C. Fadi and LCM

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该博客讨论了如何找到两个正整数a和b，使得它们的最大值max(a,b)尽可能小，同时它们的最小公倍数LCM(a,b)等于给定的整数X。给出了一些示例来解释这个问题，并要求找到满足条件的a和b的一对解。" 127244354,9640242,面向对象编程：复数类实现与测试,"['java', '面向对象', '算法']

链接：https://codeforces.com/contest/1285/problem/C

Today, Osama gave Fadi an integer XX, and Fadi was wondering about the minimum possible value of max(a,b)max(a,b) such that LCM(a,b)LCM(a,b)equals XX. Both aa and bb should be positive integers.

LCM(a,b)LCM(a,b) is the smallest positive integer that is divisible by both aa and bb. For example, LCM(6,8)=24LCM(6,8)=24, LCM(4,12)=12LCM(4,12)=12, LCM(2,3)=6LCM(2,3)=6.

Of course, Fadi immediately knew the answer. Can you be just like Fadi and find any such pair?

Input

The first and only line contains an integer XX (1≤X≤10121≤X≤1012).

Output

Print two positive integers, aa and bb, such that the value of max(a,b)max(a,b) is minimum possible and LCM(a,b)LCM(a,b) equals XX. If there are several possible such pairs, you can print any.

Examples

input

Copy

output

Copy

1 2

input

Copy

output

Copy

2 3

input

Copy

output

Copy

1 4

input

Copy

output

Copy

1 1

代码：

#include<bits/stdc++.h>
using namespace std;
long long n,k,t,mod=1e9+7,s,ans=0;
long long a[100001],b[100001];
int main()
{
	cin>>n;
	k=sqrt(n);
	if(n==1)
	{
		cout<<1<<" "<<1;
		return 0; 
	} 
	if(k*k==n)
	k--;
	for(;k>=1;k--)
	{
		if(n%k==0)
		{
			t=n/k;
			if(__gcd(t,k)==1)
			{
				cout<<k<<" "<<t;
				return 0;
			}
			
		}
	}
}