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.
The first line contains an integer *n* (1 ≤ *n* ≤ 5·10<sup class="upper-index">5</sup>) — the number of lines.

Next *n* lines contain commands. A command consists of a character that represents the operation ("&amp;", "|" or "^" for AND, OR or XOR respectively), and the constant *x*<sub class="lower-index">*i*</sub> 0 ≤ *x*<sub class="lower-index">*i*</sub> ≤ 1023.
You can read about bitwise operations in [https://en.wikipedia.org/wiki/Bitwise_operation](https://en.wikipedia.org/wiki/Bitwise_operation).

Second sample:

Let *x* be an input of the Petya's program. It's output is ((*x*&amp;1)&amp;3)&amp;5 = *x*&amp;(1&amp;3&amp;5) = *x*&amp;1. So these two programs always give the same outputs.
3
| 3
^ 2
| 1
2
| 3
^ 2
3
& 1
& 3
& 5
1
& 1
3
^ 1
^ 2
^ 3
0
2
| 999
^ 689
2
| 999
^ 689
3
& 242
^ 506
^ 522
2
| 781
^ 253
2
| 56
^ 875
2
| 56
^ 875
3
^ 125
^ 377
& 1019
2
| 4
^ 260
1
& 123
1
& 123
1
| 123
1
| 123
1
^ 123
1
^ 123
10
^ 218
& 150
| 935
& 61
| 588
& 897
| 411
| 584
^ 800
| 704
2
| 1023
^ 260
10
^ 160
& 1021
& 510
^ 470
& 1022
& 251
& 760
& 1016
| 772
| 515
2
| 775
^ 112
1
& 0
1
& 0
1
| 0
0
1
^ 0
0
1
& 1023
0
1
| 1023
1
| 1023
1
^ 1023
1
^ 1023
BITCOUNT = 10
OPCOUNT = int(input())
x = [2]*BITCOUNT
for j in range(OPCOUNT):
    token = input().split()
    if len(token) < 2:
        break
    else:
        #Assumes user didn't screw up input
        op, num = token[0], int(token[1])
        num = ('{0:0' + str(BITCOUNT) + 'b}').format(num)
        num = [int(n) for n in num]
        if op == '&':
            for i in range(len(num)):
                if num[i] == 0:
                    x[i] = 0
        elif op == '|':
            for i in range(len(num)):
                if (num[i] == 1) and (x[i] != 2):
                        x[i] = 1
        elif op == '^':
            for i in range(len(num)):
                if x[i] in {0, 1}:
                    x[i] = x[i] ^ num[i]
                else:
                    if num[i] == 1:
                        x[i] *= -1
        else:
            print("what op is this?")
            break
        # print(x)

get_x_mask = 0
flip_mask = 0
pow2 = 1
for i in reversed(range(BITCOUNT)):
    if x[i] == 2 or x[i] == -2:
        get_x_mask += pow2
    if x[i] == -2 or x[i] == -1:
        flip_mask += pow2
    pow2 *= 2

print(2)
# print("| {}".format(('{0:0' + str(BITCOUNT) + 'b}').format(get_x_mask)))
# print("| {}".format(('{0:0' + str(BITCOUNT) + 'b}').format(flip_mask)))
print("| {}".format(get_x_mask))
print("| {}".format(flip_mask))