#include <bits/stdc++.h>
using namespace std;
struct node {
  node *left = NULL;
  node *right = NULL;
  int hl = 0, hr = 0;
};
long long int A[32];
void ins(node *root, long long int x) {
  for (int i = 31; i >= 0; i--) {
    if (A[i] <= x) {
      root->hr++;
      if (root->right == NULL) {
        root->right = new node;
      }
      x -= A[i];
      root = root->right;
    } else {
      root->hl++;
      if (root->left == NULL) {
        root->left = new node;
      }
      root = root->left;
    }
  }
}
void del(node *root, long long int x) {
  for (int i = 31; i >= 0; i--) {
    if (A[i] <= x) {
      root->hr--;
      x -= A[i];
      root = root->right;
    } else {
      root->hl--;
      root = root->left;
    }
  }
}
long long int ans(node *root, long long int x) {
  long long int h = x;
  long long int temp = x;
  for (int i = 31; i >= 0; i--) {
    if (A[i] <= x) {
      if (root->hl > 0) {
        root = root->left;
      } else {
        root = root->right;
        temp -= A[i];
      }
      x -= A[i];
    } else {
      if (root->hr > 0) {
        root = root->right;
        temp += A[i];
      } else {
        root = root->left;
      }
    }
  }
  return max(x, temp);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  node *root = new node;
  A[0] = 1;
  for (int i = 1; i < 32; i++) {
    A[i] = 2 * A[i - 1];
  }
  ins(root, 0);
  int q;
  cin >> q;
  while (q--) {
    char ch;
    long long int x;
    cin >> ch >> x;
    if (ch == '+') {
      ins(root, x);
    } else if (ch == '-') {
      del(root, x);
    } else if (ch == '?') {
      cout << ans(root, x) << endl;
    }
  }
  return 0;
}

### Prompt

Your task is to create 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;
struct node {
  node *left = NULL;
  node *right = NULL;
  int hl = 0, hr = 0;
};
long long int A[32];
void ins(node *root, long long int x) {
  for (int i = 31; i >= 0; i--) {
    if (A[i] <= x) {
      root->hr++;
      if (root->right == NULL) {
        root->right = new node;
      }
      x -= A[i];
      root = root->right;
    } else {
      root->hl++;
      if (root->left == NULL) {
        root->left = new node;
      }
      root = root->left;
    }
  }
}
void del(node *root, long long int x) {
  for (int i = 31; i >= 0; i--) {
    if (A[i] <= x) {
      root->hr--;
      x -= A[i];
      root = root->right;
    } else {
      root->hl--;
      root = root->left;
    }
  }
}
long long int ans(node *root, long long int x) {
  long long int h = x;
  long long int temp = x;
  for (int i = 31; i >= 0; i--) {
    if (A[i] <= x) {
      if (root->hl > 0) {
        root = root->left;
      } else {
        root = root->right;
        temp -= A[i];
      }
      x -= A[i];
    } else {
      if (root->hr > 0) {
        root = root->right;
        temp += A[i];
      } else {
        root = root->left;
      }
    }
  }
  return max(x, temp);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  node *root = new node;
  A[0] = 1;
  for (int i = 1; i < 32; i++) {
    A[i] = 2 * A[i - 1];
  }
  ins(root, 0);
  int q;
  cin >> q;
  while (q--) {
    char ch;
    long long int x;
    cin >> ch >> x;
    if (ch == '+') {
      ins(root, x);
    } else if (ch == '-') {
      del(root, x);
    } else if (ch == '?') {
      cout << ans(root, x) << endl;
    }
  }
  return 0;
}

```