CF 483B Friends and Presents(二分)

B. Friends and Presents

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You have two friends. You want to present each of them several positive integers. You want to present cnt1 numbers to the first friend andcnt2 numbers to the second friend. Moreover, you want all presented numbers to be distinct, that also means that no number should be presented to both friends.

In addition, the first friend does not like the numbers that are divisible without remainder by prime number x. The second one does not like the numbers that are divisible without remainder by prime number y. Of course, you're not going to present your friends numbers they don't like.

Your task is to find such minimum number v, that you can form presents using numbers from a set 1, 2, ..., v. Of course you may choose not to present some numbers at all.

A positive integer number greater than 1 is called prime if it has no positive divisors other than 1 and itself.

Input

The only line contains four positive integers cnt1, cnt2, x, y (1 ≤ cnt1, cnt2 < 109; cnt1 + cnt2 ≤ 109; 2 ≤ x < y ≤ 3·104) — the numbers that are described in the statement. It is guaranteed that numbers x, y are prime.

Output

Print a single integer — the answer to the problem.

Examples

input

Copy

3 1 2 3

output

Copy

5

input

Copy

1 3 2 3

output

Copy

4

Note

In the first sample you give the set of numbers {1, 3, 5} to the first friend and the set of numbers {2} to the second friend. Note that if you give set {1, 3, 5} to the first friend, then we cannot give any of the numbers 1, 3, 5 to the second friend.

In the second sample you give the set of numbers {3} to the first friend, and the set of numbers {1, 2, 4} to the second friend. Thus, the answer to the problem is 4.

题意：
从1-n中找到至少从cnt1个数使得它不整除素数x，找到至少cnt2个数使得它不整除素数y.。求n的最小值

思路：

可以二分答案。注意同一个数不能加在cnt1和cnt2中，所以要判断这些数加起来够不够cnt1+cnt2。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
int main()
{
    ll t1,t2,x,y;
    while(cin>>t1>>t2>>x>>y)
    {
        ll l=1,r=10000000000,mid;
        while(l<r)
        {
            mid=(l+r)/2;
            if(mid-mid/(x*y)>=t1+t2&&mid-mid/x>=t1&&mid-mid/y>=t2)
                r=mid;
            else
                l=mid+1;
        }
        printf("%lld\n",l);
    }
    return 0;
}