#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
const long long inf = 1e18;
const int mod = 1e9 + 7;
int n;
int a[maxn];
struct node {
  node *ch[2];
  node *par;
  bool leaf;
  bool is;
  node() {
    is = 0;
    leaf = 0;
    par = ch[0] = ch[1] = NULL;
  }
};
struct trie {
  node *root;
  trie() { root = new node; }
  void add(int x) {
    node *v = root;
    for (int i = 30; i >= 0; i--) {
      bool bit = (x & (1 << i));
      if (v->ch[bit] == NULL) {
        v->ch[bit] = new node;
        (v->ch[bit])->par = v;
      }
      if (v->ch[1] != NULL && v->ch[0] != NULL) v->is = true;
      v = v->ch[bit];
    }
    v->leaf = true;
  }
  void del(int x) {
    node *v = root;
    for (int i = 30; i >= 0; i--) {
      bool bit = (x & (1 << i));
      v = v->ch[bit];
    }
    for (int i = 0; i <= 30; i++) {
      if (v == root) break;
      bool bit = (x & (1 << i));
      (v->par)->ch[bit] = NULL;
      if ((v->par)->is == true) break;
      v = v->par;
    }
    if (v != root) (v->par)->is = false;
  }
  int qu(int x) {
    node *v = root;
    int ans = 0;
    for (int i = 30; i >= 0; i--) {
      bool bit = (x & (1 << i));
      bit = !bit;
      if (v->ch[bit] != NULL) {
        ans += (1 << i);
        v = v->ch[bit];
      } else {
        if (v->ch[!bit] == NULL) break;
        v = v->ch[!bit];
      }
    }
    return ans;
  }
};
map<int, int> mp;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  trie t;
  while (n--) {
    char tp;
    int x;
    cin >> tp >> x;
    if (tp == '+') {
      mp[x]++;
      if (mp[x] == 1) t.add(x);
    }
    if (tp == '-') {
      mp[x]--;
      if (mp[x] == 0) t.del(x);
    }
    if (tp == '?') cout << max(x, t.qu(x)) << "\n";
  }
}

### 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;
const int maxn = 2e5 + 10;
const long long inf = 1e18;
const int mod = 1e9 + 7;
int n;
int a[maxn];
struct node {
  node *ch[2];
  node *par;
  bool leaf;
  bool is;
  node() {
    is = 0;
    leaf = 0;
    par = ch[0] = ch[1] = NULL;
  }
};
struct trie {
  node *root;
  trie() { root = new node; }
  void add(int x) {
    node *v = root;
    for (int i = 30; i >= 0; i--) {
      bool bit = (x & (1 << i));
      if (v->ch[bit] == NULL) {
        v->ch[bit] = new node;
        (v->ch[bit])->par = v;
      }
      if (v->ch[1] != NULL && v->ch[0] != NULL) v->is = true;
      v = v->ch[bit];
    }
    v->leaf = true;
  }
  void del(int x) {
    node *v = root;
    for (int i = 30; i >= 0; i--) {
      bool bit = (x & (1 << i));
      v = v->ch[bit];
    }
    for (int i = 0; i <= 30; i++) {
      if (v == root) break;
      bool bit = (x & (1 << i));
      (v->par)->ch[bit] = NULL;
      if ((v->par)->is == true) break;
      v = v->par;
    }
    if (v != root) (v->par)->is = false;
  }
  int qu(int x) {
    node *v = root;
    int ans = 0;
    for (int i = 30; i >= 0; i--) {
      bool bit = (x & (1 << i));
      bit = !bit;
      if (v->ch[bit] != NULL) {
        ans += (1 << i);
        v = v->ch[bit];
      } else {
        if (v->ch[!bit] == NULL) break;
        v = v->ch[!bit];
      }
    }
    return ans;
  }
};
map<int, int> mp;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  trie t;
  while (n--) {
    char tp;
    int x;
    cin >> tp >> x;
    if (tp == '+') {
      mp[x]++;
      if (mp[x] == 1) t.add(x);
    }
    if (tp == '-') {
      mp[x]--;
      if (mp[x] == 0) t.del(x);
    }
    if (tp == '?') cout << max(x, t.qu(x)) << "\n";
  }
}

```