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 friendable(army1, army2, k):
    diff = 0
    for i in range(len(army1)):
        if army1[i] != army2[i]:
            diff += 1

        if diff > k:
            return False
    
    return True


def bin_padded(x, n):
    s = bin(x)[2:]
    return (0 if len(s) > n else (n - len(s))) * '0' + s


n, m, k = [int(x) for x in input().split()]

armies = [bin_padded(int(input()), n) for _ in range(m)]
fedor_army = bin_padded(int(input()), n)

# for army in armies:
#     print(army)
# print(fedor_army)

potential_friends = 0
for army in armies:
    potential_friends += friendable(fedor_army, army, k)

print(potential_friends)