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:
n,m,k=map(int,input().split())
a=list()
for i in range(m+1):
    a.append(bin(int(input())))
x=0
for i in range(m):
    h=0
    j,q=len(a[m])-1,len(a[i])-1
    while a[m][j]!='b' and a[i][q]!='b':
        if(a[m][j]!=a[i][q]):
            h+=1
        j-=1
        q-=1
        if(a[m][j]=='b' and a[i][q]!='b'):
            while a[i][q]!='b':
                if(a[i][q]=='1'):
                    h+=1
                q-=1
        elif(a[m][j]!='b' and a[i][q]=='b'):
            while a[m][j]!='b':
                if(a[m][j]=='1'):
                    h+=1
                j-=1
    if(h<=k):
        x+=1
print(x)