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 = input()
num = ['x']*10

def norm(s):
    if len(s)<10:
        s = (10-len(s))*'0' + s
    return s

def conv(x):
    ans = 0
    for i in range(9,-1,-1):
        ans += (int(x[i])*(2**(9-i)))
    return ans

def op_and(num,s):
    for i in range(10):
        if s[i]=='0':
            num[i] = '0'
    return num

def op_or(num,s):
    for i in range(10):
        if s[i]=='1':
            num[i] = '1'
    return num

def op_xor(num,s):
    for i in range(10):
        if s[i]=='1':
            if num[i]=='1': num[i] = '0'
            elif num[i]=='0': num[i] = '1'
            elif num[i]=='x': num[i] = '-x'
            elif num[i]=='-x': num[i] = 'x'
        else:
            if num[i]=='1': num[i] = '1'
            elif num[i]=='0': num[i] = '0'
    return num

for x in range(n):
    a,b = raw_input().strip().split()
    b = int(b)
    b = bin(b)[2:]
    b = list(norm(b))
    if a=='^':
        num = op_xor(num,b)
    elif a=='|':
        num = op_or(num,b)
    elif a=='&':
        num = op_and(num,b)
    #print num

xor = ['0']*10
oor = ['0']*10
aand = ['1']*10
for i in range(10):
    if num[i]=='-x':
        xor[i]='1'
    elif num[i]=='1':
        oor[i] = '1'
    elif num[i]=='0':
        aand[i]='0'
#print num
print 3
print "^",conv(xor)
print '|',conv(oor)
print '&',conv(aand)