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;
bool compare(int indx);
vector<int> v[1010];
int szz;
long long n, m, k;
int main() {
  scanf("%I64d %I64d %I64d", &n, &m, &k);
  for (int i = 0; i <= m; i++) v[i].clear();
  for (int i = 1; i <= m + 1; i++) {
    long long num;
    scanf("%I64d", &num);
    long long c = num;
    while (c) {
      v[i].push_back(c % 2);
      c = c / 2;
    }
  }
  szz = v[m + 1].size();
  int res = 0;
  for (int i = 1; i <= m; i++) {
    if (compare(i)) res++;
  }
  cout << res << endl;
  return 0;
}
bool compare(int indx) {
  int sz = v[indx].size();
  int beshi;
  if (sz >= szz)
    beshi = indx;
  else
    beshi = m + 1;
  int mn = min(szz, sz);
  int cnt = 0;
  for (int i = 0; i < mn; i++) {
    if (v[m + 1][i] != v[indx][i]) cnt++;
  }
  int s = v[beshi].size();
  for (int j = mn; j < s; j++) {
    if (v[beshi][j] == 1) cnt++;
  }
  if (cnt <= k) return true;
  return false;
}