#include <bits/stdc++.h>
using namespace std;
struct Trie {
  map<bool, Trie> m;
  int n;
  int has;
  void insert(int x, int bit) {
    has++;
    if (bit == -1) {
      n = x;
      return;
    }
    m[(x >> bit) & 1].insert(x, bit - 1);
  }
  void remove(int x, int bit) {
    has--;
    if (bit == -1) {
      return;
    }
    if (m[(x >> bit) & 1].has > 1) {
      m[(x >> bit) & 1].remove(x, bit - 1);
    } else {
      m.erase((x >> bit) & 1);
    }
  }
  int query(int x, int bit) {
    if (bit == -1) {
      return n;
    }
    if ((x >> bit) & 1) {
      if (m.count(false)) {
        return m[false].query(x, bit - 1);
      } else {
        return m[true].query(x, bit - 1);
      }
    } else {
      if (m.count(true)) {
        return m[true].query(x, bit - 1);
      } else {
        return m[false].query(x, bit - 1);
      }
    }
  }
};
int main() {
  int q, x, i;
  char op;
  Trie t;
  scanf("%d%*c", &q);
  t.insert(0, 30);
  for (i = 0; i < q; i++) {
    scanf("%c%*c%d%*c", &op, &x);
    switch (op) {
      case '+':
        t.insert(x, 30);
        break;
      case '-':
        t.remove(x, 30);
        break;
      case '?':
        printf("%d\n", x ^ t.query(x, 30));
        break;
      default:
        break;
    }
  }
  return 0;
}

### Prompt

Develop a solution in cpp 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>
using namespace std;
struct Trie {
  map<bool, Trie> m;
  int n;
  int has;
  void insert(int x, int bit) {
    has++;
    if (bit == -1) {
      n = x;
      return;
    }
    m[(x >> bit) & 1].insert(x, bit - 1);
  }
  void remove(int x, int bit) {
    has--;
    if (bit == -1) {
      return;
    }
    if (m[(x >> bit) & 1].has > 1) {
      m[(x >> bit) & 1].remove(x, bit - 1);
    } else {
      m.erase((x >> bit) & 1);
    }
  }
  int query(int x, int bit) {
    if (bit == -1) {
      return n;
    }
    if ((x >> bit) & 1) {
      if (m.count(false)) {
        return m[false].query(x, bit - 1);
      } else {
        return m[true].query(x, bit - 1);
      }
    } else {
      if (m.count(true)) {
        return m[true].query(x, bit - 1);
      } else {
        return m[false].query(x, bit - 1);
      }
    }
  }
};
int main() {
  int q, x, i;
  char op;
  Trie t;
  scanf("%d%*c", &q);
  t.insert(0, 30);
  for (i = 0; i < q; i++) {
    scanf("%c%*c%d%*c", &op, &x);
    switch (op) {
      case '+':
        t.insert(x, 30);
        break;
      case '-':
        t.remove(x, 30);
        break;
      case '?':
        printf("%d\n", x ^ t.query(x, 30));
        break;
      default:
        break;
    }
  }
  return 0;
}

```