#include <bits/stdc++.h>
using namespace std;
const int N = 5e3 + 10;
const int Mod = 998244353;
long long n, t, k;
long long dp[N];
string s;
long long ans = 1;
long long fact[N], inv[N];
void Entekhab();
long long Tavan(long long, long long);
int Check(string);
long long Ent(long long, long long);
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k >> s;
  Entekhab();
  if (k == 0 or Check(s) < k) {
    cout << 1 << endl;
    exit(0);
  }
  for (int i = 0; i < n; i++) {
    int l = i, r = i, cnt1 = 0, cnt0 = 0;
    bool seen0 = false, seen1 = false;
    for (int j = i; j < n; j++) {
      if (s[j] == '1')
        cnt1++, seen1 = true;
      else
        seen0 = true, cnt0++;
      if (j == n - 1 or (cnt1 == k and s[j + 1] == '1')) {
        r = j;
        break;
      }
    }
    if (s[i] == '1' and seen0) {
      ans += Ent(cnt1, r - l);
      ans %= Mod;
    } else if (s[i] == '0' and seen1) {
      ans += Ent(cnt0, r - l);
      ans %= Mod;
    }
  }
  cout << ans << endl;
}
int Check(string s) {
  int res = 0;
  for (int i = 0; i < s.size(); i++)
    if (s[i] == '1') res++;
  return res;
}
void Entekhab() {
  fact[0] = 1, inv[0] = 1;
  for (long long i = 1; i < N; i++)
    fact[i] = (fact[i - 1] * i) % Mod, inv[i] = Tavan(fact[i], (Mod - 2));
}
long long Ent(long long x, long long y) {
  if (x > y) return 0;
  if (y < 0 or x < 0) return 0;
  return (fact[y] * ((inv[x] * inv[y - x]) % Mod)) % Mod;
}
long long Tavan(long long x, long long y) {
  if (y == 0) return 1;
  long long z = Tavan(x, y / 2);
  z = (z * z) % Mod;
  if (y & 1) return (z * x) % Mod;
  return z;
}

### Prompt

Please formulate 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 = 5e3 + 10;
const int Mod = 998244353;
long long n, t, k;
long long dp[N];
string s;
long long ans = 1;
long long fact[N], inv[N];
void Entekhab();
long long Tavan(long long, long long);
int Check(string);
long long Ent(long long, long long);
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k >> s;
  Entekhab();
  if (k == 0 or Check(s) < k) {
    cout << 1 << endl;
    exit(0);
  }
  for (int i = 0; i < n; i++) {
    int l = i, r = i, cnt1 = 0, cnt0 = 0;
    bool seen0 = false, seen1 = false;
    for (int j = i; j < n; j++) {
      if (s[j] == '1')
        cnt1++, seen1 = true;
      else
        seen0 = true, cnt0++;
      if (j == n - 1 or (cnt1 == k and s[j + 1] == '1')) {
        r = j;
        break;
      }
    }
    if (s[i] == '1' and seen0) {
      ans += Ent(cnt1, r - l);
      ans %= Mod;
    } else if (s[i] == '0' and seen1) {
      ans += Ent(cnt0, r - l);
      ans %= Mod;
    }
  }
  cout << ans << endl;
}
int Check(string s) {
  int res = 0;
  for (int i = 0; i < s.size(); i++)
    if (s[i] == '1') res++;
  return res;
}
void Entekhab() {
  fact[0] = 1, inv[0] = 1;
  for (long long i = 1; i < N; i++)
    fact[i] = (fact[i - 1] * i) % Mod, inv[i] = Tavan(fact[i], (Mod - 2));
}
long long Ent(long long x, long long y) {
  if (x > y) return 0;
  if (y < 0 or x < 0) return 0;
  return (fact[y] * ((inv[x] * inv[y - x]) % Mod)) % Mod;
}
long long Tavan(long long x, long long y) {
  if (y == 0) return 1;
  long long z = Tavan(x, y / 2);
  z = (z * z) % Mod;
  if (y & 1) return (z * x) % Mod;
  return z;
}

```