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 PYTHON):

Solution:
N   = input()
n1  = 0
n2  = 1023
for i in range(N):
    op = raw_input().split(" ")
    if op[0] == "|":
        n1|=int(op[1])
        n2|=int(op[1])
    elif op[0] == "^":
        n1^=int(op[1])
        n2^=int(op[1])
    else:#
        n1&=int(op[1])
        n2&=int(op[1])
i = 1
xor = [0]*10
orr = [0]*10
adr = [1]*10


pos = 0
while i<=512:
    if n1&i  and not n2&i :
        xor[pos] = 1
    elif n1&i  and n2&i :
        orr[pos] = 1
    elif not n1&i  and not n2&i :
        adr[pos] = 0
    elif not n1&i  and n2&i :
        pass

    i = i<<1

    pos += 1
#print xor,orr,adr
def do(arr):
    ans = 0
    for i,v in enumerate(arr):
        ans += v<<i
    return ans
print 3
print "^",do(xor)
print "|",do(orr)
print "&",do(adr)