#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
const int N = 5050, MOD = 998244353;
i64 inv[N], fac[N], inv_fac[N];
void precalc() {
  inv[1] = 1;
  for (int n = 2; n < N; n++) {
    inv[n] = (MOD - (MOD / n) * inv[MOD % n] % MOD) % MOD;
  }
  fac[0] = 1;
  for (int n = 1; n < N; n++) {
    fac[n] = n * fac[n - 1] % MOD;
  }
  inv_fac[0] = 1;
  for (int n = 1; n < N; n++) {
    inv_fac[n] = inv[n] * inv_fac[n - 1] % MOD;
  }
}
i64 binom(i64 n, i64 k) {
  if (k < 0 || k > n) return 0;
  return fac[n] * inv_fac[k] % MOD * inv_fac[n - k] % MOD;
}
int main() {
  int n, k;
  cin >> n >> k;
  precalc();
  string a;
  cin >> a;
  i64 res = 1;
  for (int i = 0; i < n; i++) {
    int o = 0, j = i;
    for (; j < n; j++) {
      if (o == k && a[j] == '1') break;
      o += a[j] == '1';
    }
    if (i == 0 && o < k) break;
    int z = j - i - o;
    res = (res + binom(j - i - 1, z - (a[i] == '1'))) % MOD;
    cerr << j - i - 1 << " " << z - (a[i] == '1') << endl;
  }
  cout << res << '\n';
}

### Prompt

Create a solution in CPP for 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;
using i64 = long long;
const int N = 5050, MOD = 998244353;
i64 inv[N], fac[N], inv_fac[N];
void precalc() {
  inv[1] = 1;
  for (int n = 2; n < N; n++) {
    inv[n] = (MOD - (MOD / n) * inv[MOD % n] % MOD) % MOD;
  }
  fac[0] = 1;
  for (int n = 1; n < N; n++) {
    fac[n] = n * fac[n - 1] % MOD;
  }
  inv_fac[0] = 1;
  for (int n = 1; n < N; n++) {
    inv_fac[n] = inv[n] * inv_fac[n - 1] % MOD;
  }
}
i64 binom(i64 n, i64 k) {
  if (k < 0 || k > n) return 0;
  return fac[n] * inv_fac[k] % MOD * inv_fac[n - k] % MOD;
}
int main() {
  int n, k;
  cin >> n >> k;
  precalc();
  string a;
  cin >> a;
  i64 res = 1;
  for (int i = 0; i < n; i++) {
    int o = 0, j = i;
    for (; j < n; j++) {
      if (o == k && a[j] == '1') break;
      o += a[j] == '1';
    }
    if (i == 0 && o < k) break;
    int z = j - i - o;
    res = (res + binom(j - i - 1, z - (a[i] == '1'))) % MOD;
    cerr << j - i - 1 << " " << z - (a[i] == '1') << endl;
  }
  cout << res << '\n';
}

```