C. Boboniu and Bit Operations
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Boboniu likes bit operations. He wants to play a game with you.

Boboniu gives you two sequences of non-negative integers ?1,?2,…,?? and ?1,?2,…,??.

For each ? (1≤?≤?), you're asked to choose a ? (1≤?≤?) and let ??=??&??, where & denotes the bitwise AND operation. Note that you can pick the same ? for different ?'s.

Find the minimum possible ?1|?2|…|??, where | denotes the bitwise OR operation.

Input
The first line contains two integers ? and ? (1≤?,?≤200).

The next line contains ? integers ?1,?2,…,?? (0≤??<29).

The next line contains ? integers ?1,?2,…,?? (0≤??<29).

Output
Print one integer: the minimum possible ?1|?2|…|??.

Examples
inputCopy
4 2
2 6 4 0
2 4
outputCopy
2
inputCopy
7 6
1 9 1 9 8 1 0
1 1 4 5 1 4
outputCopy
0
inputCopy
8 5
179 261 432 162 82 43 10 38
379 357 202 184 197
outputCopy
147
Note
For the first example, we have ?1=?1&?2=0, ?2=?2&?1=2, ?3=?3&?1=0, ?4=?4&?1=0.Thus ?1|?2|?3|?4=2, and this is the minimal answer we can get.

/*
 * 题意: 给你两个数组a长度为n(1~200), b长度为m(1~200)，定义数组c为c[i] = a[i] & b[j]，一个b[h]可以和多个a[i]配对，然后让你求c[0] | c[1] ... | c[n - 1]的最小值
 *  数组a,b的元素范围(0~2^9-1)
 *
 * 解决: 之前老是想着贪心，但是贪心是没法满足最优解的。换个思路，数组元素最大是2^9 - 1，刚好是二进制(1111 1111)，所以最后的c[0] | c[1] ... | c[n - 1]的范围
 *  也只能是[0~2^9-1]，所以只需要从0到2^9-1遍历即可，找到一个解就是最后结果
 */

#include <bits/stdc++.h>

using namespace std;

int n, m;
int a[234], b[234];

bool check(int x) {
  bool ok;
  for (int i = 0; i < n; i++) {
    ok = false;
    for (int j = 0; j < m; j++) {
      if (((a[i] & b[j]) | x) == x) {
        ok = true;
        break;
      }
    }
    if (!ok) return false;
  }
  return true;
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < m; i++) cin >> b[i];

  for (int res = 0; res < 512; res++) {
    if (check(res) == true) {
      cout << res << "\n";
      break;
    }
  }

  return 0;
}