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 double eps = 1e-9;
const int maxn = (int)1e5 + 10;
const int mod = (int)1e9 + 7;
int dx[] = {0, 0, 1, -1, 1, -1, 1, -1};
int dy[] = {1, -1, 0, 0, 1, -1, -1, 1};
long long a, b, c, d[1006], i, j, k, l, t, n, m, p, q, r, s;
char h;
double e, f, g;
vector<long long> u, v, w;
pair<long long, long long> pp;
string x, y, z;
map<string, long long> mm;
string tobin(int n) {
  string s;
  int i;
  while (n > 0) {
    if (((bool)(n & (1 << (i)))))
      s = '1' + s;
    else
      s = '0' + s;
    n /= 2;
  }
  return s;
}
int main() {
  while (cin >> a >> b >> c) {
    for (i = (0); i <= (b); i++) {
      cin >> d[i];
    }
    x = tobin(d[b]);
    l = ((x).size());
    for (i = (l); i <= (a - 1); i++) x = "0" + x;
    m = 0;
    for (i = 0; i < (b); i++) {
      y = tobin(d[i]);
      l = ((y).size());
      for (j = (l); j <= (a - 1); j++) y = "0" + y;
      n = 0;
      for (j = 0; j < (a); j++) {
        if (x[j] != y[j]) n++;
      }
      if (n <= c) m++;
    }
    cout << m << endl;
  }
  return 0;
}