#include <bits/stdc++.h>
using namespace std;
const int maxN = 200005;
const int len = 31;
struct Trie {
  int next[2];
  int count;
} tree[maxN * len];
int cnt;
void initTree() {
  cnt = 0;
  memset(tree, -1, sizeof(tree));
}
void add(int x) {
  int now = 0;
  bool k;
  for (int i = len; i >= 0; i--) {
    k = x & (1 << i);
    if (tree[now].next[k] == -1) tree[now].next[k] = ++cnt;
    now = tree[now].next[k];
    tree[now].count = tree[now].count == -1 ? 1 : tree[now].count + 1;
  }
}
void _delete(int x) {
  int now = 0;
  bool k;
  for (int i = len; i >= 0; i--) {
    k = x & (1 << i);
    now = tree[now].next[k];
    tree[now].count--;
  }
}
int query(int x) {
  int v = 0, now = 0;
  bool k;
  for (int i = len; i >= 0; i--) {
    k = x & (1 << i);
    if (tree[now].next[!k] != -1 && tree[tree[now].next[!k]].count > 0) k = !k;
    v = v | (k << i);
    now = tree[now].next[k];
  }
  return v;
}
int q;
int main() {
  cin >> q;
  string op;
  int x;
  initTree();
  add(0);
  while (q--) {
    cin >> op >> x;
    if (op == "+")
      add(x);
    else if (op == "-")
      _delete(x);
    else {
      int t = query(x);
      cout << (t ^ x) << endl;
    }
  }
  return 0;
}

### Prompt

Please formulate 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;
const int maxN = 200005;
const int len = 31;
struct Trie {
  int next[2];
  int count;
} tree[maxN * len];
int cnt;
void initTree() {
  cnt = 0;
  memset(tree, -1, sizeof(tree));
}
void add(int x) {
  int now = 0;
  bool k;
  for (int i = len; i >= 0; i--) {
    k = x & (1 << i);
    if (tree[now].next[k] == -1) tree[now].next[k] = ++cnt;
    now = tree[now].next[k];
    tree[now].count = tree[now].count == -1 ? 1 : tree[now].count + 1;
  }
}
void _delete(int x) {
  int now = 0;
  bool k;
  for (int i = len; i >= 0; i--) {
    k = x & (1 << i);
    now = tree[now].next[k];
    tree[now].count--;
  }
}
int query(int x) {
  int v = 0, now = 0;
  bool k;
  for (int i = len; i >= 0; i--) {
    k = x & (1 << i);
    if (tree[now].next[!k] != -1 && tree[tree[now].next[!k]].count > 0) k = !k;
    v = v | (k << i);
    now = tree[now].next[k];
  }
  return v;
}
int q;
int main() {
  cin >> q;
  string op;
  int x;
  initTree();
  add(0);
  while (q--) {
    cin >> op >> x;
    if (op == "+")
      add(x);
    else if (op == "-")
      _delete(x);
    else {
      int t = query(x);
      cout << (t ^ x) << endl;
    }
  }
  return 0;
}

```