#include <bits/stdc++.h>
using namespace std;
long long mod = 998244353;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long mod_add(long long a, long long b) {
  a = a % mod;
  b = b % mod;
  return (((a + b) % mod) + mod) % mod;
}
long long mod_mul(long long a, long long b) {
  a = a % mod;
  b = b % mod;
  return (((a * b) % mod) + mod) % mod;
}
long long inv(long long i) {
  if (i == 1) return 1;
  return (mod - ((mod / i) * inv(mod % i)) % mod) % mod;
}
long long maxn = 5001;
vector<long long> fact(maxn, 1);
vector<long long> ifact(maxn, 1);
void calFact() {
  for (long long i = 2; i < maxn; i++) fact[i] = mod_mul(fact[i - 1], i);
}
long long ncr(long long n, long long r) {
  long long ans = fact[n];
  ans = mod_mul(ans, inv(fact[n - r]));
  ans = mod_mul(ans, inv(fact[r]));
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  calFact();
  long long n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  vector<long long> dp(n, 0);
  dp[0] = 1;
  bool first = false;
  for (long long i = 1; i < n; i++) {
    long long ones = 0, zeroes = 0;
    dp[i] = dp[i - 1];
    for (long long j = i; j >= 0; j--) {
      if (s[j] == '1') {
        if (ones == k) break;
        ones++;
      } else
        zeroes++;
    }
    if (ones < k) continue;
    long long temp;
    if (first == false) {
      temp = (ncr(zeroes + ones, zeroes) - 1 + mod) % mod;
    } else {
      if (s[i] == '1') {
        temp = 0;
        if (zeroes > 0) {
          temp = ncr(zeroes + ones - 1, zeroes - 1);
        }
      } else {
        temp = 0;
        if (ones > 0) {
          temp = ncr(zeroes + ones - 1, zeroes);
        }
      }
    }
    first = true;
    dp[i] = mod_add(dp[i], temp);
  }
  cout << dp[n - 1] << "\n";
}

### Prompt

In cpp, your task is to solve 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;
long long mod = 998244353;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long mod_add(long long a, long long b) {
  a = a % mod;
  b = b % mod;
  return (((a + b) % mod) + mod) % mod;
}
long long mod_mul(long long a, long long b) {
  a = a % mod;
  b = b % mod;
  return (((a * b) % mod) + mod) % mod;
}
long long inv(long long i) {
  if (i == 1) return 1;
  return (mod - ((mod / i) * inv(mod % i)) % mod) % mod;
}
long long maxn = 5001;
vector<long long> fact(maxn, 1);
vector<long long> ifact(maxn, 1);
void calFact() {
  for (long long i = 2; i < maxn; i++) fact[i] = mod_mul(fact[i - 1], i);
}
long long ncr(long long n, long long r) {
  long long ans = fact[n];
  ans = mod_mul(ans, inv(fact[n - r]));
  ans = mod_mul(ans, inv(fact[r]));
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  calFact();
  long long n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  vector<long long> dp(n, 0);
  dp[0] = 1;
  bool first = false;
  for (long long i = 1; i < n; i++) {
    long long ones = 0, zeroes = 0;
    dp[i] = dp[i - 1];
    for (long long j = i; j >= 0; j--) {
      if (s[j] == '1') {
        if (ones == k) break;
        ones++;
      } else
        zeroes++;
    }
    if (ones < k) continue;
    long long temp;
    if (first == false) {
      temp = (ncr(zeroes + ones, zeroes) - 1 + mod) % mod;
    } else {
      if (s[i] == '1') {
        temp = 0;
        if (zeroes > 0) {
          temp = ncr(zeroes + ones - 1, zeroes - 1);
        }
      } else {
        temp = 0;
        if (ones > 0) {
          temp = ncr(zeroes + ones - 1, zeroes);
        }
      }
    }
    first = true;
    dp[i] = mod_add(dp[i], temp);
  }
  cout << dp[n - 1] << "\n";
}

```