Mad scientist Mike is building a time machine in his spare time. To finish the work, he needs a resistor with a certain resistance value.

However, all Mike has is lots of identical resistors with unit resistance R₀ = 1. Elements with other resistance can be constructed from these resistors. In this problem, we will consider the following as elements:

1. one resistor;
2. an element and one resistor plugged in sequence;
3. an element and one resistor plugged in parallel.

With the consecutive connection the resistance of the new element equals R = R_e + R₀. With the parallel connection the resistance of the new element equals . In this case R_e equals the resistance of the element being connected.

Mike needs to assemble an element with a resistance equal to the fraction . Determine the smallest possible number of resistors he needs to make such an element.

1 1

1

3 2

3

199 200

200

 /*************************************************************************

     > File Name: cf.cpp

     > Author:

     > Mail:

     > Created Time: 2016年07月10日 星期日 17时04分04秒

  ************************************************************************/

 #include<iostream>

 #include<bits/stdc++.h>

 using namespace std;

 typedef long long ll;

 ll solve(ll a,ll b,ll &ans)

 {

     if(a < b)

     {

         swap(a,b);

     }

     while(b)

     {

         ans += a/b;

         a = a%b;

         swap(a,b);

     }

     return ans;

 }

 int main()

 {

     ll a,b;

     cin >> a >> b;

     ll ans= ;

     ans = solve(a,b,ans);

     cout << ans << endl;

     return ;

 }