#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fll;
const int MAX = 3e6;
struct trie {
  int p;
  vector<vector<int>> t;
  vector<int> fim, pref;
  trie() : t(MAX, vector<int>(2)), fim(MAX), pref(MAX) { p = 1; }
  void insert(string second) {
    int x = 0;
    for (char c : second) {
      int &y = t[x][c - 'a'];
      if (!y) y = p++;
      x = y;
      pref[x]++;
    }
    fim[x]++;
  }
  void erase(string second) {
    int x = 0;
    vector<int> pos;
    pos.push_back(x);
    for (char c : second) {
      int &y = t[x][c - 'a'];
      x = y, pref[x]--;
      if (!pref[x]) y = 0;
    }
    fim[x]--;
  }
  bool find(string second) {
    int x = 0;
    for (char c : second) {
      x = t[x][c - 'a'];
      if (!x) return false;
    }
    return fim[x];
  }
  int query(string second) {
    int x = 0, ret = 0;
    for (int i = 0; i < second.size(); i++) {
      int c = second[i] - 'a';
      if (t[x][!c])
        ret += 1 << (second.size() - 1 - i), x = t[x][!c];
      else
        x = t[x][c];
    }
    return ret;
  }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int q;
  cin >> q;
  trie t;
  string aux;
  for (int i = 0; i <= 30; i++) aux += 'a';
  t.insert(aux);
  while (q--) {
    char c;
    int n;
    cin >> c >> n;
    string second;
    for (int i = 0; i <= 30; i++) second += 'a';
    int at = 30;
    while (n) {
      second[at] += n % 2;
      n /= 2;
      at--;
    }
    if (c == '+')
      t.insert(second);
    else if (c == '-')
      t.erase(second);
    else {
      int ans = t.query(second);
      cout << ans << '\n';
    }
  }
  exit(0);
}

### Prompt

Please create a solution in Cpp 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 INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fll;
const int MAX = 3e6;
struct trie {
  int p;
  vector<vector<int>> t;
  vector<int> fim, pref;
  trie() : t(MAX, vector<int>(2)), fim(MAX), pref(MAX) { p = 1; }
  void insert(string second) {
    int x = 0;
    for (char c : second) {
      int &y = t[x][c - 'a'];
      if (!y) y = p++;
      x = y;
      pref[x]++;
    }
    fim[x]++;
  }
  void erase(string second) {
    int x = 0;
    vector<int> pos;
    pos.push_back(x);
    for (char c : second) {
      int &y = t[x][c - 'a'];
      x = y, pref[x]--;
      if (!pref[x]) y = 0;
    }
    fim[x]--;
  }
  bool find(string second) {
    int x = 0;
    for (char c : second) {
      x = t[x][c - 'a'];
      if (!x) return false;
    }
    return fim[x];
  }
  int query(string second) {
    int x = 0, ret = 0;
    for (int i = 0; i < second.size(); i++) {
      int c = second[i] - 'a';
      if (t[x][!c])
        ret += 1 << (second.size() - 1 - i), x = t[x][!c];
      else
        x = t[x][c];
    }
    return ret;
  }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int q;
  cin >> q;
  trie t;
  string aux;
  for (int i = 0; i <= 30; i++) aux += 'a';
  t.insert(aux);
  while (q--) {
    char c;
    int n;
    cin >> c >> n;
    string second;
    for (int i = 0; i <= 30; i++) second += 'a';
    int at = 30;
    while (n) {
      second[at] += n % 2;
      n /= 2;
      at--;
    }
    if (c == '+')
      t.insert(second);
    else if (c == '-')
      t.erase(second);
    else {
      int ans = t.query(second);
      cout << ans << '\n';
    }
  }
  exit(0);
}

```