#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
int fen[N];
void add(int i, int val) {
  for (; i < N; i |= (i + 1)) fen[i] += val;
}
int sum(int i) {
  int s = 0;
  for (; i >= 0; i = (i & (i + 1)) - 1) s += fen[i];
  return s;
}
int main() {
  int Q;
  scanf("%d", &Q);
  vector<int> t(Q), a(Q), ts(Q);
  char buf[10];
  for (int i = 0; i < Q; ++i) {
    scanf("%s%d", buf, &a[i]);
    ts[i] = a[i];
    if (buf[0] == '-') t[i] = -1;
    if (buf[0] == '+') t[i] = 1;
  }
  add(0, 1);
  ts.push_back(0);
  sort(ts.begin(), ts.end());
  ts.resize(unique(ts.begin(), ts.end()) - ts.begin());
  for (int q = 0; q < Q; ++q) {
    int i = lower_bound(ts.begin(), ts.end(), a[q]) - ts.begin();
    if (t[q]) {
      add(i, t[q]);
    } else {
      int res = 0;
      for (int b = 30; b >= 0; --b) {
        int B = (1 << b);
        int l = res, r = res + B - 1;
        if (!(a[q] & B)) {
          l += B;
          r += B;
        }
        int ri = upper_bound(ts.begin(), ts.end(), r) - ts.begin();
        int li = lower_bound(ts.begin(), ts.end(), l) - ts.begin();
        if (sum(ri - 1) == sum(li - 1)) {
          res += (a[q] & B);
        } else {
          res += (B ^ (a[q] & B));
        }
      }
      printf("%d\n", a[q] ^ res);
    }
  }
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution 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 N = 200005;
int fen[N];
void add(int i, int val) {
  for (; i < N; i |= (i + 1)) fen[i] += val;
}
int sum(int i) {
  int s = 0;
  for (; i >= 0; i = (i & (i + 1)) - 1) s += fen[i];
  return s;
}
int main() {
  int Q;
  scanf("%d", &Q);
  vector<int> t(Q), a(Q), ts(Q);
  char buf[10];
  for (int i = 0; i < Q; ++i) {
    scanf("%s%d", buf, &a[i]);
    ts[i] = a[i];
    if (buf[0] == '-') t[i] = -1;
    if (buf[0] == '+') t[i] = 1;
  }
  add(0, 1);
  ts.push_back(0);
  sort(ts.begin(), ts.end());
  ts.resize(unique(ts.begin(), ts.end()) - ts.begin());
  for (int q = 0; q < Q; ++q) {
    int i = lower_bound(ts.begin(), ts.end(), a[q]) - ts.begin();
    if (t[q]) {
      add(i, t[q]);
    } else {
      int res = 0;
      for (int b = 30; b >= 0; --b) {
        int B = (1 << b);
        int l = res, r = res + B - 1;
        if (!(a[q] & B)) {
          l += B;
          r += B;
        }
        int ri = upper_bound(ts.begin(), ts.end(), r) - ts.begin();
        int li = lower_bound(ts.begin(), ts.end(), l) - ts.begin();
        if (sum(ri - 1) == sum(li - 1)) {
          res += (a[q] & B);
        } else {
          res += (B ^ (a[q] & B));
        }
      }
      printf("%d\n", a[q] ^ res);
    }
  }
  return 0;
}

```