#include <bits/stdc++.h>
using namespace std;
int ind(char c) { return c - '0'; }
int trie[32 * 200005][2], c[32 * 200005], cn;
map<int, int> m;
void insert(string s) {
  int node = 0;
  for (int i = 0; i < s.size(); i++) {
    if (trie[node][ind(s[i])])
      node = trie[node][ind(s[i])];
    else {
      cn++;
      c[node]++;
      trie[node][ind(s[i])] = cn;
      node = cn;
    }
  }
}
void erase(string s) {
  vector<int> v(1, 0);
  int node = 0;
  for (int i = 0; i < s.size(); i++) {
    if (trie[node][ind(s[i])])
      node = trie[node][ind(s[i])], v.push_back(node);
    else
      return;
  }
  for (int i = v.size() - 1; i > 0; i--) {
    if (!c[v[i]])
      trie[v[i - 1]][ind(s[i - 1])] = 0, c[v[i - 1]]--;
    else
      return;
  }
}
int solve(string s) {
  int node = 0, ans = 0;
  for (int i = 0; i < s.size(); i++) {
    if (trie[node][!ind(s[i])])
      node = trie[node][!ind(s[i])], ans += (1 << (31 - i - 1));
    else
      node = trie[node][ind(s[i])];
  }
  return ans;
}
string convert(int n, int base) {
  string ret = "";
  while (n) {
    ret += n % base + '0';
    n /= base;
  }
  for (int i = ret.size(); i < 31; i++) ret += '0';
  reverse(ret.begin(), ret.end());
  return ret;
}
int main() {
  insert(convert(0, 2));
  int q;
  scanf("%d", &q);
  while (q--) {
    char t;
    int a;
    cin >> t >> a;
    if (t == '+')
      m[a]++, insert(convert(a, 2));
    else if (t == '-') {
      m[a]--;
      if (!m[a]) erase(convert(a, 2));
    } else
      cout << solve(convert(a, 2)) << endl;
  }
}

### 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;
int ind(char c) { return c - '0'; }
int trie[32 * 200005][2], c[32 * 200005], cn;
map<int, int> m;
void insert(string s) {
  int node = 0;
  for (int i = 0; i < s.size(); i++) {
    if (trie[node][ind(s[i])])
      node = trie[node][ind(s[i])];
    else {
      cn++;
      c[node]++;
      trie[node][ind(s[i])] = cn;
      node = cn;
    }
  }
}
void erase(string s) {
  vector<int> v(1, 0);
  int node = 0;
  for (int i = 0; i < s.size(); i++) {
    if (trie[node][ind(s[i])])
      node = trie[node][ind(s[i])], v.push_back(node);
    else
      return;
  }
  for (int i = v.size() - 1; i > 0; i--) {
    if (!c[v[i]])
      trie[v[i - 1]][ind(s[i - 1])] = 0, c[v[i - 1]]--;
    else
      return;
  }
}
int solve(string s) {
  int node = 0, ans = 0;
  for (int i = 0; i < s.size(); i++) {
    if (trie[node][!ind(s[i])])
      node = trie[node][!ind(s[i])], ans += (1 << (31 - i - 1));
    else
      node = trie[node][ind(s[i])];
  }
  return ans;
}
string convert(int n, int base) {
  string ret = "";
  while (n) {
    ret += n % base + '0';
    n /= base;
  }
  for (int i = ret.size(); i < 31; i++) ret += '0';
  reverse(ret.begin(), ret.end());
  return ret;
}
int main() {
  insert(convert(0, 2));
  int q;
  scanf("%d", &q);
  while (q--) {
    char t;
    int a;
    cin >> t >> a;
    if (t == '+')
      m[a]++, insert(convert(a, 2));
    else if (t == '-') {
      m[a]--;
      if (!m[a]) erase(convert(a, 2));
    } else
      cout << solve(convert(a, 2)) << endl;
  }
}

```