#include <bits/stdc++.h>
using namespace std;
set<int> x[1048576];
map<int, int> M;
int q;
void add(int p) {
  int ret = 1;
  for (int i = 0; i < 20; i++) {
    x[ret].insert(p);
    if ((p / (1 << (30 - i))) % 2 == 0) ret = ret * 2;
    if ((p / (1 << (30 - i))) % 2 == 1) ret = ret * 2 + 1;
  }
}
void sakujyo(int p) {
  int ret = 1;
  for (int i = 0; i < 20; i++) {
    x[ret].erase(p);
    if ((p / (1 << (30 - i))) % 2 == 0) ret = ret * 2;
    if ((p / (1 << (30 - i))) % 2 == 1) ret = ret * 2 + 1;
  }
}
int max_value(int p) {
  int ret = 1;
  for (int i = 0; i < 19; i++) {
    int bit = 0;
    if ((p / (1 << (30 - i))) % 2 == 0) bit = 0;
    if ((p / (1 << (30 - i))) % 2 == 1) bit = 1;
    if (bit == 0) {
      if (x[2 * ret + 1].size() >= 1)
        ret = ret * 2 + 1;
      else {
        ret = ret * 2;
      }
    }
    if (bit == 1) {
      if (x[2 * ret].size() >= 1)
        ret = ret * 2;
      else {
        ret = ret * 2 + 1;
      }
    }
  }
  int maxn = 0;
  for (auto itr = x[ret].begin(); itr != x[ret].end(); itr++) {
    int c = *itr;
    maxn = max(maxn, c ^ p);
  }
  return maxn;
}
int main() {
  cin >> q;
  add(0);
  M[0]++;
  for (int i = 0; i < q; i++) {
    char op;
    int p;
    cin >> op >> p;
    if (op == '+') {
      add(p);
      M[p]++;
    }
    if (op == '-') {
      if (M[p] == 1) sakujyo(p);
      M[p]--;
    }
    if (op == '?') {
      cout << max_value(p) << endl;
    }
  }
  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;
set<int> x[1048576];
map<int, int> M;
int q;
void add(int p) {
  int ret = 1;
  for (int i = 0; i < 20; i++) {
    x[ret].insert(p);
    if ((p / (1 << (30 - i))) % 2 == 0) ret = ret * 2;
    if ((p / (1 << (30 - i))) % 2 == 1) ret = ret * 2 + 1;
  }
}
void sakujyo(int p) {
  int ret = 1;
  for (int i = 0; i < 20; i++) {
    x[ret].erase(p);
    if ((p / (1 << (30 - i))) % 2 == 0) ret = ret * 2;
    if ((p / (1 << (30 - i))) % 2 == 1) ret = ret * 2 + 1;
  }
}
int max_value(int p) {
  int ret = 1;
  for (int i = 0; i < 19; i++) {
    int bit = 0;
    if ((p / (1 << (30 - i))) % 2 == 0) bit = 0;
    if ((p / (1 << (30 - i))) % 2 == 1) bit = 1;
    if (bit == 0) {
      if (x[2 * ret + 1].size() >= 1)
        ret = ret * 2 + 1;
      else {
        ret = ret * 2;
      }
    }
    if (bit == 1) {
      if (x[2 * ret].size() >= 1)
        ret = ret * 2;
      else {
        ret = ret * 2 + 1;
      }
    }
  }
  int maxn = 0;
  for (auto itr = x[ret].begin(); itr != x[ret].end(); itr++) {
    int c = *itr;
    maxn = max(maxn, c ^ p);
  }
  return maxn;
}
int main() {
  cin >> q;
  add(0);
  M[0]++;
  for (int i = 0; i < q; i++) {
    char op;
    int p;
    cin >> op >> p;
    if (op == '+') {
      add(p);
      M[p]++;
    }
    if (op == '-') {
      if (M[p] == 1) sakujyo(p);
      M[p]--;
    }
    if (op == '?') {
      cout << max_value(p) << endl;
    }
  }
  return 0;
}

```