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;
const int MAXN = (int)(1e5 + 1e1);
const double PI = acos(-1.0);
long long ans, t, k, n, m, q = 1, a[MAXN], b[MAXN], c[MAXN];
bool e;
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m >> k;
  for (int i = 1; i <= m + 1; i++) {
    cin >> a[i];
  }
  while (a[m + 1] > 0) {
    t = a[m + 1] % 2;
    a[m + 1] /= 2;
    q++;
    if (t == 1) {
      if (q != n)
        b[q % n] = 1;
      else
        b[n] = 1;
    }
  }
  for (int i = 1; i <= m; i++) {
    long long w = 0;
    t = 0, q = 1;
    while (a[i] > 0) {
      t = a[i] % 2;
      a[i] /= 2;
      q++;
      if (t == 1) {
        if (q != n)
          c[q % n] = 1;
        else
          c[n] = 1;
      }
    }
    for (int j = 1; j <= n; j++) {
      if (c[j] != b[j]) w++;
      if (w > k) {
        e = 0;
        break;
      } else
        e = 1;
    }
    if (e) ans++;
    fill(c + 1, c + n + 1, 0);
  }
  cout << ans;
  return 0;
}