#include <bits/stdc++.h>
struct Trie {
  Trie* c[2];
  int st;
  Trie() : st(0) { c[0] = c[1] = NULL; }
  void add(int x, int d) {
    ++st;
    if (d == -1) return;
    if (c[((1 << d) & x) != 0] == NULL) c[((1 << d) & x) != 0] = new Trie();
    c[((1 << d) & x) != 0]->add(x, d - 1);
  }
  void del(int x, int d) {
    --st;
    if (d != -1) {
      c[((1 << d) & x) != 0]->del(x, d - 1);
      if (c[((1 << d) & x) != 0]->st == 0) {
        delete c[((1 << d) & x) != 0];
        c[((1 << d) & x) != 0] = NULL;
      }
    }
  }
  int q(int x, int d) {
    if (d == -1) return 0;
    if (c[(((1 << d) & x) != 0) ^ 1] != NULL)
      return c[(((1 << d) & x) != 0) ^ 1]->q(x, d - 1) | (1 << d);
    else
      return c[((1 << d) & x) != 0]->q(x, d - 1);
  }
};
Trie trie;
int trielen;
int main() {
  int q, x;
  char op;
  scanf("%d", &q);
  trie.add(0, 30);
  while (q--) {
    scanf(" %c %d", &op, &x);
    switch (op) {
      case '+':
        trie.add(x, 30);
        break;
      case '-':
        trie.del(x, 30);
        break;
      case '?':
        printf("%d\n", trie.q(x, 30));
        break;
    }
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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>
struct Trie {
  Trie* c[2];
  int st;
  Trie() : st(0) { c[0] = c[1] = NULL; }
  void add(int x, int d) {
    ++st;
    if (d == -1) return;
    if (c[((1 << d) & x) != 0] == NULL) c[((1 << d) & x) != 0] = new Trie();
    c[((1 << d) & x) != 0]->add(x, d - 1);
  }
  void del(int x, int d) {
    --st;
    if (d != -1) {
      c[((1 << d) & x) != 0]->del(x, d - 1);
      if (c[((1 << d) & x) != 0]->st == 0) {
        delete c[((1 << d) & x) != 0];
        c[((1 << d) & x) != 0] = NULL;
      }
    }
  }
  int q(int x, int d) {
    if (d == -1) return 0;
    if (c[(((1 << d) & x) != 0) ^ 1] != NULL)
      return c[(((1 << d) & x) != 0) ^ 1]->q(x, d - 1) | (1 << d);
    else
      return c[((1 << d) & x) != 0]->q(x, d - 1);
  }
};
Trie trie;
int trielen;
int main() {
  int q, x;
  char op;
  scanf("%d", &q);
  trie.add(0, 30);
  while (q--) {
    scanf(" %c %d", &op, &x);
    switch (op) {
      case '+':
        trie.add(x, 30);
        break;
      case '-':
        trie.del(x, 30);
        break;
      case '?':
        printf("%d\n", trie.q(x, 30));
        break;
    }
  }
  return 0;
}

```