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 int MOD = 998244353;
const int N = 5005;
int n, k;
int a[N], dem = 0;
long long C[N][N];
void pre_calc() {
  C[0][0] = 1;
  for (int i = 1; i <= n; i++) C[0][i] = 1;
  for (int j = 1; j <= n; j++) {
    for (int i = 1; i <= j; i++) {
      C[i][j] = (C[i][j - 1] + C[i - 1][j - 1]) % MOD;
    }
  }
}
long long xx = 1;
long long solve(int x) {
  if (x == 0) {
    return 1;
  }
  int st = 1, cnt = 0;
  long long res = 0;
  for (int i = x; i < dem; i++) {
    long long range = (a[i + 1] - 1) - (a[i - x] + 1) + 1;
    res += (C[x][range] - 1);
    res %= MOD;
    xx++;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  pre_calc();
  for (int i = 1; i <= n; i++) {
    char c;
    cin >> c;
    if (c == '1') a[++dem] = i;
  }
  if ((k > dem) || k == 0) {
    cout << 1;
    return 0;
  }
  a[++dem] = n + 1;
  long long res = 1;
  int t = 1;
  while (k) {
    int last = 1;
    xx = 0;
    if (t == 1) {
      res += solve(k);
      res %= MOD;
    } else {
      res -= solve(k);
      res = ((res % MOD) + MOD) % MOD;
    }
    a[dem] = a[dem - 1] + 1;
    a[0] = a[1] - 1;
    t = 1 - t;
    k--;
  }
  cout << res;
  return 0;
}
