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 PYTHON):

Solution:
def toBin(a):
	kq = []
	while(a > 0):
		kq.append(a%2)
		a = a / 2
	#kq.sort(reverse = True)
	return kq
	

n,m,k = map(int,raw_input().split())
x = []
fr = 0
dif = 0

for i in range(m+1):
	x.append(toBin(int(raw_input())))
	for j in range(n - len(x[i])):
		x[i].append(0)
	#print x[i]
for i in range(m):
	dif = 0
	for j in range(n):
		if x[i][j] != x[m][j]:
			dif += 1
	if dif <= k:
		fr += 1
		
print fr