B. Nastya Studies Informatics

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 divisor of 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

Copy

1 2 1 2

Output

Copy

2

Input

Copy

1 12 1 12

Output

Copy

4

Input

Copy

50 100 3 30

Output

Copy

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.

解释  : 

Let's consider some suitable pair

. As

, we can present number

as

, and number

as

, where we know that

and

.Let's consider too that from the restriction from the problem

we surely know the restriction for

and

, that is

.Let's remember we know that

(because

is

,

is

). Then we can get

. Dividing by

:

.

.Now if

, answer equals 0.Else as

is surely less than

, we can just sort out all possible pairs

of divisors

, such that

, and then to check that

and

are in the getting above restrictions. Complexity of this solution is

.

题意：a,b为两个正整数，l<= a,b <=r。x=gcd（a，b），y=lcm（a，b）。gcd是a，b的最大公约数，lcm是a，b的最小公倍数。给出l，r，x，y，求有多少种可能的a，b。

gcd(a,b)= x ,lcm(a,b) = y ;

那么有x*y = a*b ; 

设a = x*c;

  b = x*d ;  并且 gcd(c,d) ==   1 ;

那么 a*b = x*y = x*c*x*d ;   

即 

= c*d ;  然后枚举 d ,满足gcd(c,d) = 1  , 还有

 . 

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std ;
typedef long long LL ;
LL GCD(LL a ,LL b )
{

    return b?GCD(b,a%b):a ;

}
LL LCM(LL a ,LL b)
{
    return a*(b/GCD(a,b));

}
int main()
{
   int l, r, x, y;
    cin >> l >> r >> x >> y;
    if (y % x != 0){
        cout << 0 << '\n';
        return 0;
    }
    int ans = 0;
    int n = y / x;
    for (int d = 1; d * d <= n; ++d) {
        if (n % d == 0) {
            int c = n / d;
            if (l <= c * x && c * x <= r && l <= d * x && d * x <= r && GCD(c, d) == 1) 
            {
                if (d * d == n) 
                ++ans;
                else ans += 2;
            }
        }
    }

    cout << ans << '\n';


    return 0 ;
}