#include <bits/stdc++.h>
using namespace std;
struct node {
  int p[2];
  int cnt;
  node() {
    cnt = 0;
    p[0] = -1;
    p[1] = -1;
  }
};
vector<node> v;
string tobin(int n) {
  string ans = "";
  while (n != 0) {
    if (n % 2 == 0)
      ans += '0';
    else
      ans += '1';
    n /= 2;
  }
  for (int i = ans.size(); i < 31; i++) ans += '0';
  reverse(ans.begin(), ans.end());
  return ans;
}
void insert(string n, int ad) {
  int cur = 0;
  for (int i = 0; i < 31; ++i) {
    int b = n[i] - '0';
    if (v[cur].p[b] == -1) {
      v[cur].p[b] = v.size();
      v.push_back(node());
    }
    cur = v[cur].p[b];
    v[cur].cnt += ad;
  }
}
string ans(string n) {
  int cur = 0;
  string ans = tobin(0);
  for (int i = 0; i < 31; ++i) {
    ans[i] = '1';
    int b = n[i] - '0';
    int nod = v[cur].p[!b];
    if (v[cur].p[!b] != -1 && v[nod].cnt) {
      cur = nod;
    } else {
      ans[i] = '0';
      cur = v[cur].p[b];
    }
  }
  return ans;
}
int frombin(string n) {
  int k = 0;
  for (int i = 0; i < 31; i++) {
    k += pow(2.0, 30 - i) * (n[i] - '0');
  }
  return k;
}
int main() {
  int q;
  cin >> q;
  char c;
  int x;
  v.push_back(node());
  v[0].cnt = 0;
  string n = tobin(0);
  insert(n, 1);
  for (int i = 0; i < q; i++) {
    cin >> c >> x;
    string v = tobin(x);
    if (c == '+') {
      insert(v, 1);
    } else if (c == '-') {
      insert(v, -1);
    } else {
      string u = ans(v);
      int ans = frombin(u);
      cout << ans << endl;
    }
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
struct node {
  int p[2];
  int cnt;
  node() {
    cnt = 0;
    p[0] = -1;
    p[1] = -1;
  }
};
vector<node> v;
string tobin(int n) {
  string ans = "";
  while (n != 0) {
    if (n % 2 == 0)
      ans += '0';
    else
      ans += '1';
    n /= 2;
  }
  for (int i = ans.size(); i < 31; i++) ans += '0';
  reverse(ans.begin(), ans.end());
  return ans;
}
void insert(string n, int ad) {
  int cur = 0;
  for (int i = 0; i < 31; ++i) {
    int b = n[i] - '0';
    if (v[cur].p[b] == -1) {
      v[cur].p[b] = v.size();
      v.push_back(node());
    }
    cur = v[cur].p[b];
    v[cur].cnt += ad;
  }
}
string ans(string n) {
  int cur = 0;
  string ans = tobin(0);
  for (int i = 0; i < 31; ++i) {
    ans[i] = '1';
    int b = n[i] - '0';
    int nod = v[cur].p[!b];
    if (v[cur].p[!b] != -1 && v[nod].cnt) {
      cur = nod;
    } else {
      ans[i] = '0';
      cur = v[cur].p[b];
    }
  }
  return ans;
}
int frombin(string n) {
  int k = 0;
  for (int i = 0; i < 31; i++) {
    k += pow(2.0, 30 - i) * (n[i] - '0');
  }
  return k;
}
int main() {
  int q;
  cin >> q;
  char c;
  int x;
  v.push_back(node());
  v[0].cnt = 0;
  string n = tobin(0);
  insert(n, 1);
  for (int i = 0; i < q; i++) {
    cin >> c >> x;
    string v = tobin(x);
    if (c == '+') {
      insert(v, 1);
    } else if (c == '-') {
      insert(v, -1);
    } else {
      string u = ans(v);
      int ans = frombin(u);
      cout << ans << endl;
    }
  }
  return 0;
}

```