#include <bits/stdc++.h>
using namespace std;
static const int INF = 0x3f3f3f3f;
static const long long INFL = 0x3f3f3f3f3f3f3f3fLL;
template <typename T, typename U>
static void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static void amax(T &x, U y) {
  if (x < y) x = y;
}
struct Node {
  Node *c[2];
  int cnt;
  Node() : c{0, 0}, cnt(0) {}
};
int main() {
  int q;
  while (~scanf("%d", &q)) {
    const int B = 30;
    vector<Node> nodes((q + 1) * B + 1);
    int nNodes = 0;
    Node *root = new (&nodes[nNodes++]) Node();
    for (int ii = -1; ii < q; ++ii) {
      char ty[10];
      if (ii == -1) {
        *ty = '+';
      } else {
        scanf("%s", ty);
      }
      if (*ty == '+' || *ty == '-') {
        int x;
        if (ii == -1) {
          x = 0;
        } else {
          scanf("%d", &x);
        }
        Node *p = root;
        Node *path[B + 1];
        path[B] = root;
        for (int i = B - 1; i >= 0; --i) {
          Node *&c = p->c[x >> i & 1];
          if (!c) c = new (&nodes[nNodes++]) Node();
          p = c;
          path[i] = p;
        }
        for (Node *t : path) t->cnt += *ty == '+' ? +1 : -1;
      } else if (*ty == '?') {
        int x;
        scanf("%d", &x);
        int ans = 0;
        Node *p = root;
        for (int i = B - 1; i >= 0; --i) {
          int b = x >> i & 1;
          if (p->c[b ^ 1] && p->c[b ^ 1]->cnt > 0) {
            p = p->c[b ^ 1];
            ans |= 1 << i;
          } else {
            p = p->c[b];
          }
          assert(p != nullptr && p->cnt > 0);
        }
        printf("%d\n", ans);
      } else
        abort();
    }
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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;
static const int INF = 0x3f3f3f3f;
static const long long INFL = 0x3f3f3f3f3f3f3f3fLL;
template <typename T, typename U>
static void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static void amax(T &x, U y) {
  if (x < y) x = y;
}
struct Node {
  Node *c[2];
  int cnt;
  Node() : c{0, 0}, cnt(0) {}
};
int main() {
  int q;
  while (~scanf("%d", &q)) {
    const int B = 30;
    vector<Node> nodes((q + 1) * B + 1);
    int nNodes = 0;
    Node *root = new (&nodes[nNodes++]) Node();
    for (int ii = -1; ii < q; ++ii) {
      char ty[10];
      if (ii == -1) {
        *ty = '+';
      } else {
        scanf("%s", ty);
      }
      if (*ty == '+' || *ty == '-') {
        int x;
        if (ii == -1) {
          x = 0;
        } else {
          scanf("%d", &x);
        }
        Node *p = root;
        Node *path[B + 1];
        path[B] = root;
        for (int i = B - 1; i >= 0; --i) {
          Node *&c = p->c[x >> i & 1];
          if (!c) c = new (&nodes[nNodes++]) Node();
          p = c;
          path[i] = p;
        }
        for (Node *t : path) t->cnt += *ty == '+' ? +1 : -1;
      } else if (*ty == '?') {
        int x;
        scanf("%d", &x);
        int ans = 0;
        Node *p = root;
        for (int i = B - 1; i >= 0; --i) {
          int b = x >> i & 1;
          if (p->c[b ^ 1] && p->c[b ^ 1]->cnt > 0) {
            p = p->c[b ^ 1];
            ans |= 1 << i;
          } else {
            p = p->c[b];
          }
          assert(p != nullptr && p->cnt > 0);
        }
        printf("%d\n", ans);
      } else
        abort();
    }
  }
  return 0;
}

```