#include <bits/stdc++.h>
using namespace std;
const int OO = (int)1e9;
const double eps = 1e-10;
int q, nxt = 1;
const int MX = 2e5 * 33;
bool s[33];
int trie[MX][2];
int cnt[MX];
void build(int add) {
  int i = 0, currNode = 0;
  while (i < 33) {
    if (trie[currNode][s[i]] == -1)
      currNode = trie[currNode][s[i]] = nxt++;
    else
      currNode = trie[currNode][s[i]];
    cnt[currNode] += add;
    ++i;
  }
}
int ans() {
  int i = 0, currNode = 0, ret = 0;
  while (i < 33) {
    if (!s[i]) {
      if (trie[currNode][1] != -1 && cnt[trie[currNode][1]]) {
        ret += (1 << (32 - i));
        currNode = trie[currNode][1];
      } else
        currNode = trie[currNode][0];
    } else {
      if (trie[currNode][0] != -1 && cnt[trie[currNode][0]]) {
        ret += (1 << (32 - i));
        currNode = trie[currNode][0];
      } else
        currNode = trie[currNode][1];
    }
    ++i;
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cout.precision(10);
  for (int i = 0; i < int(MX); ++i)
    for (int j = 0; j < int(2); ++j) trie[i][j] = -1;
  build(1);
  cin >> q;
  char c;
  int x;
  for (int i = 0; i < int(q); ++i) {
    cin >> c >> x;
    memset(s, 0, sizeof(s));
    int j = 32;
    while (x) s[j--] = x & 1, x >>= 1;
    if (c == '+')
      build(1);
    else if (c == '-')
      build(-1);
    else
      cout << ans() << '\n';
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;
const int OO = (int)1e9;
const double eps = 1e-10;
int q, nxt = 1;
const int MX = 2e5 * 33;
bool s[33];
int trie[MX][2];
int cnt[MX];
void build(int add) {
  int i = 0, currNode = 0;
  while (i < 33) {
    if (trie[currNode][s[i]] == -1)
      currNode = trie[currNode][s[i]] = nxt++;
    else
      currNode = trie[currNode][s[i]];
    cnt[currNode] += add;
    ++i;
  }
}
int ans() {
  int i = 0, currNode = 0, ret = 0;
  while (i < 33) {
    if (!s[i]) {
      if (trie[currNode][1] != -1 && cnt[trie[currNode][1]]) {
        ret += (1 << (32 - i));
        currNode = trie[currNode][1];
      } else
        currNode = trie[currNode][0];
    } else {
      if (trie[currNode][0] != -1 && cnt[trie[currNode][0]]) {
        ret += (1 << (32 - i));
        currNode = trie[currNode][0];
      } else
        currNode = trie[currNode][1];
    }
    ++i;
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cout.precision(10);
  for (int i = 0; i < int(MX); ++i)
    for (int j = 0; j < int(2); ++j) trie[i][j] = -1;
  build(1);
  cin >> q;
  char c;
  int x;
  for (int i = 0; i < int(q); ++i) {
    cin >> c >> x;
    memset(s, 0, sizeof(s));
    int j = 32;
    while (x) s[j--] = x & 1, x >>= 1;
    if (c == '+')
      build(1);
    else if (c == '-')
      build(-1);
    else
      cout << ans() << '\n';
  }
  return 0;
}

```