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.
import java.util.Scanner;

public class ShortProgram {

    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int n = s.nextInt();
        int[] bits = new int[10];
        // 0 = set 0, 1 = set 1, -1 = no change, 2 = flip
        for (int i = 0; i < bits.length; i++) bits[i] = -1;
        for (int i = 0; i < n; i++) {
            String operator = s.next();
            String num = Integer.toBinaryString(s.nextInt());
            num = "0000000000".substring(num.length()) + num;
            switch (operator) {
                case "&":
                    and(bits, num);
                    break;
                case "|":
                    or(bits, num);
                    break;
                case "^":
                    xor(bits, num);
                    break;
            }
        }
        StringBuilder xor = new StringBuilder();
        StringBuilder or = new StringBuilder();
//        StringBuilder and = new StringBuilder();
        for (int bit : bits) {
//            System.out.print(bit);
            if (bit == 0) {
//                and.append("0");
                or.append("1");
                xor.append("1");
            } else if (bit == 1) {
//                and.append("1");
                or.append("1");
                xor.append("0");
            } else if (bit == 2) {
//                and.append("1");
                or.append("0");
                xor.append("1");
            } else {
//                and.append("1");
                or.append("0");
                xor.append("0");
            }
        }
        int ct = 0;
//        if (String.valueOf(and).equals("1111111111")) and = null;
//        else ct++;
        if (String.valueOf(or).equals("0000000000")) or = null;
        else ct++;
        if (String.valueOf(xor).equals("0000000000")) xor = null;
        else ct++;
        System.out.println(ct);
//        if (and != null) System.out.println("& " + Integer.parseInt(String.valueOf(and), 2));
        if (or != null) System.out.println("| " + Integer.parseInt(String.valueOf(or), 2));
        if (xor != null) System.out.println("^ " + Integer.parseInt(String.valueOf(xor), 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;
        }
    }
}
