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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const int INF = 1000001000;
const long long INFLL = INF * 1LL * INF;
const int mod = 1000 * 1000 * 1000 + 7;
const int mod9 = 1000 * 1000 * 1000 + 9;
const int modr = 99990001;
const double PI = 3.1415926535897932;
double sqr(double first) { return first * first; };
long double sqr(long double first) { return first * first; };
long long sqr(long long first) { return first * first; };
long long sqr(int first) { return first * 1LL * first; };
long long gcd(long long a, long long b) {
  while (b) a %= b, swap(a, b);
  return a;
}
long long bpm(long long a, long long n = -2, long long m = mod) {
  n = n >= 0 ? n : m + n;
  long long r = 1;
  while (n) {
    if (n & 1) r = (r * a) % m;
    a = (a * a) % m;
    n >>= 1;
  }
  return r;
}
long long dist(long long x1, long long y1, long long x2, long long y2) {
  return sqr(x1 - x2) + sqr(y1 - y2);
}
int bitc(long long first) {
  int r = 0;
  for (int i = first; i; i >>= 1) r += i & 1;
  return r;
}
std::ostream& operator<<(std::ostream& os, pair<int, int> p) {
  return os << "{" << p.first << " " << p.second << "}";
}
const int N = 100005;
int a[N];
int main() {
  int n, m, k;
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 0, ei = m - 1; i <= ei; i++) {
    scanf("%d", &a[i]);
  }
  int first;
  scanf("%d", &first);
  int res = 0;
  for (int i = 0, ei = m - 1; i <= ei; i++) {
    if (bitc(a[i] ^ first) <= k) {
      res++;
    }
  }
  printf("%d\n", res);
  return 0;
}