#include <bits/stdc++.h>
using namespace std;
template <class T>
using vv = vector<vector<T>>;
template <class T>
ostream &operator<<(ostream &os, const vector<T> &t) {
  os << "{";
  for (int(i) = 0; (i) < (t.size()); ++(i)) {
    os << t[i] << ",";
  }
  os << "}" << endl;
  return os;
}
template <class S, class T>
ostream &operator<<(ostream &os, const pair<S, T> &t) {
  return os << "(" << t.first << "," << t.second << ")";
}
template <class T>
inline bool MX(T &l, const T &r) {
  return l < r ? l = r, 1 : 0;
}
template <class T>
inline bool MN(T &l, const T &r) {
  return l > r ? l = r, 1 : 0;
}
const long long MOD = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(0);
  int n;
  cin >> n;
  multiset<int> st = {0};
  string s;
  int x;
  for (int(i) = 0; (i) < (n); ++(i)) {
    cin >> s >> x;
    if (s[0] == '+') {
      st.insert(x);
    } else if (s[0] == '-') {
      st.erase(st.find(x));
    } else {
      int l = 0;
      for (int i = 29, d = 1 << i; d; d /= 2, i--) {
        auto itl = st.lower_bound(l);
        auto itm = st.lower_bound(l + d);
        auto itr = st.lower_bound(l + d + d);
        if (x >> i & 1) {
          if (itl == itm) l += d;
        } else {
          if (itm != itr) l += d;
        }
      }
      cout << (l ^ x) << endl;
    }
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
template <class T>
using vv = vector<vector<T>>;
template <class T>
ostream &operator<<(ostream &os, const vector<T> &t) {
  os << "{";
  for (int(i) = 0; (i) < (t.size()); ++(i)) {
    os << t[i] << ",";
  }
  os << "}" << endl;
  return os;
}
template <class S, class T>
ostream &operator<<(ostream &os, const pair<S, T> &t) {
  return os << "(" << t.first << "," << t.second << ")";
}
template <class T>
inline bool MX(T &l, const T &r) {
  return l < r ? l = r, 1 : 0;
}
template <class T>
inline bool MN(T &l, const T &r) {
  return l > r ? l = r, 1 : 0;
}
const long long MOD = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(0);
  int n;
  cin >> n;
  multiset<int> st = {0};
  string s;
  int x;
  for (int(i) = 0; (i) < (n); ++(i)) {
    cin >> s >> x;
    if (s[0] == '+') {
      st.insert(x);
    } else if (s[0] == '-') {
      st.erase(st.find(x));
    } else {
      int l = 0;
      for (int i = 29, d = 1 << i; d; d /= 2, i--) {
        auto itl = st.lower_bound(l);
        auto itm = st.lower_bound(l + d);
        auto itr = st.lower_bound(l + d + d);
        if (x >> i & 1) {
          if (itl == itm) l += d;
        } else {
          if (itm != itr) l += d;
        }
      }
      cout << (l ^ x) << endl;
    }
  }
  return 0;
}

```