Problem: Petya learned a new programming language CALPAS. A program in this language always takes one non-negative integer and returns one non-negative integer as well.

In the language, there are only three commands: apply a bitwise operation AND, OR or XOR with a given constant to the current integer. A program can contain an arbitrary sequence of these operations with arbitrary constants from 0 to 1023. When the program is run, all operations are applied (in the given order) to the argument and in the end the result integer is returned.

Petya wrote a program in this language, but it turned out to be too long. Write a program in CALPAS that does the same thing as the Petya's program, and consists of no more than 5 lines. Your program should return the same integer as Petya's program for all arguments from 0 to 1023.

Input

The first line contains an integer n (1 ≤ n ≤ 5·105) — the number of lines.

Next n lines contain commands. A command consists of a character that represents the operation ("&", "|" or "^" for AND, OR or XOR respectively), and the constant xi 0 ≤ xi ≤ 1023.

Output

Output an integer k (0 ≤ k ≤ 5) — the length of your program.

Next k lines must contain commands in the same format as in the input.

Examples

Input

3
| 3
^ 2
| 1


Output

2
| 3
^ 2


Input

3
&amp; 1
&amp; 3
&amp; 5


Output

1
&amp; 1


Input

3
^ 1
^ 2
^ 3


Output

0

Note

You can read about bitwise operations in <https://en.wikipedia.org/wiki/Bitwise_operation>.

Second sample:

Let x be an input of the Petya's program. It's output is ((x&1)&3)&5 = x&(1&3&5) = x&1. So these two programs always give the same outputs.

Solution (in JAVA):

Solution:

import java.io.*;
import java.util.*;
public class Main{

    public static void main(String[] args) throws IOException {
        FastScanner sc=new FastScanner(System.in);
        int n=sc.nextInt();
        int a[]=new int [10];
        Arrays.fill(a, -1);
        for(int i=0;i<n;i++){
            
            char c=sc.next().charAt(0);
            int nu=sc.nextInt();
            String num=Integer.toBinaryString(nu);
            num="0000000000".substring(0,10-num.length())+num;
            switch(c){
                case'|':
                    or(a,num);
                    break;
                case '&':
                    and(a,num);
                    break;
                case'^':
                    xor(a,num);
                    break;
            }
            
        }
        StringBuilder or=new StringBuilder();
        StringBuilder xor=new StringBuilder();
        StringBuilder and=new StringBuilder();
        
        for(int i=0;i<10;i++){
            switch(a[i]){
                case -1:
                    or.append("0");
                    xor.append("0");
                    and.append("1");
                    break;
                case 0:
                    and.append("0");
                    or.append("0");
                    xor.append("0");
                    break;
                case 1:
                    and.append("1");
                    or.append("1");
                    xor.append("0");
                    break;
                case 2:
                    and.append("1");
                    or.append("0");
                    xor.append("1");
                    break;
            }
        }
       System.out.println(3);
        System.out.println("^ "+Integer.parseInt(xor.toString(),2));
        System.out.println("| "+Integer.parseInt(or.toString(),2));
        System.out.println("& "+Integer.parseInt(and.toString(),2));
        
    }
     private static void xor(int[] bits, String num) {
        for (int i = 0; i < bits.length; i++) {
            if (num.charAt(i) == '1') {
                // flip bits
                switch (bits[i]) {
                    case -1:
                        bits[i] = 2;
                        break;
                    case 1:
                        bits[i] = 0;
                        break;
                    case 0:
                        bits[i] = 1;
                        break;
                    case 2:
                        bits[i] = -1;
                        break;
                }
            }
        }
    }

    private static void or(int[] bits, String num) {
        for (int i = 0; i < bits.length; i++) {
            if (num.charAt(i) == '1') bits[i] = 1;
        }
    }

    private static void and(int[] bits, String num) {
        for (int i = 0; i < bits.length; i++) {
            if (num.charAt(i) == '0') bits[i] = 0;
        }
    }
    
    
    public static String dTb(int n){
        String ans="";
        while(n>0){
            ans=n%2+ans;
            n/=2;
        }
        for(int i=0;i<10-ans.length();i++){
            ans="0"+ans;
        }
        return ans;
    }
    public static int bTd(String b){
        int ans=0;
        for(int i=0;i<b.length();i++){
            if(b.charAt(b.length()-1-i)=='1')ans+=Math.pow(2, i);
        }
        return ans;
    }
     //__________________________________________________________
    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;
            public FastScanner(InputStream i) {
            br =  new BufferedReader(new InputStreamReader(i));
            st = new StringTokenizer("");
        }
        public String next() throws IOException {
            if(st.hasMoreTokens())return st.nextToken();
                else st = new StringTokenizer(br.readLine());
            return next();
        }
        public int nextInt() throws IOException {return Integer.parseInt(next());}
        public long nextLong() throws IOException {return Long.parseLong(next());}
        public double nextDouble() throws IOException {return Double.parseDouble(next());}
    }
}