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 PYTHON):

Solution:
n = int(raw_input())

one = 1023
zero = 0
for i in range(n):
	c, v = raw_input().split(' ')
	v = int(v)

	if c == '|':
		zero |= v
		one |= v
	elif c == '^':
		zero ^= v
		one ^= v
	elif c == '&':
		zero &= v
		one &= v

to_xor = 0
to_or = 0
to_and = 1023

i = 1
while i < 1024:
	if (zero & i) == 0 and (one & i) == 0:
		to_and ^= i
	if (zero & i) == i and (one & i) == 0:
		to_xor |= i
	if (zero & i) == i and (one & i) == i:
		to_or |= i

	i *= 2

print 3
print '|', to_or
print '^', to_xor
print '&', to_and