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.
#!/usr/bin/env python
#-*- coding: utf-8 -*-

from collections import defaultdict
from math import factorial as f
from fractions import gcd as g
import string
import sys

l = [2] * 10

N = int (raw_input ())
for i in xrange (N):
    # a, b = [i for i in raw_input ().split ()]
    t = sys.stdin.readline()
    a = t [0]
    b = int (t [1:])
    # b = ("0000000000" + bin (int (b)) [2:]) [-10:]
    if a == "|":
        for j in range (10):
            if ((b >> j) & 1):
                l [10 - j - 1] = 1
    if a == "&":
        for j in range (10):
            if not ((b >> j) & 1):
                l [10 - j - 1] = 0
    if a == "^":
        for j in range (10):
            if ((b >> j) & 1):
                l [10 - j - 1] ^= 1

ret = 0
a, b, c = 1023, 0, 0
for i in range (10):
    x = 10 - i - 1
    if l [i] == 2: continue
    if l [i] == 0:
        a &= ~(1 << x);
    if l [i] == 1:
        b |= 1 << x
    if l [i] == 3:
        c ^= 1 << x;

if a != 1023: ret += 1
if b != 0: ret += 1
if c != 0: ret += 1
print ret
if a != 1023: print "&", a
if b != 0: print "|", b
if c != 0: print "^", c
