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.
def n2b(n):
  b = bin(n)[2:]
  r = 10 - len(b)
  return list('0'*r + b)

def b2n(b):
  return int(''.join(b), 2)

def dump_encoded(s):
  ops = []
  op = '&'
  b = ''
  for i in range(10):
    if s[i] == '0':
      b += '0'
    else:
      b += '1'
  ops.append((op, b2n(b)))
  op = '|'
  b = ''
  for i in range(10):
    if s[i] == '1':
      b += '1'
    else:
      b += '0'
  ops.append((op, b2n(b)))
  op = '^'
  b = ''
  for i in range(10):
    if s[i] == 'y':
      b += '1'
    else:
      b += '0'
  ops.append((op, b2n(b)))

  print(len(ops))
  for op, n in ops:
    print(op,n)



# x, y, 0, 1

m = int(input())
ops = []
for i in range(m):
  op, n = input().split(' ')
  n = int(n)
  ops.append((op, n))


s = 1023
for i in range(len(ops)):
  op, n = ops[i]
  if op == '&':
    s = s & n
  elif op == '|':
    s = s | n
  elif op == '^':
    s = s ^ n
b1 = n2b(s)

s = 0
for i in range(len(ops)):
  op, n = ops[i]
  if op == '&':
    s = s & n
  elif op == '|':
    s = s | n
  elif op == '^':
    s = s ^ n
b2 = n2b(s)


g = ''
for k in range(10):
  if b1[k] == b2[k]:
    if b1[k] == '0':
      g += '0'
    else:
      g += '1'
  else:
    if b1[k] == '0':
      g += 'y'
    else:
      g += 'x'

dump_encoded(g)