Problem: After you had helped George and Alex to move in the dorm, they went to help their friend Fedor play a new computer game «Call of Soldiers 3».

The game has (m + 1) players and n types of soldiers in total. Players «Call of Soldiers 3» are numbered form 1 to (m + 1). Types of soldiers are numbered from 0 to n - 1. Each player has an army. Army of the i-th player can be described by non-negative integer xi. Consider binary representation of xi: if the j-th bit of number xi equal to one, then the army of the i-th player has soldiers of the j-th type. 

Fedor is the (m + 1)-th player of the game. He assume that two players can become friends if their armies differ in at most k types of soldiers (in other words, binary representations of the corresponding numbers differ in at most k bits). Help Fedor and count how many players can become his friends.

Input

The first line contains three integers n, m, k (1 ≤ k ≤ n ≤ 20; 1 ≤ m ≤ 1000).

The i-th of the next (m + 1) lines contains a single integer xi (1 ≤ xi ≤ 2n - 1), that describes the i-th player's army. We remind you that Fedor is the (m + 1)-th player.

Output

Print a single integer — the number of Fedor's potential friends.

Examples

Input

7 3 1
8
5
111
17


Output

0


Input

3 3 3
1
2
3
4


Output

3

Solution (in PYTHON3):

Solution:
t=list(map(int,input().split()))
n=t[0]
m=t[1]
k=t[2]

def binary_conv(n,d):
	quo=n
	arr=[]
	while True:
		if quo==0:
			break
		n=quo
		quo=n//2
		rem=n%2
		arr.append(rem)
	#arr=arr[::-1]
	if len(arr)!=d:
		yo=d-len(arr)
		for i in range(yo):
			arr.append(0)
	arr=arr[::-1]
	return arr
	
#print(binary_conv(128,10))	
game=[]
for i in range(m+1):
	x=int(input())
	game.append(binary_conv(x,n))
#print(game)

ans=0
for u in range(m):
	#print(game[u])
	count=0
	
	for j in range(n):
		#print(game[u][j],game[m][j])
		#print(count,k)
		if game[u][j]!=game[m][j]:
			count+=1
		if count>k:
			break
	if count<=k:
		ans+=1		
print(ans)