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.

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

Print f(x, y).

 1 /** 2  * Codeforces 3  * Problem#837E 4  * Accepted 5  * Time: 15ms 6  * Memory: 2048k 7  */ 8 #include <bits/stdc++.h> 9 using namespace std;10 #ifndef WIN3211 #define Auto "%lld"12 #else13 #define Auto "%I64d"14 #endif15 typedef bool boolean;16 #define smax(a, b) a = max(a, b)17 template<typename T>18 inline boolean readInteger(T& u){19     char x;20     int aFlag = 1;21     while(!isdigit((x = getchar())) && x != ‘-‘ && x != -1);22     if(x == -1) {23         ungetc(x, stdin);    24         return false;25     }26     if(x == ‘-‘){27         x = getchar();28         aFlag = -1;29     }30     for(u = x - ‘0‘; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - ‘0‘);31     ungetc(x, stdin);32     u *= aFlag;33     return true;34 }35 36 #define LL long long37 38 template<typename T>39 T gcd(T a, T b) {40     return (b == 0) ? (a) : (gcd(b, a % b));41 }42 43 LL a, b;44 45 inline void init() {46     readInteger(a);47     readInteger(b);48     LL g = gcd(a, b);49     a /= g, b /= g;50 }51 52 vector<LL> fac;53 void getFactor(LL x) {54     fac.clear();55     for(LL i = 2; i * i <= x; i++) {56         if((x % i) == 0) {57             while((x % i) == 0)    x /= i;58             fac.push_back(i);59         }60     }61     if(x > 1)    fac.push_back(x);    62 }63 64 LL f(LL a, LL b) {65     if(b <= 1)    return b;66     getFactor(a);67     LL near = 0, g;68     for(int i = 0; i < (signed)fac.size(); i++)69         smax(near, b / fac[i] * fac[i]);70     g = gcd(a, near);71     return b - near + f(a / g, near / g);72 }73 74 inline void solve() {75     printf(Auto, f(a, b));76 }77 78 int main() {79     init();80     solve();81     return 0;82 }