#include <bits/stdc++.h>
using namespace std;
struct node {
  int arr[2];
  bool cbt;
  int count;
  node() {
    arr[0] = arr[1] = -1;
    cbt = false;
    count = 0;
  }
};
vector<node> v;
vector<int> inp;
int xr = 0, ans = 0;
void insert(int a, int pos, int &x) {
  if (pos < 0) {
    v[a].count++;
    v[a].cbt = true;
    return;
  }
  int b = (x >> pos) & 1;
  if (v[a].arr[b] == -1) {
    node temp;
    v[a].arr[b] = ((int)((v).size()));
    v.emplace_back(temp);
  }
  insert(v[a].arr[b], pos - 1, x);
  if (v[a].arr[0] != -1 && v[a].arr[1] != -1) {
    if (v[v[a].arr[0]].cbt && v[v[a].arr[1]].cbt) v[a].cbt = true;
  }
}
void get(int a, int pos) {
  if (pos < 0 || a == -1) return;
  int b = (xr >> pos) & 1;
  if (v[a].arr[b] != -1)
    if (v[v[a].arr[b]].cbt) b = (b + 1) % 2;
  ans |= (b ^ ((xr >> pos) & 1)) << pos;
  get(v[a].arr[b], pos - 1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 0, n = 0, a = 0;
  node root;
  v.emplace_back(root);
  cin >> n >> t;
  inp.reserve(n);
  for (__typeof(n) i = (0) - ((0) > (n)); i != (n) - ((0) > (n));
       i += 1 - 2 * ((0) > (n))) {
    cin >> a;
    insert(0, 29, a);
    inp.emplace_back(a);
  }
  while (t--) {
    cin >> a;
    xr ^= a;
    ans = 0;
    get(0, 29);
    cout << ans << "\n";
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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;
struct node {
  int arr[2];
  bool cbt;
  int count;
  node() {
    arr[0] = arr[1] = -1;
    cbt = false;
    count = 0;
  }
};
vector<node> v;
vector<int> inp;
int xr = 0, ans = 0;
void insert(int a, int pos, int &x) {
  if (pos < 0) {
    v[a].count++;
    v[a].cbt = true;
    return;
  }
  int b = (x >> pos) & 1;
  if (v[a].arr[b] == -1) {
    node temp;
    v[a].arr[b] = ((int)((v).size()));
    v.emplace_back(temp);
  }
  insert(v[a].arr[b], pos - 1, x);
  if (v[a].arr[0] != -1 && v[a].arr[1] != -1) {
    if (v[v[a].arr[0]].cbt && v[v[a].arr[1]].cbt) v[a].cbt = true;
  }
}
void get(int a, int pos) {
  if (pos < 0 || a == -1) return;
  int b = (xr >> pos) & 1;
  if (v[a].arr[b] != -1)
    if (v[v[a].arr[b]].cbt) b = (b + 1) % 2;
  ans |= (b ^ ((xr >> pos) & 1)) << pos;
  get(v[a].arr[b], pos - 1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 0, n = 0, a = 0;
  node root;
  v.emplace_back(root);
  cin >> n >> t;
  inp.reserve(n);
  for (__typeof(n) i = (0) - ((0) > (n)); i != (n) - ((0) > (n));
       i += 1 - 2 * ((0) > (n))) {
    cin >> a;
    insert(0, 29, a);
    inp.emplace_back(a);
  }
  while (t--) {
    cin >> a;
    xr ^= a;
    ans = 0;
    get(0, 29);
    cout << ans << "\n";
  }
  return 0;
}

```