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.io.*;
import java.util.*;

public class tank {

    static final FastScanner fs = new FastScanner();

    //static PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) {
        int t = 1;
        while(t-- > 0) {
            run_case();
        }

        //out.close();
    }

    static void run_case() {
        int n = fs.nextInt(), x;

        String s;

        char[] num;

        int one = 1023, two = 0;

        for (int i = 0; i < n; i++) {
            s = fs.next();

            x = fs.nextInt();

            if(s.charAt(0) == '|') {
                one |= x;
                two |= x;
            }else if(s.charAt(0) == '&') {
                one &= x;
                two &= x;
            }else {
                one ^= x;
                two ^= x;
            }
        }

        int ob, tb;

        int[] ans = new int[3];

        ans[0] = 1023;

        for (int i = 0; i < 10; i++) {
            ob = (1 << i) & one;
            tb = (1 << i) & two;

            if(ob == 0 && tb == 0) {
                ans[0] ^= (1 << i);
            }else if(ob == 0 && tb == (1 << i)) {
                ans[1] |= (1 << i);
            }else if(ob == (1 << i) && tb == (1 << i)) {
                ans[2] |= (1 << i);
            }
        }

        System.out.println(3);
        System.out.println("& " + ans[0]);
        System.out.println("^ " + ans[1]);
        System.out.println("| " + ans[2]);
    }

    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        String next() {
            while (!st.hasMoreTokens())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        String nextLine(){
            try {
                return br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }

            return "";
        }

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

        int[] readArray(int n) {
            int[] a=new int[n];
            for (int i=0; i<n; i++) a[i]=nextInt();
            return a;
        }

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

        double nextDouble() {
            return Double.parseDouble(next());
        }
    }
}