#include <bits/stdc++.h>
int dx[] = {0, -1, 0, 1, -1, 1, -1, 1};
int dy[] = {-1, 0, 1, 0, 1, -1, -1, 1};
const int MOD = 1e9 + 7, N = 1e5 + 5, oo = 1e9;
const double pi = acos(-1);
using namespace std;
struct trie {
  int in = 0;
  trie *ptr[2];
  trie() {
    memset(ptr, 0, sizeof ptr);
    in = 0;
  }
} * root;
void add(string s) {
  trie *node = root;
  for (int i = 0; i < (int)(s).size(); i++) {
    if (node->ptr[s[i] - '0'] == 0) node->ptr[s[i] - '0'] = new trie();
    node = node->ptr[s[i] - '0'];
    node->in++;
  }
}
void remove(string s) {
  trie *node = root;
  for (int i = 0; i < (int)(s).size(); i++) {
    node = node->ptr[s[i] - '0'];
    node->in--;
  }
}
int get(string s) {
  trie *node = root;
  int ret = 0, shift = 31;
  for (int i = 0; i < (int)(s).size(); i++, shift--) {
    if (s[i] == '0') {
      if (node->ptr[1] and node->ptr[1]->in) {
        ret += (1 << shift);
        node = node->ptr[1];
      } else if (node->ptr[0])
        node = node->ptr[0];
    } else {
      if (node->ptr[0] and node->ptr[0]->in) {
        ret += (1 << shift);
        node = node->ptr[0];
      } else if (node->ptr[1])
        node = node->ptr[1];
    }
  }
  return ret;
}
string binary(int n) {
  string s = "";
  while (n) {
    s += ((n % 2) + '0');
    n /= 2;
  }
  while ((int)(s).size() < 32) s += '0';
  reverse(s.begin(), s.end());
  return s;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  root = new trie();
  add(binary(0));
  int q;
  cin >> q;
  while (q--) {
    char c;
    int n;
    cin >> c >> n;
    if (c == '+')
      add(binary(n));
    else if (c == '-')
      remove(binary(n));
    else
      cout << get(binary(n)) << '\n';
  }
  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>
int dx[] = {0, -1, 0, 1, -1, 1, -1, 1};
int dy[] = {-1, 0, 1, 0, 1, -1, -1, 1};
const int MOD = 1e9 + 7, N = 1e5 + 5, oo = 1e9;
const double pi = acos(-1);
using namespace std;
struct trie {
  int in = 0;
  trie *ptr[2];
  trie() {
    memset(ptr, 0, sizeof ptr);
    in = 0;
  }
} * root;
void add(string s) {
  trie *node = root;
  for (int i = 0; i < (int)(s).size(); i++) {
    if (node->ptr[s[i] - '0'] == 0) node->ptr[s[i] - '0'] = new trie();
    node = node->ptr[s[i] - '0'];
    node->in++;
  }
}
void remove(string s) {
  trie *node = root;
  for (int i = 0; i < (int)(s).size(); i++) {
    node = node->ptr[s[i] - '0'];
    node->in--;
  }
}
int get(string s) {
  trie *node = root;
  int ret = 0, shift = 31;
  for (int i = 0; i < (int)(s).size(); i++, shift--) {
    if (s[i] == '0') {
      if (node->ptr[1] and node->ptr[1]->in) {
        ret += (1 << shift);
        node = node->ptr[1];
      } else if (node->ptr[0])
        node = node->ptr[0];
    } else {
      if (node->ptr[0] and node->ptr[0]->in) {
        ret += (1 << shift);
        node = node->ptr[0];
      } else if (node->ptr[1])
        node = node->ptr[1];
    }
  }
  return ret;
}
string binary(int n) {
  string s = "";
  while (n) {
    s += ((n % 2) + '0');
    n /= 2;
  }
  while ((int)(s).size() < 32) s += '0';
  reverse(s.begin(), s.end());
  return s;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  root = new trie();
  add(binary(0));
  int q;
  cin >> q;
  while (q--) {
    char c;
    int n;
    cin >> c >> n;
    if (c == '+')
      add(binary(n));
    else if (c == '-')
      remove(binary(n));
    else
      cout << get(binary(n)) << '\n';
  }
  return 0;
}

```