#include <bits/stdc++.h>
using namespace std;
struct trie {
  vector<trie*> b;
  vector<int> num;
  trie() : num({0, 0}), b({NULL, NULL}) {}
};
void add(int n, trie* pt) {
  for (int i = 30; i >= 0; i--) {
    int x = (n & (1 << i));
    if (x > 0) x = 1;
    pt->num[x]++;
    if (pt->b[x]) {
      pt = pt->b[x];
    } else if (i != 0) {
      struct trie* t = new trie();
      pt->b[x] = t;
      pt = pt->b[x];
    }
  }
}
void rem(int n, trie* pt) {
  for (int i = 30; i >= 0; i--) {
    int x = (n & (1 << i));
    if (x > 0) x = 1;
    pt->num[x]--;
    if (pt->num[x] == 0) {
      pt->b[x] = NULL;
      return;
    }
    pt = pt->b[x];
  }
}
int query(int n, trie* pt) {
  int ans = 0;
  for (int i = 30; i >= 0; i--) {
    if (n & (1 << i)) {
      if (pt->num[0] > 0) {
        ans = ans + (1 << i);
        pt = pt->b[0];
      } else
        pt = pt->b[1];
    } else {
      if (pt->num[1] > 0) {
        ans = ans + (1 << i);
        pt = pt->b[1];
      } else
        pt = pt->b[0];
    }
  }
  return ans;
}
void prnt(trie* t) {
  for (int i = 0; i < 31; i++) {
    cout << t->num[0] << " " << t->num[1] << endl;
    if (t->b[1])
      t = t->b[1];
    else
      t = t->b[0];
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int q;
  cin >> q;
  char c;
  int n;
  trie obj = trie();
  obj.num[0] = 1;
  trie* pt = &obj;
  for (int i = 0; i < 30; i++) {
    struct trie* t = new trie();
    pt->b[0] = t;
    pt = pt->b[0];
    pt->num[0] = 1;
  }
  while (q--) {
    cin >> c >> n;
    pt = &obj;
    if (c == '+')
      add(n, &obj);
    else if (c == '-')
      rem(n, pt);
    else {
      cout << query(n, pt) << 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;
struct trie {
  vector<trie*> b;
  vector<int> num;
  trie() : num({0, 0}), b({NULL, NULL}) {}
};
void add(int n, trie* pt) {
  for (int i = 30; i >= 0; i--) {
    int x = (n & (1 << i));
    if (x > 0) x = 1;
    pt->num[x]++;
    if (pt->b[x]) {
      pt = pt->b[x];
    } else if (i != 0) {
      struct trie* t = new trie();
      pt->b[x] = t;
      pt = pt->b[x];
    }
  }
}
void rem(int n, trie* pt) {
  for (int i = 30; i >= 0; i--) {
    int x = (n & (1 << i));
    if (x > 0) x = 1;
    pt->num[x]--;
    if (pt->num[x] == 0) {
      pt->b[x] = NULL;
      return;
    }
    pt = pt->b[x];
  }
}
int query(int n, trie* pt) {
  int ans = 0;
  for (int i = 30; i >= 0; i--) {
    if (n & (1 << i)) {
      if (pt->num[0] > 0) {
        ans = ans + (1 << i);
        pt = pt->b[0];
      } else
        pt = pt->b[1];
    } else {
      if (pt->num[1] > 0) {
        ans = ans + (1 << i);
        pt = pt->b[1];
      } else
        pt = pt->b[0];
    }
  }
  return ans;
}
void prnt(trie* t) {
  for (int i = 0; i < 31; i++) {
    cout << t->num[0] << " " << t->num[1] << endl;
    if (t->b[1])
      t = t->b[1];
    else
      t = t->b[0];
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int q;
  cin >> q;
  char c;
  int n;
  trie obj = trie();
  obj.num[0] = 1;
  trie* pt = &obj;
  for (int i = 0; i < 30; i++) {
    struct trie* t = new trie();
    pt->b[0] = t;
    pt = pt->b[0];
    pt->num[0] = 1;
  }
  while (q--) {
    cin >> c >> n;
    pt = &obj;
    if (c == '+')
      add(n, &obj);
    else if (c == '-')
      rem(n, pt);
    else {
      cout << query(n, pt) << endl;
    }
  }
  return 0;
}

```