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:
from sys import stdin

n = int(stdin.readline())
z, o = 0, 1023
for _ in xrange(n):
	op, num = stdin.readline().split()
	num = int(num)
	if op == '&':
		z = z & num
		o = o & num
	elif op == '^':
		z = z ^ num
		o = o ^ num
	else:
		z = z | num
		o = o | num

n2, n3, n4 = 0, 0, 1023
for i in xrange(10):
	x = (1 << i)
	q = z & x
	w = o & x
	if q and (not w):
		n2 |= x
	elif q and w:
		n3 |= x
	elif (not q) and (not w):
		n4 &= ~x

print(3)
print("^ " + str(n2))
print("| " + str(n3))
print("& " + str(n4))