#include <bits/stdc++.h>
using namespace std;
const int N = 5005, mod = 998244353;
int n, k, sum, ans, tem;
int fac[N], inv[N], pre[N], suf[N];
char s[N];
int C(int i, int j) {
  if (i < j) return 0;
  return 1ll * fac[i] * inv[j] % mod * inv[i - j] % mod;
}
int qp(int x, int y) {
  int res = 1;
  while (y) {
    if (y % 2) res = 1ll * res * x % mod;
    x = 1ll * x * x % mod;
    y >>= 1;
  }
  return res;
}
void add(int& x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
int main() {
  scanf("%d%d", &n, &k);
  scanf("%s", s + 1);
  ans = 1;
  for (int i = 1; i <= n; ++i) {
    pre[i] = pre[i - 1] + s[i] - '0';
  }
  for (int i = n; i >= 1; --i) suf[i] = suf[i + 1] + s[i] - '0';
  if (pre[n] < k) {
    cout << 1;
    return 0;
  }
  fac[0] = 1;
  for (int i = 1; i <= n; ++i) fac[i] = 1ll * i * fac[i - 1] % mod;
  inv[n] = qp(fac[n], mod - 2);
  for (int i = n - 1; i >= 0; --i) inv[i] = 1ll * inv[i + 1] * (i + 1) % mod;
  for (int i = 1; i <= n; ++i) {
    for (int j = i + 1; j <= n; ++j) {
      tem = pre[j] - pre[i - 1];
      if (tem > k) continue;
      if (s[i] == '0') tem--;
      if (s[j] == '0') tem--;
      add(ans, C(j - i - 1, tem));
    }
  }
  printf("%d\n", ans);
}

### Prompt

Your task is to create 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;
const int N = 5005, mod = 998244353;
int n, k, sum, ans, tem;
int fac[N], inv[N], pre[N], suf[N];
char s[N];
int C(int i, int j) {
  if (i < j) return 0;
  return 1ll * fac[i] * inv[j] % mod * inv[i - j] % mod;
}
int qp(int x, int y) {
  int res = 1;
  while (y) {
    if (y % 2) res = 1ll * res * x % mod;
    x = 1ll * x * x % mod;
    y >>= 1;
  }
  return res;
}
void add(int& x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
int main() {
  scanf("%d%d", &n, &k);
  scanf("%s", s + 1);
  ans = 1;
  for (int i = 1; i <= n; ++i) {
    pre[i] = pre[i - 1] + s[i] - '0';
  }
  for (int i = n; i >= 1; --i) suf[i] = suf[i + 1] + s[i] - '0';
  if (pre[n] < k) {
    cout << 1;
    return 0;
  }
  fac[0] = 1;
  for (int i = 1; i <= n; ++i) fac[i] = 1ll * i * fac[i - 1] % mod;
  inv[n] = qp(fac[n], mod - 2);
  for (int i = n - 1; i >= 0; --i) inv[i] = 1ll * inv[i + 1] * (i + 1) % mod;
  for (int i = 1; i <= n; ++i) {
    for (int j = i + 1; j <= n; ++j) {
      tem = pre[j] - pre[i - 1];
      if (tem > k) continue;
      if (s[i] == '0') tem--;
      if (s[j] == '0') tem--;
      add(ans, C(j - i - 1, tem));
    }
  }
  printf("%d\n", ans);
}

```