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
import heapq
from bisect import bisect_left
from collections import Counter, deque, defaultdict
input = lambda: sys.stdin.readline().rstrip("\r\n")
sys.setrecursionlimit(300005)
LOGN = 20
INF = 10 ** 9 + 5
# MOD = 10 ** 9 + 7
MOD = 998244353
# ============================ START OF MY CODE ============================

class nCk:
    def __init__(self, maxn, mod):
        self.mod = mod
        self.fact = fact = [1 for _ in range(maxn + 1)]
        self.temp = temp = [1 for _ in range(maxn + 1)]
        self.invs = invs = [1 for _ in range(maxn + 1)]
        for i in range(2, maxn + 1):
            fact[i] = fact[i - 1] * i % mod
            temp[i] = (- temp[mod % i] * (mod // i)) % mod
            invs[i] = invs[i - 1] * temp[i] % mod

    def get(self, n, k):
        if not 0 <= k <= n:
            return 0
        return (self.fact[n] * self.invs[k] % self.mod) * self.invs[n - k] % self.mod

def solve():
    N, K = map(int, input().split())
    S = input()
    if S.count('1') < K or K == 0:
        print(1)
        return

    nck = nCk(5001, MOD)
    ans = left = 0
    cur_ones = 0
    for right in range(N):
        cur_ones += int(S[right])
        while cur_ones > K:
            cur_ones -= int(S[left])
            left += 1
        if S[right] == '1' and right - left >= cur_ones:
            ans += nck.get(right - left, cur_ones)
        if S[right] != '1' and right - left >= cur_ones - 1 >= 0:
            ans += nck.get(right - left, cur_ones - 1)
        ans %= MOD
    print((ans + 1) % MOD)

if __name__ == "__main__":
    solve()
