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 

N, K = map(int,input().split())

S = list(input().rstrip())
S = [int(s) for s in S]
mod = 998244353
  
from time import time
"""
N, K = 10 * 5*10**2, 10 * 3*10**2
S = ["10110" for i in range(K)]
S = "".join(S)
S = [int(s) for s in S]
"""
it = time()

cons = 0
prof = []
for s in S:
  if s == 0:
    cons += 1
  else:
    prof.append(cons)
    cons = 0
    
prof.append(cons)
#print(prof)  
#print("1", time() - it)
_inv_t = {}
_inv_t[1] = 1
_inv_t[0] = 0
def inv(x, mod):
  x %= mod
  if x not in _inv_t:
    _inv_t[x] = inv(mod % x, mod) * (mod - mod // x) % mod
  return _inv_t[x]

for x in range(2,N+10):
  _inv_t[x] = inv(mod % x, mod) * (mod - mod // x) % mod

comb = [[1, 0], [1, 1, 0]]
for n in range(2, N+3):
  now = 1
  arr = [1]
  for i in range(1, n + 1):
    arr.append((comb[-1][i-1] + comb[-1][i]) % mod)
  arr.append(0)
  comb.append(arr)
    
#print(comb)    
#print("2", time() - it)
import sys
if K == 0:
  print(1)
  sys.exit()
if K > S.count(1):
  print(1)
  sys.exit()

  
ans = 0
fac = -1
for k in range(K, max(-1, K-2), -1):
  fac *= -1
  n = sum(prof[K-k:K+1]) + k
  for i in range(K + 1, len(prof)-(K-k)+1):
    #print("n", n)
    ans += comb[n][k] * fac
    ans %= mod
    if i < len(prof):
      n += prof[i]
      n -= prof[i-k-1]
  #print(ans)  

#print("3", time() - it)
  
print(ans)
  