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 byte_v(a, n):
    ans = []
    while a != 0:
        ans.append(a % 2)
        a //= 2
    ans.reverse()
    
    ans = (n - len(ans))  * [0] + ans

    return ans


n, m, k = map(int, input().split())
players = []
for i in range(m + 1):
    curr_army = int(input())
    players.append(list(byte_v(curr_army, n)))
phill = players[-1]
ans = 0
for i in range(m):
    curr = 0
    
    for j in range(n):
        if phill[j] != players[i][j]:
            curr += 1
    if curr <= k:
        ans += 1
print(ans)