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.BufferedReader;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class P879C
{
	public static void main(String[] args)
	{
		FastScanner scan = new FastScanner();
		int n = scan.nextInt();
		char[] op = new char[n];
		int[] x = new int[n];
		for (int i = 0; i < n; i++)
		{
			op[i] = scan.next().charAt(0);
			x[i] = scan.nextInt();
		}
		int off = 0;
		int on = 1023;
		for (int i = 0; i < n; i++)
		{
			char c = op[i];
			int v = x[i];
			if (c == '&')
			{
				off &= v;
				on &= v;
			}
			else if (c == '|')
			{
				off |= v;
				on |= v;
			}
			else
			{
				off ^= v;
				on ^= v;
			}
		}
		int and = 1023, or = 0, xor = 0;
		for (int i = 0; i < 10; i++)
		{
			boolean offEnd = (off & (1 << i)) != 0;
			boolean onEnd = (on & (1 << i)) != 0;
			if (!offEnd && !onEnd)
				and -= (1 << i);
			else if (offEnd && onEnd)
				or |= (1 << i);
			else if (offEnd && !onEnd)
				xor |= (1 << i);
		}
		System.out.println(3);
		System.out.println("& " + and);
		System.out.println("| " + or);
		System.out.println("^ " + xor);
	}
	
	static class FastScanner
	{
		BufferedReader br;
		StringTokenizer st;

		public FastScanner()
		{
			try
			{
				br = new BufferedReader(new InputStreamReader(System.in));
				st = new StringTokenizer(br.readLine());
			} catch (Exception e)
			{
				e.printStackTrace();
			}
		}

		public String next()
		{
			if (st.hasMoreTokens())
				return st.nextToken();
			try
			{
				st = new StringTokenizer(br.readLine());
			} catch (Exception e)
			{
				e.printStackTrace();
			}
			return st.nextToken();
		}

		public int nextInt()
		{
			return Integer.parseInt(next());
		}

		public long nextLong()
		{
			return Long.parseLong(next());
		}

		public String nextLine()
		{
			String line = "";
			try
			{
				line = br.readLine();
			} catch (Exception e)
			{
				e.printStackTrace();
			}
			return line;
		}
	}
}