Codeforces Round #267 (Div. 2) B. Fedor and New Game_html/css_WEB-ITnose

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.

Sample test(s)

Input

7 3 18511117

Output

Input

3 3 31234

Output

3题意：给你m+1个数让你判断所给的数的二进制形式与第m+1个数不想同的位数是否小于等于k，累计答案思路：题意题<pre name="code" class="n">#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 1010;int num[maxn], cnt;int main() {    int n, m, k;    scanf("%d%d%d", &n, &m, &k);    for (int i = 0; i < m; i++)      scanf("%d", &num[i]);   scanf("%d", &cnt);  int ans = 0;    for (int i = 0; i < m; i++) {        int cur = 0;        for (int j = 0; j < n; j++)          if (((1<<j)&num[i]) != ((1<<j)&cnt))                cur++;      if (cur <= k)            ans++;  }   printf("%d\n", ans);    return 0;}