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:
import sys

def differ(f, c):
    df = 0
    fed = list(bin(f)[2:])
    chk = list(bin(c)[2:])
    if len(fed) > len(chk):
        for x in range(len(fed)-len(chk)):
            chk.insert(0,'0')
    else:
        for x in range(len(chk)-len(fed)):
            fed.insert(0,'0') 
    assert len(fed) == len(chk)
    for x in range(len(fed)):
        if fed[x] != chk[x]:
            df += 1
    return df

[n,m,k] = [int(x) for x in sys.stdin.readline().split()]

plist = []
for x in range(m+1):
    plist.append(int(sys.stdin.readline()))

fedor = plist[len(plist)-1]
plist = plist[:-1]

count = 0
for x in plist:
    if differ(fedor, x) <= k:
        count += 1

print count