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
n=int(input())
l=[]
for i in range(n):
    l.append(input().split())
x=89
l1=[[0,1] for i in range(10)]
for i in l:
    a=bin(int(i[1])).lstrip('0b')
    if len(a)!=10:
        a='0'*(10-len(a))+a
    if i[0]=='|':
        for j in range(10):
            l1[j][0]=l1[j][0]|int(a[j])
            l1[j][1]=l1[j][1]|int(a[j])
    elif i[0]=='&':
        for j in range(10):
            l1[j][0]=l1[j][0]&int(a[j])
            l1[j][1]=l1[j][1]&int(a[j])
    else:
        for j in range(10):
            l1[j][0]=l1[j][0]^int(a[j])
            l1[j][1]=l1[j][1]^int(a[j])
a2=1023
o=0
x=0
for i in range(10):
    if l1[i][0]==1 and l1[i][1]==1:
        o+=2**(9-i)
    elif l1[i][0]==0 and l1[i][1]==0:
        a2-=2**(9-i)
    elif l1[i][0]==1 and l1[i][1]==0:
        x+=2**(9-i)
print(3)
print('&',a2)
print('|',o)
print('^',x)

        
        
        
