codeforces 878A

A. Short Program

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 x_i 0 ≤ x_i ≤ 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.

（这是一道好题啊

题意：给一定的步骤，求一个满足同样功能的步骤（步骤数小于5，不需要从原步骤里选

解题思路：先取两个典型的例子0和1023，按照input的步骤进行下去，分别得到答案a和b，然后按位分析如何由0和1023到a和b

按位分析：

0->0  1->0   先&0再^0

0->0  1->1   先&1再^0

0->1  1->0   先&1再^1

0->1  1->1   先&0再^1

对每一位进行上述的四种情况分析就行啦。

ac代码：

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 using namespace std;
 5 int x=0,y=1023;
 6 int fun(int u,int i) {
 7     if(u&(1<<i)) return 1;
 8     return 0;
 9 }
10 int main() {
11     ios::sync_with_stdio(false);
12     cin.tie(0);cout.tie(0);
13     int n,a;
14     cin>>n;
15     string s;
16     for(int i=0;i<n;++i) {
17         cin>>s>>a;
18         if(s[0]=='|') {x|=a; y|=a;}
19         if(s[0]=='^') {x^=a; y^=a;}
20         if(s[0]=='&') {x&=a; y&=a;}
21
22     }
23     int i=0,v1=0,v2=0;
24     while(i<=10) {
25         if(!fun(x,i) && fun(y,i)) v1+=(1<<i);
26         if(fun(x,i) && fun(y,i)) v2+=(1<<i);
27         if(fun(x,i) && !fun(y,i)) {
28             v1+=(1<<i);
29             v2+=(1<<i);
30         }
31         ++i;
32     }
33     cout<<2<<endl<<"& "<<v1<<endl<<"^ "<<v2<<endl;
34     return 0;
35 }