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(input())
A = 1023
B = 0
a = 1023
b = 0
print(4)
for i in range(n):
    s = str(input())
    if s[0] == '&':
        a = a & int(s[2:])
        b = b & int(s[2:])
    elif s[0] == '|':
        a = a | int(s[2:])
        b = b | int(s[2:])
    elif s[0] == '^':
        a = a ^ int(s[2:])
        b = b ^ int(s[2:])
Al = [1,1,1,1,1,1,1,1,1,1]
Bl = [0,0,0,0,0,0,0,0,0,0]
al = [0 for kk in range(10)]
bl = [0 for kk in range(10)]
#print(a,b)
for i in range(9,-1,-1):
    if a >= 2 ** i:
        al[i] = 1
        a -= 2**i
for i in range(9,-1,-1):
    if b >= 2 ** i:
        bl[i] = 1
        b -= 2**i
#print(al,bl)
And = []
Or = []
Xor = []
for i in range(10):
    if al[i] == 0 and bl[i] == 0:
        And.append(i)
    if al[i] == 0 and bl[i] == 1:
        Xor.append(i)
    if al[i] == 1 and bl[i] == 1:
        Or.append(i)
#print(And, Or, Xor)
###
r = 0
for i in range(10):
    if i not in And:
        r += 2**i
print("&", r)
###
r = 0
if Xor == []:
    print("^ 0")
    print("^ 0")
elif Xor == [0]:
    print("^ 0")
    print("& 1")
#elif len(Xor) == 1:
    #r = 2**(max(Xor)+1) - 1
    #print("^", r)
    #print("| 1")
else:
    r = 2**(max(Xor)+1) - 1
    print("^", r)
    r = 0
    for i in range(max(Xor) + 1):
        if i not in Xor:
            r += 2**i
    print("^", r)
###
r = 0
for i in range(10):
    if i in Or:
        r += 2**i
print("|", r)
#print((1023 | 999)^689)
#print((0 | 999)^689)
#print((300 | 999)^689)
#print((((300 & 350)^31)^15)|326)