#include <bits/stdc++.h>
using namespace std;
const long long inf = 1000000000000000000;
const long long mod = 998244353;
long long fact[1000005], n, k, pr, cnt, na, a[5005], ans;
string s;
long long bin_pow(long long x, long long step) {
  if (step == 0) return 1;
  if (step == 1) return x;
  if (step % 2 == 0) {
    long long p = bin_pow(x, step / 2);
    return ((p * p) % mod);
  }
  return ((x * bin_pow(x, step - 1)) % mod);
}
long long inv(long long x) { return bin_pow(x, mod - 2); }
long long c(long long n, long long k) {
  return ((((fact[n] * inv(fact[n - k])) % mod) * inv(fact[k])) % mod);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  fact[0] = fact[1] = 1;
  for (long long i = 2; i <= 1000000; ++i) fact[i] = (fact[i - 1] * i) % mod;
  cin >> n >> k;
  cin >> s;
  s = '$' + s;
  pr = cnt = 0;
  for (long long i = 1; i <= n; ++i) {
    if (s[i] == '1') a[++na] = i;
  }
  if (k > na || k == 0) {
    cout << 1;
    return 0;
  }
  if (k == na) {
    cout << c(n, k);
    return 0;
  }
  a[0] = 0;
  a[++na] = n + 1;
  for (long long i = k + 1; i <= na; ++i) {
    ans += c(a[i] - a[i - k - 1] - 1, k);
    ans %= mod;
    if (i > k + 1) {
      ans -= c(a[i - 1] - a[i - k - 1] - 1, k - 1);
      ans += mod;
      ans %= mod;
    }
  }
  cout << ans;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 long long inf = 1000000000000000000;
const long long mod = 998244353;
long long fact[1000005], n, k, pr, cnt, na, a[5005], ans;
string s;
long long bin_pow(long long x, long long step) {
  if (step == 0) return 1;
  if (step == 1) return x;
  if (step % 2 == 0) {
    long long p = bin_pow(x, step / 2);
    return ((p * p) % mod);
  }
  return ((x * bin_pow(x, step - 1)) % mod);
}
long long inv(long long x) { return bin_pow(x, mod - 2); }
long long c(long long n, long long k) {
  return ((((fact[n] * inv(fact[n - k])) % mod) * inv(fact[k])) % mod);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  fact[0] = fact[1] = 1;
  for (long long i = 2; i <= 1000000; ++i) fact[i] = (fact[i - 1] * i) % mod;
  cin >> n >> k;
  cin >> s;
  s = '$' + s;
  pr = cnt = 0;
  for (long long i = 1; i <= n; ++i) {
    if (s[i] == '1') a[++na] = i;
  }
  if (k > na || k == 0) {
    cout << 1;
    return 0;
  }
  if (k == na) {
    cout << c(n, k);
    return 0;
  }
  a[0] = 0;
  a[++na] = n + 1;
  for (long long i = k + 1; i <= na; ++i) {
    ans += c(a[i] - a[i - k - 1] - 1, k);
    ans %= mod;
    if (i > k + 1) {
      ans -= c(a[i - 1] - a[i - k - 1] - 1, k - 1);
      ans += mod;
      ans %= mod;
    }
  }
  cout << ans;
  return 0;
}

```