::mt19937_64 myrand(
      std::chrono::high_resolution_clock::now().time_since_epoch().count());
  using namespace std;
  using LL = long long int;
  void init();
  void process();
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T = 1;
    init();
    for (int t = 1; t <= T; t++) {
      process();
    }
    cout.flush();
    return 0;
  }
  constexpr LL MOD = 998244353;
  vector<LL> fac(5000 * 2 + 1), inv(5000 * 2 + 1);
  LL fastpw(LL x, LL p) {
    if (p == 0) return 1;
    LL tmp = fastpw(x * x % MOD, p / 2);
    if (p % 2 == 0)
      return tmp;
    else
      return tmp * x % MOD;
  }
  LL modinv(LL x) { return fastpw(x, MOD - 2); }
  void init() {
    fac[0] = 1;
    for (LL i = 1; i < fac.size(); i++) {
      fac[i] = i * fac[i - 1] % MOD;
    }
    for (size_t i = 0; i < inv.size(); i++) {
      inv[i] = modinv(fac[i]);
    }
  }
  LL C(LL n, LL m) { return fac[n] * inv[m] % MOD * inv[n - m] % MOD; }
  void process() {
    int n, k;
    cin >> n >> k;
    vector<int> arr(n);
    {
      for (auto& x : arr) {
        char ch;
        cin >> ch;
        x = ch - '0';
      }
    }
    auto permute = [&](LL l, LL r, LL one) {
      LL len = r - l + 1;
      LL cnt[] = {len - one, one};
      if (len == 1 || cnt[0] == len || cnt[1] == len) return 0ll;
      cnt[1 - arr[l]] -= 1;
      return C(len - 1, cnt[0]);
    };
    LL ans = 1, r = 0, cnt = arr[0];
    for (LL i = 0; i < n; cnt -= arr[i], i += 1) {
      while (r + 1 < n && cnt + arr[r + 1] <= k) {
        cnt += arr[r + 1], r += 1;
      }
      if (cnt != k) break;
      if (r == n - 1) {
        for (int j = i; j < n; cnt -= arr[j], j += 1) {
          ans = (ans + permute(j, r, cnt)) % MOD;
        }
        break;
      } else {
        ans = (ans + permute(i, r, cnt)) % MOD;
      }
    }
    cout << ans << '\n';
  }
  