#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 6;
const int MOD = 1e9 + 7;
const int INF = 1e9 + 7;
const double eps = 1e-6;
const double Pi = acos(-1.0);
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  for (; c < '0' || '9' < c; c = getchar())
    if (c == '-') f = -1;
  for (; '0' <= c && c <= '9'; c = getchar()) x = (x << 3) + (x << 1) + c - '0';
  return x * f;
}
int ch[N * 30][2], val[N * 30], cnt;
int rev[100];
void insert(int a) {
  int now = 0;
  for (int i = 20; i >= 0; i--) {
    int bt = (a >> i) & 1;
    if (!ch[now][bt]) ch[now][bt] = ++cnt;
    now = ch[now][bt];
    val[now]++;
  }
}
int cal(int k) {
  int now = 0, a = 0;
  for (int i = 20; i >= 0; i--) {
    int bt = rev[i];
    a <<= 1;
    if (ch[now][bt]) {
      if (val[ch[now][bt]] < k)
        k -= val[ch[now][bt]];
      else {
        now = ch[now][bt];
        a |= 0;
        continue;
      }
    }
    if (ch[now][1 - bt]) {
      if (val[ch[now][1 - bt]] < k)
        return INF;
      else {
        now = ch[now][1 - bt];
        a |= 1;
        continue;
      }
    }
    return INF;
  }
  return a;
}
int n, m;
int vis[N * 4];
int main() {
  while (~scanf("%d%d", &n, &m)) {
    cnt = 0;
    memset(rev, 0, sizeof(rev));
    memset(ch, 0, sizeof(ch));
    memset(val, 0, sizeof(val));
    memset(vis, 0, sizeof(vis));
    for (int i = 1, first; i <= n; i++) {
      scanf("%d", &first);
      if (!vis[first]) insert(first);
      vis[first] = 1;
    }
    for (int i = 1, first; i <= m; i++) {
      scanf("%d", &first);
      for (int j = 20; j >= 0; j--)
        if ((first >> j) & 1) {
          rev[j] ^= 1;
        }
      int l = 1, r = n, ans = 0, mid;
      while (l <= r) {
        mid = r + l >> 1;
        if (cal(mid) + 1 <= mid)
          ans = mid, l = mid + 1;
        else
          r = mid - 1;
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
const int N = 3e5 + 6;
const int MOD = 1e9 + 7;
const int INF = 1e9 + 7;
const double eps = 1e-6;
const double Pi = acos(-1.0);
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  for (; c < '0' || '9' < c; c = getchar())
    if (c == '-') f = -1;
  for (; '0' <= c && c <= '9'; c = getchar()) x = (x << 3) + (x << 1) + c - '0';
  return x * f;
}
int ch[N * 30][2], val[N * 30], cnt;
int rev[100];
void insert(int a) {
  int now = 0;
  for (int i = 20; i >= 0; i--) {
    int bt = (a >> i) & 1;
    if (!ch[now][bt]) ch[now][bt] = ++cnt;
    now = ch[now][bt];
    val[now]++;
  }
}
int cal(int k) {
  int now = 0, a = 0;
  for (int i = 20; i >= 0; i--) {
    int bt = rev[i];
    a <<= 1;
    if (ch[now][bt]) {
      if (val[ch[now][bt]] < k)
        k -= val[ch[now][bt]];
      else {
        now = ch[now][bt];
        a |= 0;
        continue;
      }
    }
    if (ch[now][1 - bt]) {
      if (val[ch[now][1 - bt]] < k)
        return INF;
      else {
        now = ch[now][1 - bt];
        a |= 1;
        continue;
      }
    }
    return INF;
  }
  return a;
}
int n, m;
int vis[N * 4];
int main() {
  while (~scanf("%d%d", &n, &m)) {
    cnt = 0;
    memset(rev, 0, sizeof(rev));
    memset(ch, 0, sizeof(ch));
    memset(val, 0, sizeof(val));
    memset(vis, 0, sizeof(vis));
    for (int i = 1, first; i <= n; i++) {
      scanf("%d", &first);
      if (!vis[first]) insert(first);
      vis[first] = 1;
    }
    for (int i = 1, first; i <= m; i++) {
      scanf("%d", &first);
      for (int j = 20; j >= 0; j--)
        if ((first >> j) & 1) {
          rev[j] ^= 1;
        }
      int l = 1, r = n, ans = 0, mid;
      while (l <= r) {
        mid = r + l >> 1;
        if (cal(mid) + 1 <= mid)
          ans = mid, l = mid + 1;
        else
          r = mid - 1;
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}

```