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 Codeforces
{
    public static void main(String args[])throws Exception
    {
        BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb=new StringBuilder();
        int n=Integer.parseInt(bu.readLine());
        int b[][]=new int[10][2],i,j;
        for(i=0;i<10;i++) b[i][1]=1;    //our starting value

        for(i=0;i<n;i++)
        {
            String s[]=bu.readLine().split(" ");
            int x=Integer.parseInt(s[1]);
            if(s[0].equals("|"))
            for(j=0;j<10;j++)
            {
                b[j][0]|=(x>>j)&1;
                b[j][1]|=(x>>j)&1;
            }
            else if(s[0].equals("^"))
            for(j=0;j<10;j++)
            {
                b[j][0]^=(x>>j)&1;
                b[j][1]^=(x>>j)&1;
            }
            else
            for(j=0;j<10;j++)
            {
                b[j][0]&=(x>>j)&1;
                b[j][1]&=(x>>j)&1;
            }
        }

        int and=1023,or=0,xor=0;
        for(i=0;i<10;i++)
        if(b[i][0]==b[i][1])
        {
            if(b[i][0]==1)  //both places have 1 irrespective of starting, this can be made possible with or
            or+=1<<i;
            else //both places have 0 irrespective of starting, this means we have an and 0 here
            and-=1<<i;
        }
        else
        {
            if(b[i][0]==1)  //this means 0->1 and 1->0, thus xor happened
            xor+=1<<i;
            //else 0 stayed 0 and 1 stayed 1, and happened
        }
        sb.append("3\n");
        sb.append("& "+and+"\n");
        sb.append("^ "+xor+"\n");
        sb.append("| "+or+"\n");
        System.out.print(sb);
    }
}