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.
a, b = 0, 1023
# commands = []
for i in range(int(input())):
    cmd = input()
    # commands.append(cmd)
    c, x = cmd.split()
    x = int(x)
    if c == "|":
        a, b = a | x, b | x
    elif c == "&":
        a, b = a & x, b & x
    else:
        a, b = a ^ x, b ^ x
    # a, b = eval("a " + cmd), eval("b " + cmd)
x = 0
y = 1023
z = 0
for i in range(10):
    a_i = (a >> i) & 1
    b_i = (b >> i) & 1
    if a_i and b_i:
        x |= 1 << i
    if (not a_i) and (not b_i):
        y ^= 1 << i
    if a_i and (not b_i):
        z ^= 1 << i
print("3\n| {}\n& {}\n^ {}".format(x, y, z))
# for i in range(1024):
#     t1 = i
#     for cmd in commands:
#         t1 = eval("t1 " + cmd)
#     t2 = eval("(((i | x) & y) ^ z)")
#     if t1 != t2:
#         print("Neq on ", i)
