#include <bits/stdc++.h>
using namespace std;
string s;
struct node {
  node* nxt[3];
  int cnt;
};
struct node* get() {
  node* n = new node;
  for (int i = 0; i < 2; i++) n->nxt[i] = NULL;
  n->cnt = 1;
  return n;
}
node* root = get();
void add(int z) {
  string l = s;
  node* tmp = root;
  for (int i = 0; i < s.size(); i++) {
    int x = s[i] - '0';
    if (tmp->nxt[x] == NULL)
      tmp->nxt[x] = get();
    else
      tmp->nxt[x]->cnt += z;
    tmp = tmp->nxt[x];
  }
}
void conv(int x) {
  s = "";
  while (x > 0) {
    int tmp = x % 2;
    x /= 2;
    if (tmp == 0) {
      s += '0';
    } else
      s += '1';
  }
  while (s.size() < 32) {
    s += '0';
  }
  reverse(s.begin(), s.end());
}
long long q() {
  long long res = 0;
  struct node* tmp = root;
  for (int i = 0; i < s.size(); i++) {
    int x = s[i] - '0';
    if (x == 0) {
      if (tmp->nxt[1] != NULL) {
        if (tmp->nxt[1]->cnt > 0) {
          res |= (1 << s.size() - i - 1);
          tmp = tmp->nxt[1];
        } else
          tmp = tmp->nxt[0];
      } else {
        tmp = tmp->nxt[0];
      }
    } else {
      if (tmp->nxt[0] != NULL) {
        if (tmp->nxt[0]->cnt > 0) {
          res |= (1 << s.size() - i - 1);
          tmp = tmp->nxt[0];
        } else
          tmp = tmp->nxt[1];
      } else {
        tmp = tmp->nxt[1];
      }
    }
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  long long t, x;
  char c;
  cin >> t;
  int ok = 0;
  while (t--) {
    cin >> c >> x;
    conv(x);
    if (c == '+') {
      add(1);
      ok++;
    } else if (c == '-') {
      add(-1);
      ok--;
    } else {
      if (ok == 0) {
        cout << x << endl;
        continue;
      }
      cout << max(x, q()) << endl;
    }
  }
}

### Prompt

Generate 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;
string s;
struct node {
  node* nxt[3];
  int cnt;
};
struct node* get() {
  node* n = new node;
  for (int i = 0; i < 2; i++) n->nxt[i] = NULL;
  n->cnt = 1;
  return n;
}
node* root = get();
void add(int z) {
  string l = s;
  node* tmp = root;
  for (int i = 0; i < s.size(); i++) {
    int x = s[i] - '0';
    if (tmp->nxt[x] == NULL)
      tmp->nxt[x] = get();
    else
      tmp->nxt[x]->cnt += z;
    tmp = tmp->nxt[x];
  }
}
void conv(int x) {
  s = "";
  while (x > 0) {
    int tmp = x % 2;
    x /= 2;
    if (tmp == 0) {
      s += '0';
    } else
      s += '1';
  }
  while (s.size() < 32) {
    s += '0';
  }
  reverse(s.begin(), s.end());
}
long long q() {
  long long res = 0;
  struct node* tmp = root;
  for (int i = 0; i < s.size(); i++) {
    int x = s[i] - '0';
    if (x == 0) {
      if (tmp->nxt[1] != NULL) {
        if (tmp->nxt[1]->cnt > 0) {
          res |= (1 << s.size() - i - 1);
          tmp = tmp->nxt[1];
        } else
          tmp = tmp->nxt[0];
      } else {
        tmp = tmp->nxt[0];
      }
    } else {
      if (tmp->nxt[0] != NULL) {
        if (tmp->nxt[0]->cnt > 0) {
          res |= (1 << s.size() - i - 1);
          tmp = tmp->nxt[0];
        } else
          tmp = tmp->nxt[1];
      } else {
        tmp = tmp->nxt[1];
      }
    }
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  long long t, x;
  char c;
  cin >> t;
  int ok = 0;
  while (t--) {
    cin >> c >> x;
    conv(x);
    if (c == '+') {
      add(1);
      ok++;
    } else if (c == '-') {
      add(-1);
      ok--;
    } else {
      if (ok == 0) {
        cout << x << endl;
        continue;
      }
      cout << max(x, q()) << endl;
    }
  }
}

```