#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
constexpr double eps = 1e-8;
constexpr int NINF = 0xc0c0c0c0;
constexpr int INF = 0x3f3f3f3f;
constexpr ll mod = 1e9 + 7;
constexpr ll N = 1e6 + 5;
struct Trie {
  static constexpr int MAXBIT = 30;
  int next[N][2], cnt, num[N], sz[N][2];
  void init() {
    memset(next, 0, sizeof(next));
    memset(num, 0, sizeof(num));
    memset(sz, 0, sizeof(sz));
    cnt = 1;
  }
  void insert(int x) {
    int cur = 1;
    for (int i = MAXBIT; ~i; i--) {
      int bit = (x >> i) & 1;
      if (!next[cur][bit]) next[cur][bit] = ++cnt;
      sz[cur][bit]++;
      cur = next[cur][bit];
    }
    num[cur] = x;
  }
  int query(int x) {
    int cur = 1, ans = 0;
    for (int i = MAXBIT; ~i; i--) {
      int bit = (x >> i) & 1;
      if (sz[cur][bit] < (1 << i)) {
        cur = next[cur][bit];
      } else {
        cur = next[cur][bit ^ 1];
        ans |= 1 << i;
      }
    }
    return ans;
  }
} trie;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, m;
  cin >> n >> m;
  vector<bool> vis(300000 + 5, false);
  trie.init();
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    if (!vis[x]) {
      vis[x] = true;
      trie.insert(x);
    }
  }
  int sum = 0;
  while (m--) {
    int x;
    cin >> x;
    sum ^= x;
    cout << trie.query(sum) << '\n';
  }
  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;
using ll = long long;
using pii = pair<int, int>;
constexpr double eps = 1e-8;
constexpr int NINF = 0xc0c0c0c0;
constexpr int INF = 0x3f3f3f3f;
constexpr ll mod = 1e9 + 7;
constexpr ll N = 1e6 + 5;
struct Trie {
  static constexpr int MAXBIT = 30;
  int next[N][2], cnt, num[N], sz[N][2];
  void init() {
    memset(next, 0, sizeof(next));
    memset(num, 0, sizeof(num));
    memset(sz, 0, sizeof(sz));
    cnt = 1;
  }
  void insert(int x) {
    int cur = 1;
    for (int i = MAXBIT; ~i; i--) {
      int bit = (x >> i) & 1;
      if (!next[cur][bit]) next[cur][bit] = ++cnt;
      sz[cur][bit]++;
      cur = next[cur][bit];
    }
    num[cur] = x;
  }
  int query(int x) {
    int cur = 1, ans = 0;
    for (int i = MAXBIT; ~i; i--) {
      int bit = (x >> i) & 1;
      if (sz[cur][bit] < (1 << i)) {
        cur = next[cur][bit];
      } else {
        cur = next[cur][bit ^ 1];
        ans |= 1 << i;
      }
    }
    return ans;
  }
} trie;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, m;
  cin >> n >> m;
  vector<bool> vis(300000 + 5, false);
  trie.init();
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    if (!vis[x]) {
      vis[x] = true;
      trie.insert(x);
    }
  }
  int sum = 0;
  while (m--) {
    int x;
    cin >> x;
    sum ^= x;
    cout << trie.query(sum) << '\n';
  }
  return 0;
}

```