Codeforces Round #443 (Div. 1) A. Short Program

A. Short Program

link

http://codeforces.com/contest/878/problem/A

describe

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.

翻譯

如今有一個程序，輸入一個[0,1024)的數，而後通過一些位運算以後，輸出一個數。

如今讓你簡化中間的位運算的過程，使得不超過5步，使得答案和原程序同樣。

題解

考慮每一位，只會存在4種狀況：
對於輸入的數字的每一位而言，要麼是1，要麼是0。而這些每一位的數輸出以後要麼變成了0，要麼就變成了1.
(1)0->0,1->0
(2)0->1,1->0
(3)0->0,1->1
(4)0->1,1->1
對於四種狀況，咱們均可以經過^和|就能夠解決，分狀況討論輸出便可。

代碼

#include<bits/stdc++.h>
using namespace std;

int n,a,b,p;
string s;
int main(){
    cin>>n;
    a = 0,b = 1023;
    for(int i=0;i<n;i++){
        cin>>s>>p;
        if(s[0]=='|'){
            a|=p;
            b|=p;
        }else if(s[0]=='^'){
            a^=p;
            b^=p;
        }else{
            a&=p;
            b&=p;
        }
    }
    int ans1=0,ans2=0;
    for(int i=0;i<10;i++){
        int a1=a&(1<<i);
        int b1=b&(1<<i);
        if(a1&&b1){
            ans1|=(1<<i);
        }
        if(a1&&!b1){
            ans2|=(1<<i);
        }
        if(!a1&&!b1){
            ans1|=(1<<i);
            ans2|=(1<<i);
        }
    }
    cout<<"2"<<endl;
    cout<<"| "<<ans1<<endl;
    cout<<"^ "<<ans2<<endl;
}