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 operator import and_, or_, xor

n = int(input())
ops = []
for _ in range(n):
    s = input().split()
    ops.append([s[0], int(s[1])])

#print(ops)
samples = [0, 1023]
ans = []

tab = {'^': xor, '&': and_, '|': or_}
for s in samples:
    for op in ops:
        s = tab[op[0]](s, op[1])
    ans.append(s)

#print(ans)
diz, con, xxor = '', '', ''
for b1, b2 in zip('{0:0>10b}'.format(ans[0]), '{0:0>10b}'.format(ans[1])):
    if b1 == '1' and b2 == '0':
        diz += '0' # |
        con += '1' # &
        xxor += '1' # ^
    elif b1 == '0' and b2 == '1':
        diz += '0'
        con += '1'
        xxor += '0'
    elif b1 == '1' and b2 == '1':
        diz += '1'
        con += '1'
        xxor += '0'
    elif b1 == '0' and b2 == '0':
        diz += '0'
        con += '0'
        xxor += '0'
print(3)
print('|', int(diz, 2))
print('&', int(con, 2))
print('^', int(xxor, 2))

