Codeforces 837 E Vasya's Function

Discription

Vasya is studying number theory. He has denoted a function f(a, b) such that:

• f(a, 0) = 0;
• f(a, b) = 1 + f(a, b - gcd(a, b)), where gcd(a, b) is the greatest common divisor of a and b.

Vasya has two numbers x and y, and he wants to calculate f(x, y). He tried to do it by himself, but found out that calculating this function the way he wants to do that might take very long time. So he decided to ask you to implement a program that will calculate this function swiftly.

Input

The first line contains two integer numbers x and y (1 ≤ x, y ≤ 10¹²).

Output

Print f(x, y).

Example

Input

3 5

Output

3

Input

6 3

Output

1
 
水水数论，可以发现随着运算的过程，gcd单调不减，所以我们可以每次处理gcd相同的一段。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll num[21],N=0,mn;
ll a,b,d,ans=0,A,B;
 
ll gcd(ll x,ll y){
	return y?gcd(y,x%y):x;
}
 
inline void dvd(ll x){
	for(int i=2;i*(ll)i<=x;i++) if(!(x%i)){
		num[++N]=i;
		while(!(x%i)) x/=i;
		if(x==1) break;
	}
	if(x!=1) num[++N]=x;
}
 
inline void solve(){
	dvd(a);
 
	d=gcd(a,b);
	while(b){
		A=a/d,B=b/d,mn=1ll<<60;
		if(A==1){
			ans+=B;
			break;
		}
 
		for(int i=1;i<=N;i++) if(!(A%num[i])) mn=min(mn,B%num[i]);
		ans+=mn,b-=mn*d,d=gcd(b,a);
	}
 
	printf("%I64d\n",ans);
}
 
int main(){
	scanf("%I64d%I64d",&a,&b);
	solve();
	return 0;
}