Short Program CodeForces

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.

Example
Input
3
| 3
^ 2
| 1
Output
2
| 3
^ 2
Input
3
& 1
& 3
& 5
Output
1
& 1
Input
3
^ 1
^ 2
^ 3
Output
0
这题输出的时候忘了加空格啊，，，

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<queue>#include<algorithm>#define INF 0x3f3f3f3fusing namespace std;bool check(int x,int i){    return x&(1<<i);}int main(){    int n;    int x1=0;    int x2=1023;    char op[5];    int num;    scanf("%d",&n);    for(int i=0;i<n;i++)    {        scanf("%s%d",op,&num);        if(op[0]=='|')            x1|=num,x2|=num;        else            if(op[0]=='^')                x1^=num,x2^=num;            else                x1&=num,x2&=num;    }    int v1=0,v2=0;    for(int i=0;i<10;i++)    {        if(check(x1,i)&&check(x2,i))            v2+=(1<<i);        else            if(check(x1,i)&&!check(x2,i))                v1+=(1<<i),v2+=(1<<i);            else                if(!check(x1,i)&&check(x2,i))                    v1+=(1<<i);    }    printf("2\n");    cout<<"& "<<v1<<endl;    cout<<"^ "<<v2<<endl;    return 0;}