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>.
//package CF;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;


public class D{
    static class Node
    {
	Node z; Node o; int val;
    }
    static int query(Node cur)
    {
	int ans = 0;
	for (int i = 31, j = x.length-32; i >= 0; i--,j++)
	{
	    Node nxt;
	    if((j < 0?'0':x[j]) == '0'){
		nxt = cur.o;
		if(nxt == null || nxt.val == 0) nxt = cur.z;
		else ans |= 1<<i;
	    }
	    else{
		nxt = cur.z;
		if(nxt == null || nxt.val == 0) { nxt = cur.o; ans |= 1<<i;}
		
	    }
//	    System.out.println(ans);
//	    if(nxt == null) break;
	    cur = nxt;
	}
	return ans;
    }
    static void put(Node n,int c)
    {
	if(n == null) n = new Node();
	Node nxt;
	if((c < 0?'0':x[c]) == '0'){
	    nxt = n.z;
	    if(nxt == null) nxt = n.z = new Node();
	}
	else{
	    nxt = n.o;
	    if(nxt == null) nxt = n.o = new Node();
	}
	nxt.val++;
	if(c == x.length-1) 
	{
	    return;
	}
	put(nxt,c+1);
    }
    static void del(Node n, int c)
    {
	if(n == null) return;
	Node nxt;
	if((c < 0?'0':x[c]) == '0'){
	    nxt = n.z;
	    if(c == x.length-1){
		nxt.val--;
		if(nxt.val == 0) nxt = n.z = null; 
		return;
	    }
	}
	else{
	    nxt = n.o;
	    if(c == x.length-1){
		nxt.val--;
		if(nxt.val == 0) nxt = n.o = null;
		return;
	    }
	}
	nxt.val--;
	del(nxt,c+1);
    }
    static char [] x;
    public static void main(String[] args) throws IOException   {
	Scanner bf = new Scanner(System.in);
	PrintWriter out = new PrintWriter(System.out);
	Node root = new Node();
	x = "0".toCharArray();
	put(root,x.length-32);
	int q = bf.nextInt();
	for (int i = 0; i < q; i++)
	{
	    char c = bf.next().charAt(0);  
	    int n = bf.nextInt();
	    x = Integer.toBinaryString(n).toCharArray();
	    if(c == '+'){
		put(root, x.length-32);
	    }
	    else if(c == '-'){
		del(root, x.length-32);
	    }
	    else {
		out.println(n^query(root));
	    }
	}

	out.flush();
	out.close();
    }
    static class Scanner 
    {
	StringTokenizer st;
	BufferedReader br;

	public Scanner(InputStream s){	br = new BufferedReader(new InputStreamReader(s));}

	public String next() throws IOException 
	{
	    while (st == null || !st.hasMoreTokens()) 
		st = new StringTokenizer(br.readLine());
	    return st.nextToken();
	}

	public int nextInt() throws IOException {return Integer.parseInt(next());}

	public long nextLong() throws IOException {return Long.parseLong(next());}

	public String nextLine() throws IOException {return br.readLine();}

	public boolean ready() throws IOException {return br.ready();}
    }

}