#include <cstdio>
using namespace std;
int gcd(int n1,int n2)
{
if(n1%n2==0) return n2;
return gcd(n2,n1%n2);
}
int a,b;
int main()
{
scanf("%d%d",&a,&b);
printf("%d %d",gcd(a,b),a*b/gcd(a,b));
}

def gcd(a,b):
if(b==0):
return a
else:
return gcd(b,a%b)

a=int(input())
b=int(input())
ans=gcd(a,b)
ans1=a/ans*b
print(ans,int(ans1))

a=c=int(input())
b=d=int(input())
r=1
while (a!=b):
if r!=0:
if a>b:

r=a%b
a=b
b=r
elif a<b:
r=b%a
b=a
a=r
（前一半代码

import math
a=int(input())
b=int(input())
c=1

def HLsearch(a,b):
if(a%b==0):
return b
return HLsearch(b,a%b)

d=HLsearch(a,b)
e=a*b/d
print(d,'%.0f'%e)
//还有交题解这样的操作...

#include <cstdio>
using namespace std;
int gcd(int n1,int n2)
{
if(n1%n2==0) return n2;
return gcd(n2,n1%n2);
}
int a,b;
int main()
{
scanf("%d%d",&a,&b);
printf("%d %d",gcd(a,b),a*b/gcd(a,b));
}

def gcd(a,b):
if(b==0):
return a
else:
return gcd(b,a%b)

a=int(input())
b=int(input())
ans=gcd(a,b)
ans1=a/ans*b
print(ans,int(ans1))