Codeforces Round #485 (Div. 2)-B-High School: Become Human

B. High School: Become Human

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Year 2118. Androids are in mass production for decades now, and they do all the work for humans. But androids have to go to school to be able to solve creative tasks. Just like humans before.

It turns out that high school struggles are not gone. If someone is not like others, he is bullied. Vasya-8800 is an economy-class android which is produced by a little-known company. His design is not perfect, his characteristics also could be better. So he is bullied by other androids.

One of the popular pranks on Vasya is to force him to compare xyxy with yxyx. Other androids can do it in milliseconds while Vasya's memory is too small to store such big numbers.

Please help Vasya! Write a fast program to compare xyxy with yxyx for Vasya, maybe then other androids will respect him.

Input

On the only line of input there are two integers xx and yy (1≤x,y≤1091≤x,y≤109).

Output

If xy<yxxy<yx, then print '<' (without quotes). If xy>yxxy>yx, then print '>' (without quotes). If xy=yxxy=yx, then print '=' (without quotes).

Examples

input

Copy

5 8

output

Copy

>

input

Copy

10 3

output

Copy

<

input

Copy

6 6

output

Copy

=

Note

In the first example 58=5⋅5⋅5⋅5⋅5⋅5⋅5⋅5=39062558=5⋅5⋅5⋅5⋅5⋅5⋅5⋅5=390625, and 85=8⋅8⋅8⋅8⋅8=3276885=8⋅8⋅8⋅8⋅8=32768. So you should print '>'.

In the second example 103=1000<310=59049103=1000<310=59049.

In the third example 66=46656=6666=46656=66.

不知道为什么。。。取对数就莫名wa。。。比较x*logy和y*logx就是错的。。。而比较x/y和logx/logy就是对的。。。。

没考虑到一个问题就是x^y=y^x有可能x!=y....所以应该先判读大于和小于最后否则就是等于，这样就避免了考虑精度问题由于本题是1e9，因此x/y-logx/logy=exp()<=1e-12

一般精度是要高于范围三位

 #include<iostream>
#include<string.h>
#include<stdio.h>
#include<math.h>
using namespace std;
int main(){
    int x,y;
    while (~scanf("%d%d",&x,&y)){
         double a=log(x)/log(y);
         double b=x*1.0/y;
         if(a>b){
             printf(">\n");
         }else if(a<b){
             printf("<\n");
         }
         else printf("=\n");
    }
    return ;
}