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>.
import java.io.BufferedInputStream;
import java.text.DecimalFormat;
import java.util.Arrays;
import java.util.Scanner;

public class Test {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(new BufferedInputStream(System.in, 10000));



        Node root = new Node(30);
        root.add(0);

        int q = scanner.nextInt();
        scanner.nextLine();
        for (int i = 0; i < q; i++) {
            String l = scanner.nextLine();
            String[] s = l.split("\\s+");
            int v = Integer.parseInt(s[1]);
            if (s[0].equals("+")) {
                root.add(v);
            }
            if (s[0].equals("-")) {
                root.remove(v);
            }
            if (s[0].equals("?")) {
                int max = root.max(v);
                System.out.println(max ^ v);
            }
        }
    }

    private static class Node {
        Node zero;
        Node one;

        int zeroCost;
        int oneCost;
        int level;
        int mask;

        public Node(int level) {
            this.level = level;
            if (level > -1) {
                this.mask = (int) Math.pow(2, level);
            }
        }

        public void add(int value) {
            if (level == -1) {
                return;
            }
            if ((value & mask) == 0) {
                zeroCost++;
                if (zero == null) {
                    zero = new Node(level - 1);
                }
                zero.add(value);
            } else {
                if (one == null) {
                    one = new Node(level - 1);
                }
                oneCost++;
                one.add(value);
            }
        }

        public void remove(int value) {
            if (level == -1) {
                return;
            }
            if ((value & mask) == 0) {
                zeroCost--;
                zero.remove(value);
            } else {
                oneCost--;
                one.remove(value);
            }
        }

        public int max(int value) {
            if (level == -1) {
                return 0;
            }
            int current = (value & mask);
            if (current == 0) {
                if (oneCost > 0) {
                    return mask | one.max(value);
                } else {
                    return zero.max(value);
                }
            } else {
                if (zeroCost > 0) {
                    return zero.max(value);
                } else {
                    return mask | one.max(value);
                }
            }
        }
    }

}