#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fLL;
const int MAX = 5e3 + 10;
const int MOD = 998244353;
long long fat[MAX], ifat[MAX];
long long choose(int a, int b) {
  if (a < b) return 0;
  return fat[a] * ifat[b] % MOD * ifat[a - b] % MOD;
}
long long fexp(long long a, long long e) {
  long long ans = 1;
  while (e) {
    if (e & 1) ans = ans * a % MOD;
    a = a * a % MOD;
    e /= 2;
  }
  return ans;
}
void compute() {
  fat[0] = 1;
  ifat[0] = 1;
  for (int i = 1; i < MAX; i++) {
    fat[i] = i * fat[i - 1] % MOD;
    ifat[i] = fexp(fat[i], MOD - 2);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  compute();
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  if (k == 0 || count(s.begin(), s.end(), '1') < k) {
    cout << 1 << '\n';
    return 0;
  }
  long long ans = 1;
  for (int l = 0; l < n; l++) {
    int cnt = 0, r = l;
    while (r < n && cnt + (s[r] == '1') <= k) cnt += s[r++] == '1';
    long long add = 0;
    if (s[l] == '0' && cnt > 0) add = choose((r - l) - 1, cnt - 1);
    if (s[l] == '1' && ((r - l) - cnt) > 0) add = choose((r - l) - 1, cnt);
    ans = (ans + add) % MOD;
  }
  cout << ans << '\n';
  return 0;
}

### 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;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fLL;
const int MAX = 5e3 + 10;
const int MOD = 998244353;
long long fat[MAX], ifat[MAX];
long long choose(int a, int b) {
  if (a < b) return 0;
  return fat[a] * ifat[b] % MOD * ifat[a - b] % MOD;
}
long long fexp(long long a, long long e) {
  long long ans = 1;
  while (e) {
    if (e & 1) ans = ans * a % MOD;
    a = a * a % MOD;
    e /= 2;
  }
  return ans;
}
void compute() {
  fat[0] = 1;
  ifat[0] = 1;
  for (int i = 1; i < MAX; i++) {
    fat[i] = i * fat[i - 1] % MOD;
    ifat[i] = fexp(fat[i], MOD - 2);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  compute();
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  if (k == 0 || count(s.begin(), s.end(), '1') < k) {
    cout << 1 << '\n';
    return 0;
  }
  long long ans = 1;
  for (int l = 0; l < n; l++) {
    int cnt = 0, r = l;
    while (r < n && cnt + (s[r] == '1') <= k) cnt += s[r++] == '1';
    long long add = 0;
    if (s[l] == '0' && cnt > 0) add = choose((r - l) - 1, cnt - 1);
    if (s[l] == '1' && ((r - l) - cnt) > 0) add = choose((r - l) - 1, cnt);
    ans = (ans + add) % MOD;
  }
  cout << ans << '\n';
  return 0;
}

```