#include <bits/stdc++.h>
using namespace std;
const int maxN = 1e4;
long long F[maxN + 1];
void fact() {
  F[0] = F[1] = 1;
  for (int i = 2; i <= maxN; i++) F[i] = (1LL * F[i - 1] * i) % 998244353;
}
long long power(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = (1LL * res * a) % 998244353, b--;
    } else {
      a = (1LL * a * a) % 998244353, b >>= 1;
    }
  }
  return res;
}
long long C(long long n, long long r) {
  if (r > n) return 0;
  long long res = F[n];
  res = (1LL * res * power(F[r], 998244353 - 2)) % 998244353;
  res = (1LL * res * power(F[n - r], 998244353 - 2)) % 998244353;
  return res;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  string second;
  cin >> second;
  if (!k) {
    cout << "1\n";
    return;
  }
  long long ans = 1, l = 0, r = 0, cnt = 0;
  while (r < n) {
    if (second[r] == '1') {
      if (cnt == k) {
        (ans += (C(r - l, k) - 1 + 998244353) % 998244353) %= 998244353;
        while (l < n && second[l] != '1') l++;
        l++, cnt--;
        ans = (ans - C(r - l, k - 1) + 1 + 998244353) % 998244353;
      }
      cnt++;
    }
    r++;
  }
  if (cnt == k) (ans += (C(r - l, k) - 1 + 998244353) % 998244353) %= 998244353;
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0), fact();
  solve();
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxN = 1e4;
long long F[maxN + 1];
void fact() {
  F[0] = F[1] = 1;
  for (int i = 2; i <= maxN; i++) F[i] = (1LL * F[i - 1] * i) % 998244353;
}
long long power(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = (1LL * res * a) % 998244353, b--;
    } else {
      a = (1LL * a * a) % 998244353, b >>= 1;
    }
  }
  return res;
}
long long C(long long n, long long r) {
  if (r > n) return 0;
  long long res = F[n];
  res = (1LL * res * power(F[r], 998244353 - 2)) % 998244353;
  res = (1LL * res * power(F[n - r], 998244353 - 2)) % 998244353;
  return res;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  string second;
  cin >> second;
  if (!k) {
    cout << "1\n";
    return;
  }
  long long ans = 1, l = 0, r = 0, cnt = 0;
  while (r < n) {
    if (second[r] == '1') {
      if (cnt == k) {
        (ans += (C(r - l, k) - 1 + 998244353) % 998244353) %= 998244353;
        while (l < n && second[l] != '1') l++;
        l++, cnt--;
        ans = (ans - C(r - l, k - 1) + 1 + 998244353) % 998244353;
      }
      cnt++;
    }
    r++;
  }
  if (cnt == k) (ans += (C(r - l, k) - 1 + 998244353) % 998244353) %= 998244353;
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0), fact();
  solve();
  return 0;
}

```