#include <bits/stdc++.h>
using namespace std;
long long INV2 = 500000004;
long long INV6 = 166666668;
long long power(long long a, long long b, long long c) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b & 1) x = (x * y) % c;
    y = (y * y) % c;
    b /= 2;
  }
  return x % c;
}
int dx[] = {0, -1, 0, 1};
int dy[] = {-1, 0, 1, 0};
multiset<int>::iterator it;
int main() {
  int q;
  scanf("%d", &q);
  multiset<int> s;
  s.insert(0);
  char ch;
  int val;
  while (q--) {
    cin >> ch;
    scanf("%d", &val);
    if (ch == '+')
      s.insert(val);
    else if (ch == '-')
      s.erase(s.find(val));
    else {
      long long ans = 0;
      for (int i = 30; i >= 0; --i) {
        int idx = 1 << i;
        if (val & idx) {
          it = s.lower_bound(ans);
          if (it != s.end() && *it & idx) ans |= idx;
        } else {
          long long temp = ans | (1 << i);
          it = s.lower_bound(ans | (1 << i));
          if (it != s.end() && (*it) >> i == temp >> i) ans = temp;
        }
      }
      ans = ans ^ val;
      printf("%lld\n", ans);
    }
  }
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to 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;
long long INV2 = 500000004;
long long INV6 = 166666668;
long long power(long long a, long long b, long long c) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b & 1) x = (x * y) % c;
    y = (y * y) % c;
    b /= 2;
  }
  return x % c;
}
int dx[] = {0, -1, 0, 1};
int dy[] = {-1, 0, 1, 0};
multiset<int>::iterator it;
int main() {
  int q;
  scanf("%d", &q);
  multiset<int> s;
  s.insert(0);
  char ch;
  int val;
  while (q--) {
    cin >> ch;
    scanf("%d", &val);
    if (ch == '+')
      s.insert(val);
    else if (ch == '-')
      s.erase(s.find(val));
    else {
      long long ans = 0;
      for (int i = 30; i >= 0; --i) {
        int idx = 1 << i;
        if (val & idx) {
          it = s.lower_bound(ans);
          if (it != s.end() && *it & idx) ans |= idx;
        } else {
          long long temp = ans | (1 << i);
          it = s.lower_bound(ans | (1 << i));
          if (it != s.end() && (*it) >> i == temp >> i) ans = temp;
        }
      }
      ans = ans ^ val;
      printf("%lld\n", ans);
    }
  }
  return 0;
}

```