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 no_diff_bits (a,b):
    maxi = max(a,b)
    mini = min(a,b)
    counter = 0
    while maxi != 0 :
        r1 = maxi % 2
        maxi = maxi // 2
        r2 = mini % 2
        mini = mini // 2
        if r1 != r2:
            counter = counter + 1
    #print(counter)
    return counter
n , m , k = [int(x) for x in input().split()]
c = []
for i in range(m+1):
    x = int(input())
    c.append(x)
counter = 0
for i in range(m):
    diff = no_diff_bits(c[i],c[m])
    if diff <= k :
        counter = counter +1
print(counter)