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.
n = int(input())

mask_ans1 = 1023
mask_ans2 = 0

for i in range(n):
	com, x = input().split()
	x = int(x)
	if com == '&':
		mask_ans1 &= x
		mask_ans2 &= x
	elif com == '|':
		mask_ans1 |= x
		mask_ans2 |= x
	elif com == '^':
		mask_ans1 ^= x
		mask_ans2 ^= x

ans11 = 1023 & ~mask_ans2
ans12 = mask_ans1
ans13 = 1023&(1023^mask_ans1)

ans1 = 1023^(1023 & ~mask_ans1) & ans11
ans2 = mask_ans2 & ans12
ans3 = 1023&(mask_ans2^0) & ans13

lst = []
cnt = 3

if ans1 == 1023:
	cnt -= 1
if ans2 == 0:
	cnt -= 1
if ans3 == 0:
	cnt -= 1

print(cnt)
if ans1 != 1023:
	print('&', ans1)
if ans2 != 0:
	print('|', ans2)
if ans3 != 0:
	print('^', ans3)
