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

from collections import deque, defaultdict, Counter

import random, math
import sys
import os
from io import BytesIO

sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))
# sys.setrecursionlimit()

MOD = 998244353

fac = [1] * 6000
inv = [1] * 6000
for i in range(2, 6000):
    fac[i] = i * fac[i-1] % MOD
inv[5999] = pow(fac[5999], -1, MOD)
for i in range(5999, 2, -1):
    inv[i-1] = inv[i] * i % MOD

def comb(n, k):
    if k > n: return 0
    return fac[n] * inv[k] % MOD * inv[n-k] % MOD

def calc(n, k, s):
    if k <= 0: return 1
    
    ONE = ord("1")

    ind = [-1]
    for i in range(n):
        if s[i] == ONE:
            ind.append(i)
    ind.append(n)

    #if len(ind) - 2 < k: return 1

    ans = 1
    prev = k-1
    
    m = len(ind)
    last = -1
    for i in range(k, m-1):
        cur = ind[i+1] - ind[i-k] - 1
        ans += (comb(cur, k) - comb(prev, k-1) ) % MOD
        ans %= MOD
        # print(f"{ans = :}, {cur = :}, {prev = :}")
        prev = ind[i] - ind[i-k]
        
    return ans

def solve():
    global tcases
    # n = int(sys.stdin.readline())
    n, k = map(int, sys.stdin.readline().split())
    #a = list(map(int, sys.stdin.readline().split()))
    s = sys.stdin.readline()[:-1]
    # p = sys.stdin.readline()[:-1]
    
    print(calc(n, k, s))


tcases = 1 # int(sys.stdin.readline())

for t in range(tcases):
    solve()


