CF987B

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 xy with yx. 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 yx for Vasya, maybe then other androids will respect him.

Input

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

Output

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

Examples

Input

5 8

Output

>

Input

10 3

Output

<

Input

6 6

Output

=

Note

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

In the second example 10 3=1000<3 10=59049.

In the third example 6 6=46656=6 6.

这道题，看题意首先想到快速幂，再看数据范围，显然会炸，那么最简单粗暴的方法就是，比较 x ^ y 与 y ^ x 的大小关系。（如此简洁明了的题面 >_<）

我们要先在两式旁取对数，就是比较 ln x ^ y 与 ln y ^ x 的大小关系，先假设左式小于右式：（前方高能，请注意）

ln x ^ y < ln y & x; 即 y * ln x < x * lny;

所以 ln x / x < ln y / y;

那么通过归纳我们可以设 f (n) = ln n / n;

取这个函数的导数，即 f'（n） = ( ln n - 1 )/ n ^ 2;

那么当 f'（n）> 0 时， ln n > 1, 所以当 x, y > e (因为是整数，相当于大于等于3)时， 若 x > y, 则 x ^ y < y ^ x;

证明完成之后，我们就可以知道，当给出的 x , y 大于 3 的时候，只需要判断 x 和 y 的大小关系即可，其他的只要特判就可以了

#include<bits/stdc++.h>
using namespace std;
int main()
{
	int x,y,i;
	cin>>x>>y;
	if(x == y){
		cout<<"="<<endl;
		return 0;
	}
	else{
		if(x == 1){
			cout<<"<";
			return 0;
		}
		if(y == 1){
			cout<<">";
			return 0;
		}
		int h = max(x,y);
		if(h <= 4){
			long long sum1 = 1,sum2 = 1;
			for(i = 1; i <= y; i++){
				sum1 *= x;
			}
			for(i = 1; i <= x; i++){
				sum2 *= y;
			}
			if(sum1 < sum2){
				cout<<"<";
			}
			else if(sum1 > sum2){
				cout<<">";
			}
			else{
				cout<<"=";
			}
		}
		else{
			if(x > y){
				cout<<"<";
			}
			else{
				cout<<">";
			}
		}
	}
	return 0;
}