#include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
struct node {
  node* nxt[2];
  int cnt;
};
node* root;
node* build() {
  node* temp = new node();
  temp->nxt[0] = NULL;
  temp->nxt[1] = NULL;
  temp->cnt = 0;
  return temp;
}
void insert(long long x) {
  node* temp = root;
  (root->cnt)++;
  (temp->cnt)++;
  for (int i = 30; i >= 0; i--) {
    int cur = (x >> i) & 1;
    if (temp->nxt[cur] == NULL) temp->nxt[cur] = build();
    temp = temp->nxt[cur];
    (temp->cnt)++;
  }
}
void del(node*& temp, int i, long long x) {
  (temp->cnt)--;
  if (i != 0) {
    int cur = (x >> (i - 1)) & 1;
    del(temp->nxt[cur], i - 1, x);
  }
  if ((temp->cnt) == 0) temp = NULL;
}
long long query(long long x) {
  long long ans = 0;
  node* temp = root;
  for (int i = 30; i >= 0; --i) {
    int cur = (x >> i) & 1;
    if (temp->nxt[cur ^ 1] != NULL) {
      ans += 1ll << i;
      temp = temp->nxt[cur ^ 1];
    } else
      temp = temp->nxt[cur];
  }
  return ans;
}
int main() {
  root = build();
  node* temp = root;
  (root->cnt)++;
  (temp->cnt)++;
  for (int i = 30; i >= 0; i--) {
    int cur = (0 >> i) & 1;
    if (temp->nxt[cur] == NULL) temp->nxt[cur] = build();
    temp = temp->nxt[cur];
    (temp->cnt)++;
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    string s;
    long long x;
    cin >> s;
    scanf("%lld", &x);
    if (s[0] == '+')
      insert(x);
    else if (s[0] == '-')
      del(root, 31, x);
    else
      printf("%lld\n", query(x));
  }
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
struct node {
  node* nxt[2];
  int cnt;
};
node* root;
node* build() {
  node* temp = new node();
  temp->nxt[0] = NULL;
  temp->nxt[1] = NULL;
  temp->cnt = 0;
  return temp;
}
void insert(long long x) {
  node* temp = root;
  (root->cnt)++;
  (temp->cnt)++;
  for (int i = 30; i >= 0; i--) {
    int cur = (x >> i) & 1;
    if (temp->nxt[cur] == NULL) temp->nxt[cur] = build();
    temp = temp->nxt[cur];
    (temp->cnt)++;
  }
}
void del(node*& temp, int i, long long x) {
  (temp->cnt)--;
  if (i != 0) {
    int cur = (x >> (i - 1)) & 1;
    del(temp->nxt[cur], i - 1, x);
  }
  if ((temp->cnt) == 0) temp = NULL;
}
long long query(long long x) {
  long long ans = 0;
  node* temp = root;
  for (int i = 30; i >= 0; --i) {
    int cur = (x >> i) & 1;
    if (temp->nxt[cur ^ 1] != NULL) {
      ans += 1ll << i;
      temp = temp->nxt[cur ^ 1];
    } else
      temp = temp->nxt[cur];
  }
  return ans;
}
int main() {
  root = build();
  node* temp = root;
  (root->cnt)++;
  (temp->cnt)++;
  for (int i = 30; i >= 0; i--) {
    int cur = (0 >> i) & 1;
    if (temp->nxt[cur] == NULL) temp->nxt[cur] = build();
    temp = temp->nxt[cur];
    (temp->cnt)++;
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    string s;
    long long x;
    cin >> s;
    scanf("%lld", &x);
    if (s[0] == '+')
      insert(x);
    else if (s[0] == '-')
      del(root, 31, x);
    else
      printf("%lld\n", query(x));
  }
  return 0;
}

```