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:
def binary(n,k):
    l=[]
    quotient=n//2
    rem=n%2
    while quotient!=0:
        l.append(rem)
        rem=quotient%2
        quotient = quotient//2
    l.append(rem)
    for i in range(k-len(l)):
        l.append(0)
    l.reverse()
    lis=list(map(str,l))
    ans=''.join(lis)
    return ans

def comparison(s1,s2):
    ansc=0
    for i in range(len(s1)):
        if s1[i]!=s2[i]:
            ansc+=1
    return ansc

inp=input().split()
n=int(inp[0])
m=int(inp[1])
k=int(inp[2])
l=[]
ansq=0
for i in range(m+1):
   xi=int(input())
   l.append(xi)
for j in range(m):
    if comparison(binary(l[j],n),binary(l[-1],n))<=k:
        ansq+=1
print(ansq)