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 PYTHON3):

Solution:
import sys
input=sys.stdin.readline
n=int(input())
start1=0
start2=1023
for _ in range(n):
  a,b=map(str,input().split())
  b=int(b)
  if a=='|':
    start1=b|start1
    start2=b|start2
  elif a=='^':
    start1=b^start1
    start2=b^start2
  else:
    start1=b&start1
    start2=b&start2
ans=[]
ands=1023
xors=0
ors=0
for i in range(10):
  k=1<<i
  if start1&k and start2&k:
    ors+=k
  elif start1&k and not start2&k:
    xors+=k
  elif (not start1&k) and (not start2&k):
    ands-=k
print(3)
print('&',ands)
print('^',xors)
print('|',ors)