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 dec_to_bin(x):
    return bin(x)[2:]

def compare(x, y):
    charX = list(dec_to_bin(x))
    charY = list(dec_to_bin(y))
    differ = 0
    while (len(charX) < len(charY)):
        charX.insert(0, '0')
    while (len(charY) < len(charX)):
        charY.insert(0, '0')
    for i in range(0, len(charX)):
        if charX[i] != charY[i]:
            differ += 1
#     print(charX)
#     print(charY)
#     print(differ)
    return differ

n, m, k = [int(x) for x in input().split()]
numbers = []
for i in range (0, m + 1):
    numbers.append(int(input()))
res = 0
for i in range(0, m):
    if compare(numbers[i], numbers[m]) <= k:
        res += 1
print(res)