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.util.*;

public class Main {

    public static void main(String[] args) {

        Scanner in = new Scanner(System.in);
        int lines = in.nextInt();
        int min = 0;
        int max = 1023;
        String code = "";
        int number = 0;

        //There are only four options: bit doesn't change, bit always changes, 
        //bit is set to 0, bit is set to 1. For each bit it's easy to find which 
        //of these options happens to it.
        //Now let's write a program of three lines:
        //Use XOR with a number that has ones in bits that must be reversed
        //Use OR with a number that has ones in bits which must be set to 1.
        //Use AND with a number that has zeroes in bits that must be set to 0.
        for (int i = 0; i < lines; i++) {
            code = in.next();
            number = in.nextInt();

            if (code.charAt(0) == '&') {
                min &= number;
                max &= number;

            }

            if (code.charAt(0) == '|') {
                min |= number;
                max |= number;

            }

            if (code.charAt(0) == '^') {
                min ^= number;
                max ^= number;

            }
//            System.out.println(code);
        } // end for

        int a = 0;
        int b = 0;
        int c = 1;

        for (int i = 0; i < 10; i++) {
            if (min % 2 == 0 && max % 2 == 0) {
                
            }
            if (min % 2 == 1 && max % 2 == 0) {
                a += c;
                b += c;
            }
            if (min % 2 == 0 && max % 2 == 1) {
                a += c;
            }
            if (min % 2 == 1 && max % 2 == 1) {
                b += c;
            }
            c *= 2;
            min /= 2;
            max /= 2;
        }// end for

        System.out.println("2");
        System.out.println("& " + a);
        // System.out.println("| " + c);
        System.out.println("^ " + b);
    } // end main
} // end class