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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int arr1[100000], arr2[100000];
int different(int x, int y) {
  memset(arr1, -1, sizeof(arr1));
  memset(arr2, -1, sizeof(arr2));
  int temp1 = 0, temp2 = 0;
  while (x != 0) {
    arr1[temp1] = x % 2;
    temp1++;
    x /= 2;
  }
  while (y != 0) {
    arr2[temp2] = y % 2;
    temp2++;
    y /= 2;
  }
  if (temp1 < temp2) {
    for (int i = temp1; i < temp2; i++) {
      arr1[i] = 0;
      temp1++;
    }
  } else if (temp2 < temp1) {
    for (int i = temp2; i < temp1; i++) {
      arr2[i] = 0;
      temp2++;
    }
  }
  int ans = 0;
  for (int i = 0; i < temp1; i++) {
    if (arr1[i] != arr2[i]) {
      ans++;
    }
  }
  return ans;
}
int main() {
  int n, m, k;
  cin >> n >> m >> k;
  int arr[m + 1];
  for (int i = 0; i < m + 1; i++) {
    cin >> arr[i];
  }
  int ans = 0;
  for (int i = 0; i < m; i++) {
    if (different(arr[m], arr[i]) <= k) ans++;
  }
  cout << ans << endl;
}