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.
5 2
00011
 5 4
11011
 81 5
100000101000000011001101010100001010111000000110000000001000101101111000001000110
 4 3
1011
 47 25
11110011111001110100101010011000110111011011101
 291 159
011001110111110110001111000110001001111100010001011000110100110010001101100110111010101111100111110000110111001100111010001000011001010011011100101000101011010110010011010100100101111010111000001011110101001011101110100100111110011001101010001110110101010101100011101110111000011111011001110
 459 15
101010101010101010101010101010101010101010101111111111111111111111010101010101010101010101011111111111111111111101010101010101010101111111111111111111111101010101010101010111111111111111111111111101010101010101111111111111111111111111110110101010101011111111111111111111111111111001000111111111111111111111111111110000011111111111111111111111111111000001111111111111111111111111111100001111111111111111111111111111100111111111111111111111111111110111111111111
 105 58
110101000111100000011011011101110100001000110011010101100010111001011111010010111101011011010110011011110
 52 11
0110011111001000010000100010010001010100000100110100

10
 5
 187538
 4
 314445058
 1
 0
 132732930
 478323746

#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e18;
struct hash_pair {
  template <class T1, class T2>
  size_t operator()(const pair<T1, T2>& p) const {
    auto hash1 = hash<T1>{}(p.first);
    auto hash2 = hash<T2>{}(p.second);
    return hash1 ^ hash2;
  }
};
long long power(long long x, unsigned long long y, unsigned long long m) {
  if (y == 0) return 1;
  long long p = power(x, y / 2, m) % m;
  p = (p * p) % m;
  return (y % 2 == 0) ? p : (x * p) % m;
}
long long modInverse(long long a, long long m) { return power(a, m - 2, m); }
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, k;
  string str;
  cin >> n >> k >> str;
  vector<long long> prefix(n + 1);
  prefix[0] = 0;
  for (long long x = 0; x < n; x++) {
    prefix[x + 1] = prefix[x] + (str[x] == '1');
  }
  long long mod = 998244353;
  vector<long long> factorials(n + 1);
  factorials[0] = 1;
  for (long long x = 1; x <= n; x++) {
    factorials[x] = (factorials[x - 1] * x) % mod;
  }
  unordered_map<pair<long long, long long>, long long, hash_pair> map3;
  map3[make_pair(1, 2)] = 2;
  long long total = 0;
  for (long long x = 0; x < n; x++) {
    for (long long y = x + 1; y < n; y++) {
      long long ones = prefix[y + 1] - prefix[x];
      long long zeros = y - x + 1 - ones;
      if (ones <= k) {
        if (str[x] == '1') {
          zeros--;
        }
        if (str[x] == '0') {
          ones--;
        }
        if (str[y] == '1') {
          zeros--;
        }
        if (str[y] == '0') {
          ones--;
        }
        if (ones >= 0 && zeros >= 0) {
          if (zeros < ones) {
            swap(zeros, ones);
          }
          map3[make_pair(zeros, ones)]++;
        }
      }
    }
  }
  while (map3.size() > 0) {
    pair<pair<long long, long long>, long long> curr = *map3.begin();
    total = (total + curr.second *
                         factorials[curr.first.first + curr.first.second] *
                         modInverse(factorials[curr.first.first], mod) *
                         modInverse(factorials[curr.first.second], mod)) %
            mod;
    map3.erase(curr.first);
  }
  if (prefix[n] < k) {
    total = 0;
  }
  total++;
  cout << total << "\n";
  return 0;
}
