#include <bits/stdc++.h>
using namespace std;
inline int read() {
  register int res = 0, c;
  while (c = getchar(), c < '0' || c > '9')
    ;
  do {
    res = (res << 3) + (res << 1) + (c ^ 48);
  } while (c = getchar(), c >= '0' && c <= '9');
  return res;
}
int trie[10000010][2], sum[10000010];
map<int, int> M;
int tot;
int main() {
  int N = read(), Q = read();
  for (int i = 1; i <= N; i++) {
    int n = read(), now = 0;
    if (M[n]) continue;
    M[n] = 1;
    for (int t = 20; t >= 0; t--) {
      int x = !!(n & (1 << t));
      if (!trie[now][x]) trie[now][x] = ++tot;
      now = trie[now][x];
      sum[now]++;
    }
  }
  int X = 0;
  for (int i = 1; i <= Q; i++) {
    X ^= read();
    int ans = 0, now = 0;
    for (int t = 20; t >= 0; t--) {
      int x = !!(X & (1 << t));
      if (!trie[now][x]) {
        break;
      }
      if (sum[trie[now][x]] != (1 << t))
        now = trie[now][x];
      else
        now = trie[now][x ^ 1], ans += (1 << t);
    }
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

Today at the lesson Vitya learned a very interesting function — mex. Mex of a sequence of numbers is the minimum non-negative number that is not present in the sequence as element. For example, mex([4, 33, 0, 1, 1, 5]) = 2 and mex([1, 2, 3]) = 0.

Vitya quickly understood all tasks of the teacher, but can you do the same?

You are given an array consisting of n non-negative integers, and m queries. Each query is characterized by one number x and consists of the following consecutive steps:

  * Perform the bitwise addition operation modulo 2 (xor) of each array element with the number x. 
  * Find mex of the resulting array. 



Note that after each query the array changes.

Input

First line contains two integer numbers n and m (1 ≤ n, m ≤ 3·105) — number of elements in array and number of queries.

Next line contains n integer numbers ai (0 ≤ ai ≤ 3·105) — elements of then array.

Each of next m lines contains query — one integer number x (0 ≤ x ≤ 3·105).

Output

For each query print the answer on a separate line.

Examples

Input

2 2
1 3
1
3


Output

1
0


Input

4 3
0 1 5 6
1
2
4


Output

2
0
0


Input

5 4
0 1 5 6 7
1
1
4
5


Output

2
2
0
2

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
inline int read() {
  register int res = 0, c;
  while (c = getchar(), c < '0' || c > '9')
    ;
  do {
    res = (res << 3) + (res << 1) + (c ^ 48);
  } while (c = getchar(), c >= '0' && c <= '9');
  return res;
}
int trie[10000010][2], sum[10000010];
map<int, int> M;
int tot;
int main() {
  int N = read(), Q = read();
  for (int i = 1; i <= N; i++) {
    int n = read(), now = 0;
    if (M[n]) continue;
    M[n] = 1;
    for (int t = 20; t >= 0; t--) {
      int x = !!(n & (1 << t));
      if (!trie[now][x]) trie[now][x] = ++tot;
      now = trie[now][x];
      sum[now]++;
    }
  }
  int X = 0;
  for (int i = 1; i <= Q; i++) {
    X ^= read();
    int ans = 0, now = 0;
    for (int t = 20; t >= 0; t--) {
      int x = !!(X & (1 << t));
      if (!trie[now][x]) {
        break;
      }
      if (sum[trie[now][x]] != (1 << t))
        now = trie[now][x];
      else
        now = trie[now][x ^ 1], ans += (1 << t);
    }
    printf("%d\n", ans);
  }
  return 0;
}

```