#include <bits/stdc++.h>
using namespace std;
const int Maxn = 200005;
struct node {
  node *first, *second;
  int cnt;
  node(node *a, node *b) {
    first = a;
    second = b;
    cnt = 0;
  }
};
node *root = new node(NULL, NULL);
void insert(node *x, int bits, int idx) {
  (x->cnt)++;
  if (idx == -1) return;
  if ((1 << idx) & bits) {
    if (x->second == NULL) x->second = new node(NULL, NULL);
    insert(x->second, bits, idx - 1);
  } else {
    if (x->first == NULL) x->first = new node(NULL, NULL);
    insert(x->first, bits, idx - 1);
  }
}
void update(node *x, int bits, int idx) {
  (x->cnt)--;
  if (idx == -1) return;
  if ((1 << idx) & bits) {
    update(x->second, bits, idx - 1);
  } else {
    update(x->first, bits, idx - 1);
  }
}
void query(node *t, int a, int idx, int &ans) {
  if (idx == -1) return;
  if ((1 << idx) & a) {
    if (t->first == NULL)
      query(t->second, a, idx - 1, ans);
    else if ((t->first)->cnt <= 0)
      query(t->second, a, idx - 1, ans);
    else {
      ans += (1 << idx);
      query(t->first, a, idx - 1, ans);
    }
  } else {
    if (t->second == NULL)
      query(t->first, a, idx - 1, ans);
    else if ((t->second)->cnt <= 0)
      query(t->first, a, idx - 1, ans);
    else {
      ans += (1 << idx);
      query(t->second, a, idx - 1, ans);
    }
  }
}
int a[Maxn];
char s[5];
unordered_map<int, int> m;
int main() {
  int n;
  scanf("%d", &n);
  insert(root, 0, 29);
  for (int i = 0; i < n; i++) {
    scanf("%s", s);
    scanf("%d", &a[i]);
    if (s[0] == '+') {
      m[a[i]]++;
      if (m[a[i]] == 1) {
        insert(root, a[i], 29);
      }
    } else if (s[0] == '-') {
      m[a[i]]--;
      if (m[a[i]] == 0) update(root, a[i], 29);
    } else {
      int ans = 0;
      query(root, a[i], 29, ans);
      printf("%d\n", ans);
    }
  }
  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 Maxn = 200005;
struct node {
  node *first, *second;
  int cnt;
  node(node *a, node *b) {
    first = a;
    second = b;
    cnt = 0;
  }
};
node *root = new node(NULL, NULL);
void insert(node *x, int bits, int idx) {
  (x->cnt)++;
  if (idx == -1) return;
  if ((1 << idx) & bits) {
    if (x->second == NULL) x->second = new node(NULL, NULL);
    insert(x->second, bits, idx - 1);
  } else {
    if (x->first == NULL) x->first = new node(NULL, NULL);
    insert(x->first, bits, idx - 1);
  }
}
void update(node *x, int bits, int idx) {
  (x->cnt)--;
  if (idx == -1) return;
  if ((1 << idx) & bits) {
    update(x->second, bits, idx - 1);
  } else {
    update(x->first, bits, idx - 1);
  }
}
void query(node *t, int a, int idx, int &ans) {
  if (idx == -1) return;
  if ((1 << idx) & a) {
    if (t->first == NULL)
      query(t->second, a, idx - 1, ans);
    else if ((t->first)->cnt <= 0)
      query(t->second, a, idx - 1, ans);
    else {
      ans += (1 << idx);
      query(t->first, a, idx - 1, ans);
    }
  } else {
    if (t->second == NULL)
      query(t->first, a, idx - 1, ans);
    else if ((t->second)->cnt <= 0)
      query(t->first, a, idx - 1, ans);
    else {
      ans += (1 << idx);
      query(t->second, a, idx - 1, ans);
    }
  }
}
int a[Maxn];
char s[5];
unordered_map<int, int> m;
int main() {
  int n;
  scanf("%d", &n);
  insert(root, 0, 29);
  for (int i = 0; i < n; i++) {
    scanf("%s", s);
    scanf("%d", &a[i]);
    if (s[0] == '+') {
      m[a[i]]++;
      if (m[a[i]] == 1) {
        insert(root, a[i], 29);
      }
    } else if (s[0] == '-') {
      m[a[i]]--;
      if (m[a[i]] == 0) update(root, a[i], 29);
    } else {
      int ans = 0;
      query(root, a[i], 29, ans);
      printf("%d\n", ans);
    }
  }
  return 0;
}

```