#include <bits/stdc++.h>
using namespace std;
const long long INF = (long long)2e18 + 7;
const int MXN = (int)4e6 + 7;
const int MOD = (int)1e9 + 7;
inline void boost() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
pair<int, int> t[MXN][2];
int q, ind;
bool bit(int mask, int id) { return mask & (1 << id); }
inline void add(int n) {
  int v = 0;
  for (int i = 29; i >= 0; i--) {
    bool id = bit(n, i);
    t[v][id].first++;
    if (!t[v][id].second) {
      t[v][id].second = ++ind;
    }
    v = t[v][id].second;
  }
}
inline int get(int n) {
  long long res = n;
  int v = 0;
  for (int i = 29; i >= 0; i--) {
    if (bit(n, i)) {
      if (!t[v][0].first && t[v][1].first) {
        res ^= (1LL << i);
        v = t[v][1].second;
      } else {
        v = t[v][0].second;
      }
    } else {
      if (t[v][1].first) {
        v = t[v][1].second;
        res ^= (1 << i);
      } else {
        v = t[v][0].second;
      }
    }
  }
  return res;
}
inline void remove(int n) {
  int v = 0;
  for (int i = 29; i >= 0; i--) {
    bool id = bit(n, i);
    t[v][id].first--;
    v = t[v][id].second;
  }
}
int main() {
  boost();
  cin >> q;
  add(0);
  while (q--) {
    int n;
    char ch;
    cin >> ch >> n;
    if (ch == '+') {
      add(n);
    } else if (ch == '-') {
      remove(n);
      add(0);
    } else if (ch == '?') {
      cout << get(n) << endl;
    }
  }
  return 0;
}

### 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 long long INF = (long long)2e18 + 7;
const int MXN = (int)4e6 + 7;
const int MOD = (int)1e9 + 7;
inline void boost() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
pair<int, int> t[MXN][2];
int q, ind;
bool bit(int mask, int id) { return mask & (1 << id); }
inline void add(int n) {
  int v = 0;
  for (int i = 29; i >= 0; i--) {
    bool id = bit(n, i);
    t[v][id].first++;
    if (!t[v][id].second) {
      t[v][id].second = ++ind;
    }
    v = t[v][id].second;
  }
}
inline int get(int n) {
  long long res = n;
  int v = 0;
  for (int i = 29; i >= 0; i--) {
    if (bit(n, i)) {
      if (!t[v][0].first && t[v][1].first) {
        res ^= (1LL << i);
        v = t[v][1].second;
      } else {
        v = t[v][0].second;
      }
    } else {
      if (t[v][1].first) {
        v = t[v][1].second;
        res ^= (1 << i);
      } else {
        v = t[v][0].second;
      }
    }
  }
  return res;
}
inline void remove(int n) {
  int v = 0;
  for (int i = 29; i >= 0; i--) {
    bool id = bit(n, i);
    t[v][id].first--;
    v = t[v][id].second;
  }
}
int main() {
  boost();
  cin >> q;
  add(0);
  while (q--) {
    int n;
    char ch;
    cin >> ch >> n;
    if (ch == '+') {
      add(n);
    } else if (ch == '-') {
      remove(n);
      add(0);
    } else if (ch == '?') {
      cout << get(n) << endl;
    }
  }
  return 0;
}

```