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:
x=input()
n=int(x.split()[0])
m=int(x.split()[1])
k=int(x.split()[2])
a=[]
friends=0
def leneq(a,b):
    if(len(b)<len(a)):
        while(len(b)<len(a)):
            b='0'+b
        return b
    else:
        while(len(a)<len(b)):
            a='0'+a
        return a 
for i in range(m+1):
    a.append(bin(int(input())))
    a[i]=a[i][2:]
for i in range(m):
    count=0
    if(len(a[i])<len(a[m])):
        a[i]=leneq(a[i],a[m])
    elif(len(a[i])>len(a[m])):
        a[m]=leneq(a[i],a[m])
    for j in range(len(a[m])):
        if(a[i][j]!=a[m][j]):
            count+=1
    if(count<=k):
        friends+=1
print(friends)