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>.
#include <bits/stdc++.h>
using namespace std;
const int CMXS = 30, MAXS = 200000 * (30 + 3) + 3;
int q, a[2][MAXS], w[MAXS], nxt = 0;
inline void add(bitset<CMXS> b) {
  for (int i = CMXS - 1, v = 0; (~i ? true : (w[v]++, false));
       (~a[b[i]][v] ? v = a[b[i]][v] : v = a[b[i]][v] = ++nxt), i--)
    w[v]++;
}
inline void sub(bitset<CMXS> b) {
  for (int i = CMXS - 1, v = 0; (~i ? true : (w[v]--, false));
       v = a[b[i]][v], i--)
    w[v]--;
}
inline void go(bitset<CMXS> b) {
  bitset<CMXS> f;
  for (int i = CMXS - 1, v = 0; ~i; i--)
    if (~a[1 - b[i]][v] && w[a[1 - b[i]][v]] > 0) {
      v = a[1 - b[i]][v];
      f[i] = 1;
    } else
      v = a[b[i]][v];
  cout << f.to_ulong() << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  cin >> q;
  memset(a, -1, sizeof a);
  add(0);
  while (q--) {
    char c;
    int x;
    cin >> c >> x;
    if (c == '+') add(x);
    if (c == '-') sub(x);
    if (c == '?') go(x);
  }
  return 0;
}
