codeforces C—Short Program（数学）

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
& 1
& 3
& 5

Output

1
& 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.

 

题意：有一个未知的数，现在对这个数进行n次为运算，每次输入一个位运算符 和一个数 x，表示这个未知的数与x进行该位运算。但是，这些操作是繁琐的，你是否可以将这些操作简化到5步以内。即n可能很大，你是否可以找出一系列位运算，（小于5次）使得这一系列运算的结果和n次操作完全相同。

 

思路：首先，不管有多少次同类的操作，我们都可以简化为一次同类运算。然后，我们并不知道我们的操作的数是什么，但是，我们可以通过对0（二进制10位全是0） 和1023（二进值10位全是1）进行同类操作，然后逐位比较这两个数的差别，分别经过什么操作，然后将这些操作加起来就可以了，这样就可以用恒定的2种操作，或，亦或，代替n次操作。

 

推荐博客 https://blog.youkuaiyun.com/zhelong3205/article/details/78363630?locationNum=9&fps=1。