Nastya Studies Informatics CodeForces

B. Nastya Studies Informatics

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Today on Informatics class Nastya learned about GCD and LCM (see links below). Nastya is very intelligent, so she solved all the tasks momentarily and now suggests you to solve one of them as well.

We define a pair of integers (a, b) good, if GCD(a, b) = x and LCM(a, b) = y, where GCD(a, b) denotes the greatest common divisorof a and b, and LCM(a, b) denotes the least common multiple of a and b.

You are given two integers x and y. You are to find the number of good pairs of integers (a, b) such that l ≤ a, b ≤ r. Note that pairs (a, b) and (b, a) are considered different if a ≠ b.

Input

The only line contains four integers l, r, x, y (1 ≤ l ≤ r ≤ 109, 1 ≤ x ≤ y ≤ 109).

Output

In the only line print the only integer — the answer for the problem.

Examples

input

1 2 1 2

output

2

input

1 12 1 12

output

4

input

50 100 3 30

output

0

Note

In the first example there are two suitable good pairs of integers (a, b): (1, 2) and (2, 1).

In the second example there are four suitable good pairs of integers (a, b): (1, 12), (12, 1), (3, 4) and (4, 3).

In the third example there are good pairs of integers, for example, (3, 30), but none of them fits the condition l ≤ a, b ≤ r.

题意：在一个区间里找有序对数，使得最大公约数和最小公倍数为题目所给

题解：大数 爆搜会超时，先化个简

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
const int maxn=200005;

int gcd(int a,int b)
{
    if(b==0)
        return a;
    return gcd(b,a%b);
}
int main()
{
    int l,r,x,y,i,num,ans=0;
    cin>>l>>r>>x>>y;
    num=y/x;
    if(y%x!=0)
    {
        cout<<"0"<<endl;
        return 0;
        
    }//num=n*z
    for(i=1;i*i<=num;i++)
    {
        int j=num/i;
        if(num%i==0&&gcd(i,j)==1&&l<=i*x&&i*x<=r&&l<=j*x&&j*x<=r)
        {
            if(i==j)
                ans=ans+1;
            else
                ans=ans+2;
        }
    }     
    cout<<ans<<endl;
}