Problem:
Score : 400 points

Problem Statement

Given are positive integers A and B.

Let us choose some number of positive common divisors of A and B.

Here, any two of the chosen divisors must be coprime.

At most, how many divisors can we choose?

Definition of common divisor

An integer d is said to be a common divisor of integers x and y when d divides both x and y.

Definition of being coprime

Integers x and y are said to be coprime when x and y have no positive common divisors other than 1.

Definition of dividing

An integer x is said to divide another integer y when there exists an integer \alpha such that y = \alpha x.

Constraints

All values in input are integers.

1 \leq A, B \leq 10^{12}

Input

Input is given from Standard Input in the following format:

A B

Output

Print the maximum number of divisors that can be chosen to satisfy the condition.

Sample Input 1

12 18

Sample Output 1

3

12 and 18 have the following positive common divisors: 1, 2, 3, and 6.

1 and 2 are coprime, 2 and 3 are coprime, and 3 and 1 are coprime, so we can choose 1, 2, and 3, which achieve the maximum result.

Sample Input 2

420 660

Sample Output 2

4

Sample Input 3

1 2019

Sample Output 3

1

1 and 2019 have no positive common divisors other than 1.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

class Main{
	public static void main(String[] $){
		System.err.println();
		Scanner s=new Scanner(System.in);
		PrimeIterator pi=new PrimeIterator(10_000_000);
		long a=s.nextLong(),b=s.nextLong(),r=1;
		for(long i:pi) {
			if(a%i==0&&b%i==0)
				++r;
			if(i*i>a||i*i>b)
				break;
		}
		System.out.println(r);
	}
}

class PrimeIterator implements Iterable<Integer>,PrimitiveIterator.OfInt{
	BitSet		bitset;
	private int	currentvalue,limit;

	public PrimeIterator(int limit){
		currentvalue=0;
		bitset=new BitSet(limit+1);
		this.limit=limit;
		construct();
	}
	/** Atkin's sieve */
	private final void construct(){
		int sqLim=(int)Math.sqrt(limit+0.5);
		{
			int n;
			for(int z=1;z<=5;z+=4){
				for(int y=z;y<=sqLim;y+=6){
					int ysq=y*y;
					for(int x=1;x<=sqLim&&(n=4*x*x+ysq)<=limit;++x)
						bitset.flip(n);
					for(int x=y+1;x<=sqLim&&(n=3*x*x-ysq)<=limit;x+=2)
						bitset.flip(n);
				}
			}
			for(int z=2;z<=4;z+=2){
				for(int y=z;y<=sqLim;y+=6){
					int ysq=y*y;
					for(int x=1;x<=sqLim&&(n=3*x*x+ysq)<=limit;x+=2)
						bitset.flip(n);
					for(int x=y+1;x<=sqLim&&(n=3*x*x-ysq)<=limit;x+=2)
						bitset.flip(n);
				}
			}
			for(int y=3;y<=sqLim;y+=6){
				int ysq=y*y;
				for(int z=1;z<=2;++z)
					for(int x=z;x<=sqLim&&(n=4*x*x+ysq)<=limit;x+=3)
						bitset.flip(n);
			}
		}
		{
			//remove square
			for(int n=5;n<=sqLim&&n!=-1;n=bitset.nextSetBit(n+1)){
				int nsq=n*n;
				for(int k=nsq;k<=limit;k+=nsq)
					bitset.clear(k);
			}
		}
		bitset.set(2);
		bitset.set(3);
	}

	@Override
	public boolean hasNext(){
		int r=getNextValue();
		return 0<=r&&r<=limit;
	}
	private int getNextValue(){
		return bitset.nextSetBit(currentvalue+1);
	}
	@Override
	public int nextInt(){
		int r=getNextValue();
		if(r<0||limit<r) throw new NoSuchElementException();
		return currentvalue=r;
	}
	@Override
	public OfInt iterator(){
		return this;
	}
	public BitSet getClonedBitSet(){
		return (BitSet)bitset.clone();
	}
}