#include <bits/stdc++.h>
using namespace std;
int ch[2101000][2], tot, siz[2210101];
void ins(int x) {
  int u = 0;
  for (int i = 19; i >= 0; i--) {
    int chi = x >> i & 1;
    if (!ch[u][chi]) ch[u][chi] = ++tot;
    u = ch[u][chi];
  }
}
void dfs(int u) {
  if (u) siz[u] = 1;
  if (ch[u][0]) dfs(ch[u][0]);
  if (ch[u][1]) dfs(ch[u][1]);
  siz[u] += siz[ch[u][1]] + siz[ch[u][0]];
}
int query(int x) {
  int u = 0, res = 0;
  for (int i = 19; i >= 0; i--) {
    int chi = x >> i & 1;
    if (siz[ch[u][chi]] != (1 << (i + 1)) - 1) {
      res |= chi << i;
      u = ch[u][chi];
    } else {
      res |= (chi ^ 1) << i;
      u = ch[u][chi ^ 1];
    }
  }
  return res;
}
int main() {
  int n, QAQ;
  scanf("%d%d", &n, &QAQ);
  for (int i = 1; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    ins(x);
  }
  dfs(0);
  int tmp = 0;
  while (QAQ--) {
    int res = 0, x;
    scanf("%d", &x);
    tmp ^= x;
    res = query(tmp);
    printf("%d\n", res ^ tmp);
  }
}

### Prompt

Your task is to create a cpp solution to the following problem:

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;
int ch[2101000][2], tot, siz[2210101];
void ins(int x) {
  int u = 0;
  for (int i = 19; i >= 0; i--) {
    int chi = x >> i & 1;
    if (!ch[u][chi]) ch[u][chi] = ++tot;
    u = ch[u][chi];
  }
}
void dfs(int u) {
  if (u) siz[u] = 1;
  if (ch[u][0]) dfs(ch[u][0]);
  if (ch[u][1]) dfs(ch[u][1]);
  siz[u] += siz[ch[u][1]] + siz[ch[u][0]];
}
int query(int x) {
  int u = 0, res = 0;
  for (int i = 19; i >= 0; i--) {
    int chi = x >> i & 1;
    if (siz[ch[u][chi]] != (1 << (i + 1)) - 1) {
      res |= chi << i;
      u = ch[u][chi];
    } else {
      res |= (chi ^ 1) << i;
      u = ch[u][chi ^ 1];
    }
  }
  return res;
}
int main() {
  int n, QAQ;
  scanf("%d%d", &n, &QAQ);
  for (int i = 1; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    ins(x);
  }
  dfs(0);
  int tmp = 0;
  while (QAQ--) {
    int res = 0, x;
    scanf("%d", &x);
    tmp ^= x;
    res = query(tmp);
    printf("%d\n", res ^ tmp);
  }
}

```