Problem:
Score : 300 points

Problem Statement

You are given four integers A, B, C, and D. Find the number of integers between A and B (inclusive) that can be evenly divided by neither C nor D.

Constraints

1\leq A\leq B\leq 10^{18}

1\leq C,D\leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B C D

Output

Print the number of integers between A and B (inclusive) that can be evenly divided by neither C nor D.

Sample Input 1

4 9 2 3

Sample Output 1

2

5 and 7 satisfy the condition.

Sample Input 2

10 40 6 8

Sample Output 2

23

Sample Input 3

314159265358979323 846264338327950288 419716939 937510582

Sample Output 3

532105071133627368

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <cmath>
#include <numeric>
#include <iostream>
#include <sstream>
#include <string>
#include <algorithm>
#include <vector>
#include <map>
#include <queue>
#include <stack>
#include <deque>
#include <list>
#include <unordered_map>
#include <tuple>
#include <complex>


#define FOR(i, a, b)		for(int i = (int)(a); i < (int)(b); ++i)
#define FORE(i, a, b)		for(int i = (int)(a); i <= (int)(b); ++i)
#define REP(i, n)			FOR(i, 0, n)
#define REPE(i, n)			FORE(i, 0, n)
#define ALL(x)				(x).begin(),(x).end()
#define DEBUG_DC(x)			cout << #x << ": " << x << endl
#define VEC_TYPE			int
#define INF					1e5+1

using namespace std;
typedef long long ll;
typedef vector<VEC_TYPE> vdef;
typedef vector<vdef> vvdef;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmaxE(T& a, T b) { if (a <= b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
template<class T> inline bool chminE(T& a, T b) { if (a >= b) { a = b; return 1; } return 0; }
template<typename T> inline T mygcd(T a, T b) { return b ? mygcd(b, a%b) : a; }
template<typename T> inline T mylcm(T a, T b) { return a * b / mygcd(a, b); }


int main()
{
	ll a, b, c, d;
	cin >> a >> b >> c >> d;
	ll ans = (ll)0;

	ll lcmcd = mylcm(c, d);
	ll cNUM = b / c - a / c;
	if (a%c == 0) ++cNUM;
	ll dNUM = b / d - a / d;
	if (a%d == 0) ++dNUM;
	ll lcmNUM = b / lcmcd - a / lcmcd;
	if (a%lcmcd == 0) ++lcmNUM;
	
	ll sum = b - a + (ll)1;
	ans=sum - cNUM - dNUM + lcmNUM;
	cout << ans << endl;

	return 0;
}