#include <bits/stdc++.h>
using namespace std;
const int MOD = 998244353;
const int N = 5005;
int n, k;
int a[N], dem = 0;
long long C[N][N];
void pre_calc() {
  C[0][0] = 1;
  for (int i = 1; i <= n; i++) C[0][i] = 1;
  for (int j = 1; j <= n; j++) {
    for (int i = 1; i <= j; i++) {
      C[i][j] = (C[i][j - 1] + C[i - 1][j - 1]) % MOD;
    }
  }
}
long long solve(int x, int l, int r) {
  int st = 1, cnt = 0;
  long long res = 0;
  for (int i = x; i < r; i++) {
    if (i - x + 1 < l) continue;
    long long range = (a[i + 1] - 1) - (a[i - x] + 1) + 1;
    res += C[x][range];
    res %= MOD;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  pre_calc();
  for (int i = 1; i <= n; i++) {
    char c;
    cin >> c;
    if (c == '1') a[++dem] = i;
  }
  if ((k > dem) || k == 0) {
    cout << 1;
    return 0;
  }
  a[++dem] = n + 1;
  long long res = 0;
  int t = 1;
  res += solve(k, 1, dem);
  res -= solve(k - 1, 2, dem - 1);
  res = ((res % MOD) + MOD) % MOD;
  cout << res;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 MOD = 998244353;
const int N = 5005;
int n, k;
int a[N], dem = 0;
long long C[N][N];
void pre_calc() {
  C[0][0] = 1;
  for (int i = 1; i <= n; i++) C[0][i] = 1;
  for (int j = 1; j <= n; j++) {
    for (int i = 1; i <= j; i++) {
      C[i][j] = (C[i][j - 1] + C[i - 1][j - 1]) % MOD;
    }
  }
}
long long solve(int x, int l, int r) {
  int st = 1, cnt = 0;
  long long res = 0;
  for (int i = x; i < r; i++) {
    if (i - x + 1 < l) continue;
    long long range = (a[i + 1] - 1) - (a[i - x] + 1) + 1;
    res += C[x][range];
    res %= MOD;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  pre_calc();
  for (int i = 1; i <= n; i++) {
    char c;
    cin >> c;
    if (c == '1') a[++dem] = i;
  }
  if ((k > dem) || k == 0) {
    cout << 1;
    return 0;
  }
  a[++dem] = n + 1;
  long long res = 0;
  int t = 1;
  res += solve(k, 1, dem);
  res -= solve(k - 1, 2, dem - 1);
  res = ((res % MOD) + MOD) % MOD;
  cout << res;
  return 0;
}

```