#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
typedef struct data {
  data* bit[2];
  int cnt = 0;
} trie;
trie* head;
void insert(int x) {
  trie* node = head;
  for (int i = 30; i >= 0; i--) {
    int curbit = (x >> i) & 1;
    if (node->bit[curbit] == NULL) {
      node->bit[curbit] = new trie();
    }
    node = node->bit[curbit];
    node->cnt++;
  }
}
void remove(int x) {
  trie* node = head;
  for (int i = 30; i >= 0; i--) {
    int curbit = (x >> i) & 1;
    node = node->bit[curbit];
    node->cnt--;
  }
}
int maxxor(int x) {
  trie* node = head;
  int ans = 0;
  for (int i = 30; i >= 0; i--) {
    int curbit = (x >> i) & 1;
    if (node->bit[curbit ^ 1] != NULL && node->bit[curbit ^ 1]->cnt > 0) {
      ans += (1 << i);
      node = node->bit[curbit ^ 1];
    } else {
      node = node->bit[curbit];
    }
  }
  return ans;
}
void solve() {
  head = new trie();
  insert(0);
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    char c;
    int x;
    cin >> c >> x;
    if (c == '+')
      insert(x);
    else if (c == '-')
      remove(x);
    else
      cout << maxxor(x) << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;
long long mod = 1e9 + 7;
typedef struct data {
  data* bit[2];
  int cnt = 0;
} trie;
trie* head;
void insert(int x) {
  trie* node = head;
  for (int i = 30; i >= 0; i--) {
    int curbit = (x >> i) & 1;
    if (node->bit[curbit] == NULL) {
      node->bit[curbit] = new trie();
    }
    node = node->bit[curbit];
    node->cnt++;
  }
}
void remove(int x) {
  trie* node = head;
  for (int i = 30; i >= 0; i--) {
    int curbit = (x >> i) & 1;
    node = node->bit[curbit];
    node->cnt--;
  }
}
int maxxor(int x) {
  trie* node = head;
  int ans = 0;
  for (int i = 30; i >= 0; i--) {
    int curbit = (x >> i) & 1;
    if (node->bit[curbit ^ 1] != NULL && node->bit[curbit ^ 1]->cnt > 0) {
      ans += (1 << i);
      node = node->bit[curbit ^ 1];
    } else {
      node = node->bit[curbit];
    }
  }
  return ans;
}
void solve() {
  head = new trie();
  insert(0);
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    char c;
    int x;
    cin >> c >> x;
    if (c == '+')
      insert(x);
    else if (c == '-')
      remove(x);
    else
      cout << maxxor(x) << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}

```