#include <bits/stdc++.h>
using namespace std;
const int N = 1e4 + 4;
long long int a[N][20], pre[N][20], ans;
struct node {
  node *left, *right;
  long long int cnt;
  node() {
    left = NULL;
    right = NULL;
    cnt = 0;
  }
};
struct trie {
  node *x;
};
void insert(node *t, long long int x) {
  for (long long int i = 31; i >= 0; i--) {
    if (x & (1 << i)) {
      if (t->right == NULL) t->right = new node();
      t = t->right;
      t->cnt++;
    } else {
      if (t->left == NULL) t->left = new node();
      t = t->left;
      t->cnt++;
    }
  }
}
void del(node *t, long long int x) {
  for (long long int i = 31; i >= 0; i--) {
    if (x & (1 << i)) {
      t = t->right;
      t->cnt--;
    } else {
      t = t->left;
      t->cnt--;
    }
  }
}
long long int query(node *t, long long int x) {
  long long int ret = 0;
  for (long long int i = 31; i >= 0; i--) {
    if (x & (1 << i)) {
      if (t->left && t->left->cnt != 0) {
        ret |= (1 << i);
        t = t->left;
      } else
        t = t->right;
    } else {
      if (t->right && t->right->cnt != 0) {
        ret |= (1 << i);
        t = t->right;
      } else
        t = t->left;
    }
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  trie t;
  t.x = new node();
  long long int n;
  cin >> n;
  insert(t.x, 0);
  for (long long int i = 1; i <= n; i++) {
    char ch;
    cin >> ch;
    long long int x;
    cin >> x;
    if (ch == '+') insert(t.x, x);
    if (ch == '-') del(t.x, x);
    if (ch == '?') cout << query(t.x, x) << endl;
  }
  return 0;
}

### Prompt

Develop a solution in CPP 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;
const int N = 1e4 + 4;
long long int a[N][20], pre[N][20], ans;
struct node {
  node *left, *right;
  long long int cnt;
  node() {
    left = NULL;
    right = NULL;
    cnt = 0;
  }
};
struct trie {
  node *x;
};
void insert(node *t, long long int x) {
  for (long long int i = 31; i >= 0; i--) {
    if (x & (1 << i)) {
      if (t->right == NULL) t->right = new node();
      t = t->right;
      t->cnt++;
    } else {
      if (t->left == NULL) t->left = new node();
      t = t->left;
      t->cnt++;
    }
  }
}
void del(node *t, long long int x) {
  for (long long int i = 31; i >= 0; i--) {
    if (x & (1 << i)) {
      t = t->right;
      t->cnt--;
    } else {
      t = t->left;
      t->cnt--;
    }
  }
}
long long int query(node *t, long long int x) {
  long long int ret = 0;
  for (long long int i = 31; i >= 0; i--) {
    if (x & (1 << i)) {
      if (t->left && t->left->cnt != 0) {
        ret |= (1 << i);
        t = t->left;
      } else
        t = t->right;
    } else {
      if (t->right && t->right->cnt != 0) {
        ret |= (1 << i);
        t = t->right;
      } else
        t = t->left;
    }
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  trie t;
  t.x = new node();
  long long int n;
  cin >> n;
  insert(t.x, 0);
  for (long long int i = 1; i <= n; i++) {
    char ch;
    cin >> ch;
    long long int x;
    cin >> x;
    if (ch == '+') insert(t.x, x);
    if (ch == '-') del(t.x, x);
    if (ch == '?') cout << query(t.x, x) << endl;
  }
  return 0;
}

```