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
'''
n = int(input())
a = []
for i in range(n):
    sign, x = map(str, input().split())
    a.append( (sign, int(x)) )
for X in range(1024):
    res = X
    for j in range(n):
        if a[j][0] == '|':
            res |= a[j][1]
        elif a[j][0] == '^':
            res ^= a[j][1]
        else:
            res &= a[j][1]
    print("X equal to " + str(X) + "  " + str(res))
        
'''

# consider all value in statement are in bunary representation
# X = X9  X8 ... X1 X0
# | & ^
# a = a9  a8 ... a1 a0
#----------------------
# 0 or 1 or X or X'
'''
0 -- > 0
1 -- > 1
x -- > 2
x' -- > 3
'''
from sys import stdin
input = stdin.readline
def check(x, b, c):
    if b == '|':
        if c == '1':
            return 1
        else:
            return x
    elif b == '&':
        if c == '1':
            return x
        else:
            return 0
    else:
        if c == '1':
            if x == 3: return 2
            if x == 2: return 3
            if x == 1: return 0
            if x == 0: return 1
        else:
            return x

n = int(input())
a = []
op = []
for i in range(n):
    sign, x = map(str, input().split())
    op.append(sign)
    a.append( bin(int(x))[2:].rjust(10, '0') )
#print(a)
And = 1023
Or = 0
Xor = 0
for i in range(10):
    result = 2
    for j in range(n):
        result = check(result , op[j], a[j][i] )
    #print(result)
    if result == 0: # bit i alwase 0  # and with 1023 - 2**(9-i)
        And-= ( 1 << (9 - i) )
    elif result == 1: # bit i alwase 1
        Or+= ( 1 << (9 - i) )
    elif result == 3: # bit i alwase x'
        Xor+= ( 1 << (9 - i) )
      
print(3)
print('&', And)
print('|', Or)
print('^', Xor)