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 __future__ import division, print_function
import math
import sys
import os
from io import BytesIO, IOBase
#from collections import deque, Counter, OrderedDict, defaultdict
#import heapq
#ceil,floor,log,sqrt,factorial,pow,pi,gcd
#import bisect
#from bisect import bisect_left,bisect_right

BUFSIZE = 8192


class FastIO(IOBase):
	newlines = 0

	def __init__(self, file):
		self._fd = file.fileno()
		self.buffer = BytesIO()
		self.writable = "x" in file.mode or "r" not in file.mode
		self.write = self.buffer.write if self.writable else None

	def read(self):
		while True:
			b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
			if not b:
				break
			ptr = self.buffer.tell()
			self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
		self.newlines = 0
		return self.buffer.read()

	def readline(self):
		while self.newlines == 0:
			b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
			self.newlines = b.count(b"\n") + (not b)
			ptr = self.buffer.tell()
			self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
		self.newlines -= 1
		return self.buffer.readline()

	def flush(self):
		if self.writable:
			os.write(self._fd, self.buffer.getvalue())
			self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
	def __init__(self, file):
		self.buffer = FastIO(file)
		self.flush = self.buffer.flush
		self.writable = self.buffer.writable
		self.write = lambda s: self.buffer.write(s.encode("ascii"))
		self.read = lambda: self.buffer.read().decode("ascii")
		self.readline = lambda: self.buffer.readline().decode("ascii")


def print(*args, **kwargs):
	"""Prints the values to a stream, or to sys.stdout by default."""
	sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
	at_start = True
	for x in args:
		if not at_start:
			file.write(sep)
		file.write(str(x))
		at_start = False
	file.write(kwargs.pop("end", "\n"))
	if kwargs.pop("flush", False):
		file.flush()


if sys.version_info[0] < 3:
	sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
	sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

input = lambda: sys.stdin.readline().rstrip("\r\n")

def inp():
    return(int(input()))
def inps():
    return input().strip()
def inlt():
    return(list(map(int,input().split())))
def insr():
    s = input().strip()
    return(list(s[:len(s)]))
def invr():
    return(map(int,input().split()))
N=10000

factorialNumInverse = [None] * (N + 1)
naturalNumInverse = [None] * (N + 1)
fact = [None] * (N + 1)
def InverseofNumber(p):
    naturalNumInverse[0] = naturalNumInverse[1] = 1
    for i in range(2, N + 1, 1):
        naturalNumInverse[i] = (naturalNumInverse[p % i] *
                                   (p - int(p / i)) % p)
def InverseofFactorial(p):
    factorialNumInverse[0] = factorialNumInverse[1] = 1
    for i in range(2, N + 1, 1):
        factorialNumInverse[i] = (naturalNumInverse[i] *
                                  factorialNumInverse[i - 1]) % p
 
def factorial(p):
    fact[0] = 1
    for i in range(1, N + 1):
        fact[i] = (fact[i - 1] * i) % p

mod=998244353
InverseofNumber(mod)
InverseofFactorial(mod)
factorial(mod)
F=fact
I=factorialNumInverse
n,k=invr()
l=insr()
arr=[-1]
for i in range(n):
	if l[i]=='1':
		arr.append(i)
arr.append(n)
if k==0 or k>len(arr)-2:
	print(1)
else:
	v=len(arr)
	ans=0
	for i in range(v-k-1):
		l,r=arr[i],arr[i+k+1]
		d=r-l-1
		ans+=(F[d]*I[k]*I[d-k])
		ans=ans%mod
		if i>0:
			e=arr[i+k]-l-1
			ans-=(F[e]*I[k-1]*I[e-k+1])
			ans=ans%mod
	print(ans)