#include <bits/stdc++.h>
using namespace std;
const int N = 200010;
char op[5];
struct Node {
  struct Node *next[2];
  int num;
};
int length;
Node *head;
int bit[N][35], tmp[35];
void Init() {
  head = new Node;
  for (int i = 0; i < 2; i++) {
    head->next[i] = NULL;
    head->num = 0;
  }
}
void Insert(int x) {
  Node *t, *s = head;
  for (int i = length - 1; i >= 0; i--) {
    tmp[i] = x & 1;
    x >>= 1;
  }
  for (int i = 0; i < length; i++) {
    int id = tmp[i];
    if (s->next[id] == NULL) {
      t = new Node;
      for (int j = 0; j < 2; j++) t->next[j] = NULL;
      t->num = 0;
      s->next[id] = t;
    }
    s = s->next[id];
    s->num++;
  }
}
long long Cal(long long x) {
  long long res = 0;
  Node *s = head;
  for (int i = length - 1; i >= 0; i--) {
    tmp[i] = x & 1;
    x >>= 1;
  }
  for (int i = 0; i < length; i++) {
    int id = tmp[i];
    if (s->next[id ^ 1] != NULL && s->next[id ^ 1]->num > 0) {
      s = s->next[id ^ 1];
      res += (long long)(1 << (length - 1 - i));
      continue;
    }
    if (s->next[id] != NULL && s->next[id]->num > 0) {
      s = s->next[id];
    }
  }
  return res;
}
void Delete(int x) {
  Node *s = head, *pre;
  for (int i = length - 1; i >= 0; i--) {
    tmp[i] = x & 1;
    x >>= 1;
  }
  for (int i = 0; i < length; i++) {
    int id = tmp[i];
    pre = s;
    s = s->next[id];
    s->num--;
  }
}
int main() {
  int n;
  scanf("%d", &n);
  Init();
  length = 30;
  Insert(0);
  for (int i = 1; i <= n; i++) {
    long long x;
    scanf("%s%I64d", op, &x);
    if (op[0] == '+') Insert(x);
    if (op[0] == '-') Delete(x);
    if (op[0] == '?') printf("%I64d\n", Cal(x));
  }
  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 = 200010;
char op[5];
struct Node {
  struct Node *next[2];
  int num;
};
int length;
Node *head;
int bit[N][35], tmp[35];
void Init() {
  head = new Node;
  for (int i = 0; i < 2; i++) {
    head->next[i] = NULL;
    head->num = 0;
  }
}
void Insert(int x) {
  Node *t, *s = head;
  for (int i = length - 1; i >= 0; i--) {
    tmp[i] = x & 1;
    x >>= 1;
  }
  for (int i = 0; i < length; i++) {
    int id = tmp[i];
    if (s->next[id] == NULL) {
      t = new Node;
      for (int j = 0; j < 2; j++) t->next[j] = NULL;
      t->num = 0;
      s->next[id] = t;
    }
    s = s->next[id];
    s->num++;
  }
}
long long Cal(long long x) {
  long long res = 0;
  Node *s = head;
  for (int i = length - 1; i >= 0; i--) {
    tmp[i] = x & 1;
    x >>= 1;
  }
  for (int i = 0; i < length; i++) {
    int id = tmp[i];
    if (s->next[id ^ 1] != NULL && s->next[id ^ 1]->num > 0) {
      s = s->next[id ^ 1];
      res += (long long)(1 << (length - 1 - i));
      continue;
    }
    if (s->next[id] != NULL && s->next[id]->num > 0) {
      s = s->next[id];
    }
  }
  return res;
}
void Delete(int x) {
  Node *s = head, *pre;
  for (int i = length - 1; i >= 0; i--) {
    tmp[i] = x & 1;
    x >>= 1;
  }
  for (int i = 0; i < length; i++) {
    int id = tmp[i];
    pre = s;
    s = s->next[id];
    s->num--;
  }
}
int main() {
  int n;
  scanf("%d", &n);
  Init();
  length = 30;
  Insert(0);
  for (int i = 1; i <= n; i++) {
    long long x;
    scanf("%s%I64d", op, &x);
    if (op[0] == '+') Insert(x);
    if (op[0] == '-') Delete(x);
    if (op[0] == '?') printf("%I64d\n", Cal(x));
  }
  return 0;
}

```