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

import io, os, sys
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline

MOD, M = 998244353, 5002
F = [1]*(M+1)
for i in range(1, M+1):
    F[i] = (F[i-1] * i) % MOD

I = [1]*M + [pow(F[M], MOD - 2, MOD)]
for i in range(M-1, 0, -1):
    I[i] = I[i+1]*(i+1) % MOD


def C(a, b):
    return (F[a]*I[b]*I[a-b]) % MOD


if __name__ == '__main__':
    n, k = [int(i) for i in input().split()]
    s = [int(i) for i in input().rstrip()]
    ans = 0
    if k == 0 or s.count(1) < k:
        print(1)
    else:
        acc = [0]*(n+1)
        for i in range(1, n+1):
            acc[i] = acc[i-1] + s[i-1]

        for t in range(1, n + 1):  # real change
            for p in range(0, n - t + 1):
                p1 = s[p]
                p2 = s[p + t - 1]
                total = acc[p + t] - acc[p]
                if (acc[p + t] + acc[-1] - acc[p] >= k) and total <= k:
                    if p1 != p2:
                        ans += C(t - 2, total - 1)
                        ans %= MOD
                    elif p1 == p2 and p1 == 1:
                        if total <= t - 2:
                            ans += C(t - 2, total)
                            ans %= MOD
                    else:
                        if total >= 2:
                            ans += C(t - 2, total - 2)
                            ans %= MOD

        print((ans + 1) % MOD)