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.
from sys import stdin
input = stdin.readline
range = xrange
n = int(input())
a = 0
b = 1023
for i in range(n):
    op, d = input().split(); d = int(d)
    if op == "^": a, b = a^d, b^d
    elif op == "&": a, b = a&d, b&d
    else: a, b = a|d, b|d
qxor = 0
qand = 0
qor = 0
d = 512
for i in range(10):
    qxor*=2; qand*=2; qor*=2
    ai = a//d; bi = b//d
    if ai and bi: qor+= 1
    elif ai and not bi: qxor+= 1; qand+= 1
    elif bi and not ai: qand+= 1
    a-= d*ai; b-= d*bi
    d//= 2
print 3
print "^", qxor
print "&", qand
print "|", qor