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.
import io, os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# input = io.StringIO(os.read(0, os.fstat(0).st_size).decode()).readline
ii=lambda:int(input())
kk=lambda:map(int, input().split())
ll=lambda:list(kk())


# xor + 0 --> unchanged(2)
# xor + 1 --> flip(3)
# or + 0 --> unchanged(2)
# or + 1 --> 1(1)
# and + 0 --> 0(0)
# and + 1 --> unchanged(2)
arr = [2]*10
for _ in range(ii()):
	s,v=input().split()
	v = int(v)
	if s == b"|":
		for i in range(10):
			if v&(2**i):
				arr[i]= 1
	elif s == b"^":
		for i in range(10):
			if v&(2**i):
				arr[i]^=1
	elif s == b"&":
		for i in range(10):
			if not (v&(2**i)):
				arr[i]=0
	else:
	    print(s)
print(3)
x=o=a= 0
for i in range(10):
	if arr[i]==3: x += 2**i
	elif arr[i]==0: a+=2**i
	elif arr[i]==1:o+=2**i
print('|',o)
print('^',x)
print('&',2**10 - a - 1)