#include <bits/stdc++.h>
using namespace std;
struct node {
  int cl, cr, h;
  node *l, *r;
  node() { l = NULL, r = NULL, cl = 0, cr = 0; }
};
void add(node *v, int num) {
  if ((num >> v->h) & 1) {
    ++v->cr;
    if (v->h == -1) return;
    if (v->r == NULL) v->r = new node();
    v->r->h = v->h - 1;
    add(v->r, num);
  } else {
    ++v->cl;
    if (v->h == -1) return;
    if (v->l == NULL) v->l = new node();
    v->l->h = v->h - 1;
    add(v->l, num);
  }
}
void del(node *v, int num) {
  if ((num >> v->h) & 1) {
    --v->cr;
    if (v->h == -1) return;
    del(v->r, num);
  } else {
    --v->cl;
    if (v->h == -1) return;
    del(v->l, num);
  }
}
int ans(node *v, int num) {
  if ((num >> v->h) & 1) {
    if (v->cl) {
      if (v->h >= 0)
        return ans(v->l, num) + (1 << v->h);
      else
        return (1 << v->h);
    } else {
      if (v->h >= 0)
        return ans(v->r, num);
      else
        return 0;
    }
  } else {
    if (v->cr) {
      if (v->h >= 0)
        return ans(v->r, num) + (1 << v->h);
      else
        return (1 << v->h);
    } else {
      if (v->h >= 0)
        return ans(v->l, num);
      else
        return 0;
    }
  }
}
int n;
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  node *x = new node();
  x->h = 29;
  add(x, 0);
  for (int i = 0; i < n; ++i) {
    char type;
    int num;
    cin >> type >> num;
    if (type == '+')
      add(x, num);
    else if (type == '-')
      del(x, num);
    else
      cout << ans(x, num) << endl;
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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 node {
  int cl, cr, h;
  node *l, *r;
  node() { l = NULL, r = NULL, cl = 0, cr = 0; }
};
void add(node *v, int num) {
  if ((num >> v->h) & 1) {
    ++v->cr;
    if (v->h == -1) return;
    if (v->r == NULL) v->r = new node();
    v->r->h = v->h - 1;
    add(v->r, num);
  } else {
    ++v->cl;
    if (v->h == -1) return;
    if (v->l == NULL) v->l = new node();
    v->l->h = v->h - 1;
    add(v->l, num);
  }
}
void del(node *v, int num) {
  if ((num >> v->h) & 1) {
    --v->cr;
    if (v->h == -1) return;
    del(v->r, num);
  } else {
    --v->cl;
    if (v->h == -1) return;
    del(v->l, num);
  }
}
int ans(node *v, int num) {
  if ((num >> v->h) & 1) {
    if (v->cl) {
      if (v->h >= 0)
        return ans(v->l, num) + (1 << v->h);
      else
        return (1 << v->h);
    } else {
      if (v->h >= 0)
        return ans(v->r, num);
      else
        return 0;
    }
  } else {
    if (v->cr) {
      if (v->h >= 0)
        return ans(v->r, num) + (1 << v->h);
      else
        return (1 << v->h);
    } else {
      if (v->h >= 0)
        return ans(v->l, num);
      else
        return 0;
    }
  }
}
int n;
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  node *x = new node();
  x->h = 29;
  add(x, 0);
  for (int i = 0; i < n; ++i) {
    char type;
    int num;
    cin >> type >> num;
    if (type == '+')
      add(x, num);
    else if (type == '-')
      del(x, num);
    else
      cout << ans(x, num) << endl;
  }
  return 0;
}

```