#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
const long long nax = 5000 + 10, mod = 998244353, inf = 1e18;
long long first[nax], invf[nax];
long long mul(long long x, long long y) { return x * y % mod; }
long long expo(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = mul(res, a);
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
long long inv(long long x) { return expo(x, mod - 2); }
long long nck(long long n, long long r) {
  if (r < 0 or r > n) return 0;
  return mul(first[n], mul(invf[n - r], invf[r]));
}
long long sub(long long x, long long y) {
  x -= y;
  if (x < 0) x += mod;
  return x;
}
long long add(long long x, long long y) {
  x += y;
  if (x >= mod) x -= mod;
  return x;
}
void solve() {
  long long n, res = 1, k;
  cin >> n >> k;
  string second;
  cin >> second;
  vector<long long> a(n), pref(n + 1, 0);
  for (long long i = 0; i < n; i++) a[i] = (second[i] - '0');
  for (long long i = 0; i < n; i++) pref[i + 1] = pref[i] + a[i];
  if (pref[n] >= k) {
    for (long long i = 0; i < n; i++)
      for (long long j = i + 1; j < n; j++) {
        long long ok = pref[j + 1] - pref[i];
        if (ok <= k) {
          ok -= (a[i] ^ 1) + (a[j] ^ 1);
          long long val = j - i - 1;
          res += nck(val, ok);
          res %= mod;
        }
      }
  }
  cout << res;
  cout << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  first[0] = 1;
  for (long long i = 1; i < nax; i++) first[i] = mul(first[i - 1], i);
  invf[nax - 1] = inv(first[nax - 1]);
  for (long long i = nax - 1; i; i--) invf[i - 1] = mul(invf[i], i);
  long long tt = 1;
  for (long long i = 1; i <= tt; i++) {
    solve();
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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 double PI = acos(-1);
const long long nax = 5000 + 10, mod = 998244353, inf = 1e18;
long long first[nax], invf[nax];
long long mul(long long x, long long y) { return x * y % mod; }
long long expo(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = mul(res, a);
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
long long inv(long long x) { return expo(x, mod - 2); }
long long nck(long long n, long long r) {
  if (r < 0 or r > n) return 0;
  return mul(first[n], mul(invf[n - r], invf[r]));
}
long long sub(long long x, long long y) {
  x -= y;
  if (x < 0) x += mod;
  return x;
}
long long add(long long x, long long y) {
  x += y;
  if (x >= mod) x -= mod;
  return x;
}
void solve() {
  long long n, res = 1, k;
  cin >> n >> k;
  string second;
  cin >> second;
  vector<long long> a(n), pref(n + 1, 0);
  for (long long i = 0; i < n; i++) a[i] = (second[i] - '0');
  for (long long i = 0; i < n; i++) pref[i + 1] = pref[i] + a[i];
  if (pref[n] >= k) {
    for (long long i = 0; i < n; i++)
      for (long long j = i + 1; j < n; j++) {
        long long ok = pref[j + 1] - pref[i];
        if (ok <= k) {
          ok -= (a[i] ^ 1) + (a[j] ^ 1);
          long long val = j - i - 1;
          res += nck(val, ok);
          res %= mod;
        }
      }
  }
  cout << res;
  cout << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  first[0] = 1;
  for (long long i = 1; i < nax; i++) first[i] = mul(first[i - 1], i);
  invf[nax - 1] = inv(first[nax - 1]);
  for (long long i = nax - 1; i; i--) invf[i - 1] = mul(invf[i], i);
  long long tt = 1;
  for (long long i = 1; i <= tt; i++) {
    solve();
  }
  return 0;
}

```