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 process(n_m_k, soldiers, fedor):
    n = n_m_k[0]
    k = n_m_k[2]
    friends = 0
    for soldier in soldiers:
        num_diff = 0
        curr_fedor = fedor
        curr_friend = soldier
        for _ in xrange(n):
            fedor_bit = curr_fedor % 2
            friend_bit = curr_friend % 2
            if fedor_bit != friend_bit:
                num_diff+=1
            curr_fedor /= 2
            curr_friend /= 2
        if num_diff <= k:
            friends+=1
    print friends


def main():
    n_m_k = map(int, raw_input().split())
    soldiers = []
    for _ in xrange(n_m_k[1]):
        soldiers.append(int(raw_input()))
    fedor = int(raw_input())
    process(n_m_k, soldiers, fedor)

main()