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;
double sqr(double x) { return x * x; }
double distsqr(double x, double y, double a, double b) {
  return sqr((x - a)) + sqr((y - b));
}
double cosine_angle(double a, double b, double c) {
  return (double)(sqr(a) + sqr(b) - sqr(c)) / ((double)2.0 * a * b);
}
bool pr[1000050];
int prime[1000050 / 10];
void PrimeGen() {
  int i, j, k, l = sqrt(1000050) + 1;
  pr[0] = pr[1] = 1;
  for (i = 4; i < 1000050; i += 2) pr[i] = 1;
  for (i = 3; i < l; i += 2) {
    if (!pr[i])
      for (j = i * i, k = i << 1; j < 1000050; j += k) pr[j] = 1;
  }
  for (i = j = 0; i < 1000050; i++)
    if (!pr[i]) prime[j++] = i;
}
bool Friend(int x, int y, int k) {
  int i;
  x = x ^ y;
  while (x) {
    if (x & 1) k--;
    x >>= 1;
  }
  if (k > -1)
    return 1;
  else
    return 0;
}
int num[1001];
int main() {
  int i, n, m, k, f;
  scanf("%d %d %d", &n, &m, &k);
  for (i = 0; i < m; i++) scanf("%d", &num[i]);
  scanf("%d", &f);
  for (i = n = 0; i < m; i++) {
    if (Friend(num[i], f, k)) n++;
  }
  printf("%d\n", n);
  return 0;
}