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.
n=int(raw_input())
#a=[int(k) for k in raw_input().split(" ")]

ts0=[0,1023]#range(1024)#
for k in range(n):
    ope,c=raw_input().split(" ")
    c=int(c)
    for jj in range(len(ts0)):
        if ope=="&":
            ts0[jj]&=c
        elif ope=="|":
            ts0[jj]|=c
        elif ope=="^":
            ts0[jj]^=c

ope1=1023
ope2=0
ope3=0

i=1
types=[0]*10
for j in range(10):
    if ts0[0]&i:
        types[j]+=2
    if ts0[-1]&i:
        types[j]+=1
    if types[j]==0:
        ope1-=i
    elif types[j]==2:
        ope2+=i
    elif types[j]==3:
        ope3+=i
    i*=2

print 3
print "&",ope1
print "^",ope2
print "|",ope3

##test : 23 min into je construis, 7 min to build test 
#ts1=range(1024) #must be initialized like ts0
#for k in range(len(ts1)):
#    ts1[k]=(((ts1[k]&ope1)^ope2)|ope3)
#    if ts1[k]!=ts0[k]:
#        print k,ts1[k],ts0[k]