#include <bits/stdc++.h>
using namespace std;
int q;
bitset<50> foo;
long long pot[50], ans, lef[10000000], righ[10000000], k, x, pos, ile[10000000];
char znak;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> q;
  pot[0] = 1;
  for (int a = 1; a <= 42; a++) pot[a] = 2 * pot[a - 1];
  pos = 0;
  for (int a = 40; a >= 0; a--) {
    righ[pos] = ++k;
    pos = k;
    ile[pos]++;
  }
  for (int a = 1; a <= q; a++) {
    cin >> znak >> x;
    if (znak == '+') {
      foo = x;
      pos = 0;
      for (int a = 40; a >= 0; a--) {
        if (foo[a] == 1) {
          if (!lef[pos]) lef[pos] = ++k;
          pos = lef[pos];
        } else {
          if (!righ[pos]) righ[pos] = ++k;
          pos = righ[pos];
        }
        ile[pos]++;
      }
    }
    if (znak == '-') {
      foo = x;
      pos = 0;
      for (int a = 40; a >= 0; a--) {
        if (foo[a] == 1) {
          if (!lef[pos]) lef[pos] = ++k;
          pos = lef[pos];
        } else {
          if (!righ[pos]) righ[pos] = ++k;
          pos = righ[pos];
        }
        ile[pos]--;
      }
    }
    if (znak == '?') {
      foo = x;
      ans = 0;
      pos = 0;
      for (int a = 40; a >= 0; a--) {
        if (foo[a]) {
          if (righ[pos] && ile[righ[pos]])
            pos = righ[pos], ans += pot[a];
          else if (lef[pos] && ile[lef[pos]])
            pos = lef[pos];
          else
            break;
        } else {
          if (lef[pos] && ile[lef[pos]])
            pos = lef[pos], ans += pot[a];
          else if (righ[pos] && ile[righ[pos]])
            pos = righ[pos];
          else
            break;
        }
      }
      cout << ans << "\n";
    }
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
int q;
bitset<50> foo;
long long pot[50], ans, lef[10000000], righ[10000000], k, x, pos, ile[10000000];
char znak;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> q;
  pot[0] = 1;
  for (int a = 1; a <= 42; a++) pot[a] = 2 * pot[a - 1];
  pos = 0;
  for (int a = 40; a >= 0; a--) {
    righ[pos] = ++k;
    pos = k;
    ile[pos]++;
  }
  for (int a = 1; a <= q; a++) {
    cin >> znak >> x;
    if (znak == '+') {
      foo = x;
      pos = 0;
      for (int a = 40; a >= 0; a--) {
        if (foo[a] == 1) {
          if (!lef[pos]) lef[pos] = ++k;
          pos = lef[pos];
        } else {
          if (!righ[pos]) righ[pos] = ++k;
          pos = righ[pos];
        }
        ile[pos]++;
      }
    }
    if (znak == '-') {
      foo = x;
      pos = 0;
      for (int a = 40; a >= 0; a--) {
        if (foo[a] == 1) {
          if (!lef[pos]) lef[pos] = ++k;
          pos = lef[pos];
        } else {
          if (!righ[pos]) righ[pos] = ++k;
          pos = righ[pos];
        }
        ile[pos]--;
      }
    }
    if (znak == '?') {
      foo = x;
      ans = 0;
      pos = 0;
      for (int a = 40; a >= 0; a--) {
        if (foo[a]) {
          if (righ[pos] && ile[righ[pos]])
            pos = righ[pos], ans += pot[a];
          else if (lef[pos] && ile[lef[pos]])
            pos = lef[pos];
          else
            break;
        } else {
          if (lef[pos] && ile[lef[pos]])
            pos = lef[pos], ans += pot[a];
          else if (righ[pos] && ile[righ[pos]])
            pos = righ[pos];
          else
            break;
        }
      }
      cout << ans << "\n";
    }
  }
  return 0;
}

```