#include <bits/stdc++.h>
using namespace std;
const int twopower = 524288;
bool tree[525000], leaves[525000];
int main() {
  ios::sync_with_stdio(false);
  int n, m, a;
  int xorval = 0;
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    cin >> a;
    if (leaves[a] == false) {
      leaves[a] = true;
      int iter = (twopower / 2) - 1 + (a / 2);
      int ctwopower = twopower / 2;
      if (leaves[a + (a % 2 == 0 ? 1 : -1)] == true) {
        tree[iter] = true;
        while (iter != 0) {
          if (tree[iter + (iter % 2 == 1 ? 1 : -1)] == true) {
            ctwopower /= 2;
            iter = (iter - 1) / 2;
            tree[iter] = true;
          } else {
            break;
          }
        }
      }
    }
  }
  for (int i = 0; i < m; ++i) {
    int currval;
    cin >> currval;
    xorval = xorval ^ currval;
    int ctpower = twopower / 2;
    currval = 0;
    while (currval < 524288) {
      int pval = ((xorval & ctpower) == 0 ? 1 : 2);
      currval *= 2;
      if (ctpower != 1) {
        if (tree[currval + pval] == true) {
          currval += (3 - pval);
        } else {
          currval += pval;
        }
        ctpower /= 2;
      } else {
        if (leaves[(currval - 524287) + pval] == true) {
          currval += (3 - pval);
        } else {
          currval += pval;
        }
        break;
      }
    }
    cout << ((currval - 524287) ^ xorval) << endl;
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
const int twopower = 524288;
bool tree[525000], leaves[525000];
int main() {
  ios::sync_with_stdio(false);
  int n, m, a;
  int xorval = 0;
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    cin >> a;
    if (leaves[a] == false) {
      leaves[a] = true;
      int iter = (twopower / 2) - 1 + (a / 2);
      int ctwopower = twopower / 2;
      if (leaves[a + (a % 2 == 0 ? 1 : -1)] == true) {
        tree[iter] = true;
        while (iter != 0) {
          if (tree[iter + (iter % 2 == 1 ? 1 : -1)] == true) {
            ctwopower /= 2;
            iter = (iter - 1) / 2;
            tree[iter] = true;
          } else {
            break;
          }
        }
      }
    }
  }
  for (int i = 0; i < m; ++i) {
    int currval;
    cin >> currval;
    xorval = xorval ^ currval;
    int ctpower = twopower / 2;
    currval = 0;
    while (currval < 524288) {
      int pval = ((xorval & ctpower) == 0 ? 1 : 2);
      currval *= 2;
      if (ctpower != 1) {
        if (tree[currval + pval] == true) {
          currval += (3 - pval);
        } else {
          currval += pval;
        }
        ctpower /= 2;
      } else {
        if (leaves[(currval - 524287) + pval] == true) {
          currval += (3 - pval);
        } else {
          currval += pval;
        }
        break;
      }
    }
    cout << ((currval - 524287) ^ xorval) << endl;
  }
  return 0;
}

```