#include <bits/stdc++.h>
using namespace std;
class node {
 public:
  int v;
  node *l, *r;
  node() : v(0), l(NULL), r(NULL) {}
};
int p[32];
int n;
node *root;
void init() {
  cin >> n;
  p[0] = 1;
  for (int i = 1; i <= 31; i++) p[i] = p[i - 1] << 1;
  root = new node;
}
int getsum(node *x) {
  if (x == NULL) return 0;
  return x->v;
}
void add(node *x, int v, int depth) {
  if (depth < 0) {
    x->v++;
    return;
  }
  if ((v & p[depth]) == 0) {
    if (x->l == NULL) x->l = new node;
    add(x->l, v, depth - 1);
  } else {
    if (x->r == NULL) x->r = new node;
    add(x->r, v, depth - 1);
  }
  x->v = getsum(x->l) + getsum(x->r);
}
void inc(node *x, int v, int depth) {
  if (x == NULL)
    while (1)
      ;
  if (depth < 0) {
    x->v--;
    return;
  }
  if ((v & p[depth]) == 0)
    inc(x->l, v, depth - 1);
  else
    inc(x->r, v, depth - 1);
  x->v = getsum(x->l) + getsum(x->r);
}
int query(node *x, int v, int depth) {
  if (depth < 0) {
    return 0;
  }
  if (v & p[depth]) {
    if (getsum(x->l) == 0)
      return p[depth] + query(x->r, v, depth - 1);
    else
      return query(x->l, v, depth - 1);
  } else {
    if (getsum(x->r) == 0)
      return query(x->l, v, depth - 1);
    else
      return p[depth] + query(x->r, v, depth - 1);
  }
}
void work() {
  char ch;
  int x;
  add(root, 0, 30);
  for (int i = 1; i <= n; i++) {
    cin >> ch >> x;
    switch (ch) {
      case '+':
        add(root, x, 30);
        break;
      case '-':
        inc(root, x, 30);
        break;
      case '?':
        cout << (x ^ query(root, x, 30)) << endl;
        break;
    }
  }
}
void print() {}
int main() {
  ios::sync_with_stdio(false);
  init();
  work();
  print();
  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;
class node {
 public:
  int v;
  node *l, *r;
  node() : v(0), l(NULL), r(NULL) {}
};
int p[32];
int n;
node *root;
void init() {
  cin >> n;
  p[0] = 1;
  for (int i = 1; i <= 31; i++) p[i] = p[i - 1] << 1;
  root = new node;
}
int getsum(node *x) {
  if (x == NULL) return 0;
  return x->v;
}
void add(node *x, int v, int depth) {
  if (depth < 0) {
    x->v++;
    return;
  }
  if ((v & p[depth]) == 0) {
    if (x->l == NULL) x->l = new node;
    add(x->l, v, depth - 1);
  } else {
    if (x->r == NULL) x->r = new node;
    add(x->r, v, depth - 1);
  }
  x->v = getsum(x->l) + getsum(x->r);
}
void inc(node *x, int v, int depth) {
  if (x == NULL)
    while (1)
      ;
  if (depth < 0) {
    x->v--;
    return;
  }
  if ((v & p[depth]) == 0)
    inc(x->l, v, depth - 1);
  else
    inc(x->r, v, depth - 1);
  x->v = getsum(x->l) + getsum(x->r);
}
int query(node *x, int v, int depth) {
  if (depth < 0) {
    return 0;
  }
  if (v & p[depth]) {
    if (getsum(x->l) == 0)
      return p[depth] + query(x->r, v, depth - 1);
    else
      return query(x->l, v, depth - 1);
  } else {
    if (getsum(x->r) == 0)
      return query(x->l, v, depth - 1);
    else
      return p[depth] + query(x->r, v, depth - 1);
  }
}
void work() {
  char ch;
  int x;
  add(root, 0, 30);
  for (int i = 1; i <= n; i++) {
    cin >> ch >> x;
    switch (ch) {
      case '+':
        add(root, x, 30);
        break;
      case '-':
        inc(root, x, 30);
        break;
      case '?':
        cout << (x ^ query(root, x, 30)) << endl;
        break;
    }
  }
}
void print() {}
int main() {
  ios::sync_with_stdio(false);
  init();
  work();
  print();
  return 0;
}

```