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: Wrong Answer)


//include
//------------------------------------------
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <fstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <string>
#include <cstring>
#include <ctime>
#include <climits>
#include <limits>

using namespace std;

//conversion
//------------------------------------------
inline int toInt(string s) { int v; istringstream sin(s); sin >> v; return v; }
template<class T> inline string toString(T x) { ostringstream sout; sout << x; return sout.str(); }

//math
//-------------------------------------------
template<class T> inline T sqr(T x) { return x * x; }

//typedef
//------------------------------------------
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef long long LL;

//container util
//------------------------------------------
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define MP make_pair
#define SZ(a) int((a).size())
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define EXISTch(s,c) ((((s).find_first_of(c)) != std::string::npos)? 1 : 0)//cがあれば1 if(1)
#define SORT(c) sort((c).begin(),(c).end())

#define REP(i,n) for(int i=0;i<(int)n;++i)
#define FOR(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();++i)

//constant
//--------------------------------------------
const double EPS = 1e-10;
const double PI = acos(-1.0);
const int INF = (int)1000000007;
const LL MOD = (LL)1000000007;//10^9+7
const LL INF2 = (LL)100000000000000000;//10^18

LL gcd(LL a, LL b) {
	return b != 0 ? gcd(b, a % b) : a;
}

LL lcm(LL a, LL b) {
	return a * b / gcd(a, b);
}



LL solve(LL n, LL c, LL d) {
	LL val = 0;
	val+=n / c + n / d;
	val -= lcm(c, d);

	return val;
}

int main() {
	LL l, r, c, d; cin >> l >> r >> c >> d;

	LL ans = solve(r, c, d) - solve(l-1, c, d);
	cout << ans << endl;

	return 0;
}