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 sys
input = sys.stdin.readline
mod = 998244353


def power(num, ex):
    pw, res = num, 1
    while ex:
        if ex & 1:
            res = res * pw
        pw = pw * pw
        ex >>= 1
    return res


def power_with_modulo(num, ex):
    pw, res = num % mod, 1
    while ex:
        if ex & 1:
            res = res * pw % mod
        pw = pw * pw % mod
        ex >>= 1
    return res


n, k = map(int, input().split())
s = str(input())
ans = 1
prefix_sum = [0 for i in range(n)]
fac = [1 for i in range(n+1)]


for i in range(1, n+1):
    fac[i] = (fac[i-1]*i) % mod

prefix_sum[0] = 1 if s[0] == '1' else 0
for i in range(1, n):
    prefix_sum[i] = prefix_sum[i-1] + int(s[i] == '1')

if prefix_sum[n-1] < k:
    print(1)
    exit()

C = [[0 for i in range(n+1)] for i in range(n+1)]
C[0][0] = 1
for i in range(1, n+1):
    C[i][0] = 1
    for j in range(1, i+1):
        C[i][j] = (C[i-1][j] + C[i-1][j-1]) % mod

#print(C)
for i in range(n):
    for j in range(i+1, n):
        cnt = (j-i+1)
        cnt1 = prefix_sum[j] - (prefix_sum[i-1] if i >= 1 else 0)
        if cnt1 > k:
            continue
        cnt -= 2
        if s[i] == '0':
            cnt1 -= 1
        if s[j] == '0':
            cnt1 -= 1
        if (cnt1 >= 0) and (cnt >= cnt1):
            ans = (ans+C[cnt][cnt1]) % mod

print(ans)





