#include <bits/stdc++.h>
const int START_H = 536870912;
std::vector<int> cnt, l, r;
void add(int node, int h, int k) {
  cnt[node]++;
  if (h == 0) {
    return;
  }
  if (l[node] == -1) {
    l[node] = l.size();
    cnt.push_back(0);
    l.push_back(-1);
    r.push_back(-1);
  }
  if (r[node] == -1) {
    r[node] = l.size();
    cnt.push_back(0);
    l.push_back(-1);
    r.push_back(-1);
  }
  if (k < h) {
    add(l[node], int(h / 2), k);
  } else {
    add(r[node], int(h / 2), k - h);
  }
}
void remove(int node, int h, int k) {
  cnt[node]--;
  if (h == 0) {
    return;
  }
  if (k < h) {
    remove(l[node], int(h / 2), k);
  } else {
    remove(r[node], int(h / 2), k - h);
  }
}
void request(int node, int h, int k, int ans) {
  if (h == 0) {
    printf("%d\n", ans);
    return;
  }
  int new_ans;
  if ((k < h && cnt[r[node]] > 0) || (cnt[l[node]] == 0)) {
    if (k < h) {
      new_ans = 2 * ans + 1;
    } else {
      k -= h;
      new_ans = 2 * ans;
    }
    request(r[node], int(h / 2), k, new_ans);
  } else {
    if (k < h) {
      new_ans = 2 * ans;
    } else {
      k -= h;
      new_ans = 2 * ans + 1;
    }
    request(l[node], int(h / 2), k, new_ans);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  cnt.push_back(0);
  l.push_back(-1);
  r.push_back(-1);
  add(0, START_H, 0);
  for (int i = 0; i < n; i++) {
    char t;
    int x;
    scanf("\n%c %d", &t, &x);
    if (t == '+') {
      add(0, START_H, x);
    } else if (t == '-') {
      remove(0, START_H, x);
    } else if (t == '?') {
      request(0, START_H, x, 0);
    }
  }
  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>
const int START_H = 536870912;
std::vector<int> cnt, l, r;
void add(int node, int h, int k) {
  cnt[node]++;
  if (h == 0) {
    return;
  }
  if (l[node] == -1) {
    l[node] = l.size();
    cnt.push_back(0);
    l.push_back(-1);
    r.push_back(-1);
  }
  if (r[node] == -1) {
    r[node] = l.size();
    cnt.push_back(0);
    l.push_back(-1);
    r.push_back(-1);
  }
  if (k < h) {
    add(l[node], int(h / 2), k);
  } else {
    add(r[node], int(h / 2), k - h);
  }
}
void remove(int node, int h, int k) {
  cnt[node]--;
  if (h == 0) {
    return;
  }
  if (k < h) {
    remove(l[node], int(h / 2), k);
  } else {
    remove(r[node], int(h / 2), k - h);
  }
}
void request(int node, int h, int k, int ans) {
  if (h == 0) {
    printf("%d\n", ans);
    return;
  }
  int new_ans;
  if ((k < h && cnt[r[node]] > 0) || (cnt[l[node]] == 0)) {
    if (k < h) {
      new_ans = 2 * ans + 1;
    } else {
      k -= h;
      new_ans = 2 * ans;
    }
    request(r[node], int(h / 2), k, new_ans);
  } else {
    if (k < h) {
      new_ans = 2 * ans;
    } else {
      k -= h;
      new_ans = 2 * ans + 1;
    }
    request(l[node], int(h / 2), k, new_ans);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  cnt.push_back(0);
  l.push_back(-1);
  r.push_back(-1);
  add(0, START_H, 0);
  for (int i = 0; i < n; i++) {
    char t;
    int x;
    scanf("\n%c %d", &t, &x);
    if (t == '+') {
      add(0, START_H, x);
    } else if (t == '-') {
      remove(0, START_H, x);
    } else if (t == '?') {
      request(0, START_H, x, 0);
    }
  }
  return 0;
}

```