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:
// practice with kaiboy
import java.io.*;
import java.util.*;

public class CF879C extends PrintWriter {
	CF879C() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF879C o = new CF879C(); o.main(); o.flush();
	}

	void main() {
		int n = sc.nextInt();
		char[] op = new char[n];
		int[] aa = new int[n];
		for (int i = 0; i < n; i++) {
			op[i] = sc.next().charAt(0);
			aa[i] = sc.nextInt();
		}
		int o = 0, x = 0;
		for (int b = 1; b < 1024; b <<= 1) {
			boolean fixed = false;
			int a = 0;
			for (int i = 0; i < n; i++) {
				boolean one = (aa[i] & b) != 0;
				if (op[i] == '&') {
					if (!one) {
						fixed = true;
						a = 0;
					}
				} else if (op[i] == '|') {
					if (one) {
						fixed = true;
						a = 1;
					}
				} else {
					if (one)
						a ^= 1;
				}
			}
			if (fixed) {
				o |= b;
				if (a == 0)
					x |= b;
			} else {
				if (a == 1)
					x |= b;
			}
		}
		println(2);
		println("| " + o);
		println("^ " + x);
	}
}