#include <bits/stdc++.h>
using namespace std;
long long fact[5003], ifact[5003];
long long mod = 998244353;
long long powwwer(long long i, long long j) {
  if (j == 0) return 1;
  long long w = powwwer(i, j / 2);
  w = (w * w) % mod;
  if (j % 2 == 1) {
    w = (w * i) % mod;
  }
  return w;
}
int main() {
  fact[0] = 1;
  ifact[0] = powwwer(fact[0], mod - 2);
  for (int i = 1; i <= 5001; i++) {
    fact[i] = (fact[i - 1] * 1LL * i) % mod;
    ifact[i] = powwwer(fact[i], mod - 2);
  }
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  int cnt = 0;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == '1') cnt++;
  }
  if (cnt < k) {
    cout << 1 << endl;
    return 0;
  }
  long long ans = 1;
  for (int i = 0; i < s.size(); i++) {
    int cnt1 = 0;
    int cnt0 = 0;
    int j = i;
    while (j >= 0 && cnt1 <= k) {
      if (s[j] == '1')
        cnt1++;
      else
        cnt0++;
      j--;
    }
    if (cnt1 > k) cnt1--;
    if (s[i] == '1') {
      if (cnt0 > 0) {
        long long z = (fact[cnt1 + cnt0 - 1] * ifact[cnt0 - 1]) % mod;
        z = (z * ifact[cnt1]) % mod;
        ans = (ans + z) % mod;
      }
    } else {
      if (cnt1 > 0) {
        long long z = (fact[cnt1 + cnt0 - 1] * ifact[cnt1 - 1]) % mod;
        z = (z * ifact[cnt0]) % mod;
        ans = (ans + z) % mod;
      }
    }
  }
  cout << ans << endl;
}

### Prompt

Please formulate a cpp solution to the following problem:

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;
long long fact[5003], ifact[5003];
long long mod = 998244353;
long long powwwer(long long i, long long j) {
  if (j == 0) return 1;
  long long w = powwwer(i, j / 2);
  w = (w * w) % mod;
  if (j % 2 == 1) {
    w = (w * i) % mod;
  }
  return w;
}
int main() {
  fact[0] = 1;
  ifact[0] = powwwer(fact[0], mod - 2);
  for (int i = 1; i <= 5001; i++) {
    fact[i] = (fact[i - 1] * 1LL * i) % mod;
    ifact[i] = powwwer(fact[i], mod - 2);
  }
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  int cnt = 0;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == '1') cnt++;
  }
  if (cnt < k) {
    cout << 1 << endl;
    return 0;
  }
  long long ans = 1;
  for (int i = 0; i < s.size(); i++) {
    int cnt1 = 0;
    int cnt0 = 0;
    int j = i;
    while (j >= 0 && cnt1 <= k) {
      if (s[j] == '1')
        cnt1++;
      else
        cnt0++;
      j--;
    }
    if (cnt1 > k) cnt1--;
    if (s[i] == '1') {
      if (cnt0 > 0) {
        long long z = (fact[cnt1 + cnt0 - 1] * ifact[cnt0 - 1]) % mod;
        z = (z * ifact[cnt1]) % mod;
        ans = (ans + z) % mod;
      }
    } else {
      if (cnt1 > 0) {
        long long z = (fact[cnt1 + cnt0 - 1] * ifact[cnt1 - 1]) % mod;
        z = (z * ifact[cnt0]) % mod;
        ans = (ans + z) % mod;
      }
    }
  }
  cout << ans << endl;
}

```