#include<bits/stdc++.h>
using namespace std;
long long a,b,ans;
int main(){
	cin>>a>>b;
	long long k=b/__gcd(a,b);
	cout<<k<<'\n';
	for(int i=a;i<=b;i+=a){
		long long aa=i,bb=k/(i/a)*a;
		if(__gcd(aa,bb)==a&&aa/__gcd(aa,bb)*(bb/__gcd(aa,bb))*__gcd(aa,bb)==b){
			cout<<aa<<' '<<bb<<'\n';
			ans++;
		}
	}
	cout<<ans;
	return 0;
}
//O(b-a)

个人认为是没问题的，求hack或调或解释