#include <bits/stdc++.h>
using namespace std;
struct trie {
  trie *child[2];
  int cnt[2];
  trie() {
    child[0] = child[1] = 0;
    cnt[0] = cnt[1] = 0;
  }
  void add(int x, int i) {
    if (i == -1) return;
    int cur = ((1 << i) & x) ? 1 : 0;
    if (child[cur] == 0) child[cur] = new trie();
    cnt[cur]++;
    child[cur]->add(x, i - 1);
  }
  void delet(int x, int i) {
    if (i == -1) return;
    int cur = ((1 << i) & x) ? 1 : 0;
    cnt[cur]--;
    child[cur]->delet(x, i - 1);
  }
  int solve(int x, int i) {
    if (i == -1) return 0;
    int cur = ((1 << i) & x) ? 1 : 0;
    if (cur == 0) {
      if (child[1] != 0 && cnt[1] > 0)
        return (1 << i) + child[1]->solve(x, i - 1);
      else
        return child[0]->solve(x, i - 1);
    } else {
      if (child[0] != 0 && cnt[0] > 0)
        return child[0]->solve(x, i - 1);
      else
        return (1 << i) + child[1]->solve(x, i - 1);
    }
  }
};
int q, x;
char c;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  trie tree;
  cin >> q;
  tree.add(0, 30);
  while (q--) {
    cin >> c >> x;
    if (c == '+')
      tree.add(x, 30);
    else if (c == '-')
      tree.delet(x, 30);
    else
      cout << max(x, x ^ (tree.solve(x, 30))) << endl;
  }
}

### Prompt

In cpp, your task is to solve the following problem:

Author has gone out of the stories about Vasiliy, so here is just a formal task description.

You are given q queries and a multiset A, initially containing only integer 0. There are three types of queries:

  1. "+ x" — add integer x to multiset A.
  2. "- x" — erase one occurrence of integer x from multiset A. It's guaranteed that at least one x is present in the multiset A before this query.
  3. "? x" — you are given integer x and need to compute the value <image>, i.e. the maximum value of bitwise exclusive OR (also know as XOR) of integer x and some integer y from the multiset A.



Multiset is a set, where equal elements are allowed.

Input

The first line of the input contains a single integer q (1 ≤ q ≤ 200 000) — the number of queries Vasiliy has to perform.

Each of the following q lines of the input contains one of three characters '+', '-' or '?' and an integer xi (1 ≤ xi ≤ 109). It's guaranteed that there is at least one query of the third type.

Note, that the integer 0 will always be present in the set A.

Output

For each query of the type '?' print one integer — the maximum value of bitwise exclusive OR (XOR) of integer xi and some integer from the multiset A.

Example

Input

10
+ 8
+ 9
+ 11
+ 6
+ 1
? 3
- 8
? 3
? 8
? 11


Output

11
10
14
13

Note

After first five operations multiset A contains integers 0, 8, 9, 11, 6 and 1.

The answer for the sixth query is integer <image> — maximum among integers <image>, <image>, <image>, <image> and <image>.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
struct trie {
  trie *child[2];
  int cnt[2];
  trie() {
    child[0] = child[1] = 0;
    cnt[0] = cnt[1] = 0;
  }
  void add(int x, int i) {
    if (i == -1) return;
    int cur = ((1 << i) & x) ? 1 : 0;
    if (child[cur] == 0) child[cur] = new trie();
    cnt[cur]++;
    child[cur]->add(x, i - 1);
  }
  void delet(int x, int i) {
    if (i == -1) return;
    int cur = ((1 << i) & x) ? 1 : 0;
    cnt[cur]--;
    child[cur]->delet(x, i - 1);
  }
  int solve(int x, int i) {
    if (i == -1) return 0;
    int cur = ((1 << i) & x) ? 1 : 0;
    if (cur == 0) {
      if (child[1] != 0 && cnt[1] > 0)
        return (1 << i) + child[1]->solve(x, i - 1);
      else
        return child[0]->solve(x, i - 1);
    } else {
      if (child[0] != 0 && cnt[0] > 0)
        return child[0]->solve(x, i - 1);
      else
        return (1 << i) + child[1]->solve(x, i - 1);
    }
  }
};
int q, x;
char c;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  trie tree;
  cin >> q;
  tree.add(0, 30);
  while (q--) {
    cin >> c >> x;
    if (c == '+')
      tree.add(x, 30);
    else if (c == '-')
      tree.delet(x, 30);
    else
      cout << max(x, x ^ (tree.solve(x, 30))) << endl;
  }
}

```