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.
The first line contains an integer *n* (1 ≤ *n* ≤ 5·10<sup class="upper-index">5</sup>) — the number of lines.

Next *n* lines contain commands. A command consists of a character that represents the operation ("&amp;", "|" or "^" for AND, OR or XOR respectively), and the constant *x*<sub class="lower-index">*i*</sub> 0 ≤ *x*<sub class="lower-index">*i*</sub> ≤ 1023.
You can read about bitwise operations in [https://en.wikipedia.org/wiki/Bitwise_operation](https://en.wikipedia.org/wiki/Bitwise_operation).

Second sample:

Let *x* be an input of the Petya's program. It's output is ((*x*&amp;1)&amp;3)&amp;5 = *x*&amp;(1&amp;3&amp;5) = *x*&amp;1. So these two programs always give the same outputs.
3
| 3
^ 2
| 1
2
| 3
^ 2
3
& 1
& 3
& 5
1
& 1
3
^ 1
^ 2
^ 3
0
2
| 999
^ 689
2
| 999
^ 689
3
& 242
^ 506
^ 522
2
| 781
^ 253
2
| 56
^ 875
2
| 56
^ 875
3
^ 125
^ 377
& 1019
2
| 4
^ 260
1
& 123
1
& 123
1
| 123
1
| 123
1
^ 123
1
^ 123
10
^ 218
& 150
| 935
& 61
| 588
& 897
| 411
| 584
^ 800
| 704
2
| 1023
^ 260
10
^ 160
& 1021
& 510
^ 470
& 1022
& 251
& 760
& 1016
| 772
| 515
2
| 775
^ 112
1
& 0
1
& 0
1
| 0
0
1
^ 0
0
1
& 1023
0
1
| 1023
1
| 1023
1
^ 1023
1
^ 1023
from bisect import *
from collections import *
import sys
import io, os
import math
import random
import operator
from heapq import *
gcd = math.gcd
sqrt = math.sqrt
maxint=10**21
def ceil(a, b):
    if(b==0):
        return maxint
    a = -a
    k = a // b
    k = -k
    return k
# arr=list(map(int, input().split()))
# n,m=map(int,input().split())

input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def strinp(testcases):
    k = 5
    if (testcases == -1 or testcases == 1):
        k = 1
    f = str(input())
    f = f[2:len(f) - k]
    return f
    
def main():
    n=int(input())
    o=[1]*10
    z=[0]*10
    for i in range(n):
        s=strinp(90).split()
        op=s[0]
        no=int(s[1])
        if(op=="|"):
            for i in range(10):
                rem=no%2
                no=no//2
                o[i]=(o[i]|rem)
                z[i]=(z[i]|rem)
        elif(op=="&"):
            for i in range(10):
                rem=no%2
                no=no//2
                o[i]=(o[i]&rem)
                z[i]=(z[i]&rem)
        else:
            for i in range(10):
                rem=no%2
                no=no//2
                o[i]=(o[i]^rem)
                z[i]=(z[i]^rem)
    a1=1023
    a2=0
    a3=0
    p=2^9
    for i in range(10):
        a=o[i]
        b=z[i]
        if(a==0 and b==0):
            a1-=p
        elif(a==1 and b==1):
            a2+=p
        elif(b==1 and a==0):
            a3+=p
        p=p//2
    print(3)
    print("&",a1)
    print("|",a2)
    print("^",a3)
main()