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 = []
for i in range(n):
    sign, x = map(str, input().split())
    a.append( (sign, int(x)) )
for X in range(1024):
    res = X
    for j in range(n):
        if a[j][0] == '|':
            res |= a[j][1]
        elif a[j][0] == '^':
            res ^= a[j][1]
        else:
            res &= a[j][1]
    print("X equal to " + str(X) + "  " + str(res))
        
'''

# consider all value in statement are in bunary representation
# X = X9  X8 ... X1 X0
# | & ^
# a = a9  a8 ... a1 a0
#----------------------
# 0 or 1 or X or X'
'''
0 -- > 0
1 -- > 1
x -- > 2
x' -- > 3
'''
from sys import stdin
input = stdin.readline
def check(x, b, c):
    if b == '|':
        if c == '1':
            return 1
        else:
            return x
    elif b == '&':
        if c == '1':
            return x
        else:
            return 0
    else:
        if c == '1':
            if x == 3: return 2
            if x == 2: return 3
            if x == 1: return 0
            if x == 0: return 1
        else:
            return x

n = int(input())
a = []
op = []
for i in range(n):
    sign, x = map(str, input().split())
    op.append(sign)
    a.append( bin(int(x))[2:].rjust(10, '0') )
#print(a)
And = 1023
Or = 0
Xor = 0
for i in range(10):
    result = 2
    for j in range(n):
        result = check(result , op[j], a[j][i] )
    #print(result)
    if result == 0: # bit i alwase 0  # and with 1023 - 2**(9-i)
        And-= ( 1 << (9 - i) )
    elif result == 1: # bit i alwase 1
        Or+= ( 1 << (9 - i) )
    elif result == 3: # bit i alwase x'
        Xor+= ( 1 << (9 - i) )
      
print(3)
print('&', And)
print('|', Or)
print('^', Xor)