#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long int llINF = 0x3f3f3f3f3f3f3f;
const int MOD = 1e9 + 7;
long long int val;
struct Trie {
  int wrd, cnt;
  map<int, Trie> m;
  Trie() {
    cnt = wrd = 0;
    m.clear();
  }
  void add(long long int x, int i) {
    cnt++;
    if (i == -1) {
      wrd++;
      return;
    }
    long long int bit = (1ll << i) & x;
    if (bit) bit = 1;
    if (!m.count(bit)) m[bit] = Trie();
    m[bit].add(x, i - 1);
  }
  int remove(long long int x, int i) {
    if (i == -1) {
      if (wrd) {
        wrd--;
        cnt--;
        return 1;
      } else
        return 0;
    }
    long long int bit = (1ll << i) & x;
    if (bit) bit = 1;
    if (!m.count(bit)) return 0;
    if (m[bit].remove(x, i - 1)) {
      cnt--;
      return 1;
    }
    return 0;
  }
  void dfs(long long int x, int i, long long int resp) {
    if (i == -1) {
      val = resp;
      return;
    }
    long long int bit = (1ll << i) & x;
    if (bit) bit = 1;
    if (bit == 1) {
      if (m.count(0) && m[0].cnt)
        m[0].dfs(x, i - 1, resp);
      else
        m[1].dfs(x, i - 1, resp | (1ll << i));
    } else {
      if (m.count(1) && m[1].cnt)
        m[1].dfs(x, i - 1, resp | (1ll << i));
      else
        m[0].dfs(x, i - 1, resp);
    }
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int q;
  cin >> q;
  Trie T = Trie();
  T.add(0, 33);
  for (int i = 0; i < q; i++) {
    char op;
    long long int x;
    cin >> op >> x;
    if (op == '+')
      T.add(x, 33);
    else if (op == '-')
      T.remove(x, 33);
    else {
      T.dfs(x, 33, 0);
      val ^= x;
      cout << val << endl;
    }
  }
}

### Prompt

Your task is to create 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;
const int INF = 0x3f3f3f3f;
const long long int llINF = 0x3f3f3f3f3f3f3f;
const int MOD = 1e9 + 7;
long long int val;
struct Trie {
  int wrd, cnt;
  map<int, Trie> m;
  Trie() {
    cnt = wrd = 0;
    m.clear();
  }
  void add(long long int x, int i) {
    cnt++;
    if (i == -1) {
      wrd++;
      return;
    }
    long long int bit = (1ll << i) & x;
    if (bit) bit = 1;
    if (!m.count(bit)) m[bit] = Trie();
    m[bit].add(x, i - 1);
  }
  int remove(long long int x, int i) {
    if (i == -1) {
      if (wrd) {
        wrd--;
        cnt--;
        return 1;
      } else
        return 0;
    }
    long long int bit = (1ll << i) & x;
    if (bit) bit = 1;
    if (!m.count(bit)) return 0;
    if (m[bit].remove(x, i - 1)) {
      cnt--;
      return 1;
    }
    return 0;
  }
  void dfs(long long int x, int i, long long int resp) {
    if (i == -1) {
      val = resp;
      return;
    }
    long long int bit = (1ll << i) & x;
    if (bit) bit = 1;
    if (bit == 1) {
      if (m.count(0) && m[0].cnt)
        m[0].dfs(x, i - 1, resp);
      else
        m[1].dfs(x, i - 1, resp | (1ll << i));
    } else {
      if (m.count(1) && m[1].cnt)
        m[1].dfs(x, i - 1, resp | (1ll << i));
      else
        m[0].dfs(x, i - 1, resp);
    }
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int q;
  cin >> q;
  Trie T = Trie();
  T.add(0, 33);
  for (int i = 0; i < q; i++) {
    char op;
    long long int x;
    cin >> op >> x;
    if (op == '+')
      T.add(x, 33);
    else if (op == '-')
      T.remove(x, 33);
    else {
      T.dfs(x, 33, 0);
      val ^= x;
      cout << val << endl;
    }
  }
}

```