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
rr = lambda: stdin.readline().strip()
rri = lambda: int(rr())
rrm = lambda: map(int, rr().split())
def rry(N = None, f = rri):
    for i in xrange(N or rri()):
        yield f()

debug=0
if debug:
    fi = open('t.txt','r')
    rr = lambda: fi.readline().replace('\n','')

import operator
N = rri()
commands = []
for _ in xrange(N):
    a, b = rr().split()
    if a == '|': c = operator.or_
    elif a == '^': c = operator.xor
    elif a == '&': c = operator.and_
    commands.append((c, int(b)))

global_and = 2**10 - 1
global_or = 0
global_xor = 0

for bit in xrange(10):
    r0, r1 = 0, 1
    for cmd, x in commands:
        b = (x >> bit) & 1
        r0 = cmd(r0, b)
        r1 = cmd(r1, b)
    if (r0, r1) == (0, 1):
        continue
    elif (r0, r1) == (1, 1):
        #or 1
        global_or |= 1 << bit
    elif (r0, r1) == (0, 0):
        global_and ^= 1<<bit
        assert ((global_and >> bit) & 1) == 0
    elif (r0, r1) == (1, 0):
        global_xor ^= 1<<bit
        assert ((global_xor >> bit) & 1) == 1

print 3
print '|', global_or
print '^', global_xor
print '&', global_and
