Problem: In an electric circuit, when two resistors R_1 and R_2 are connected in parallel, the equivalent resistance R_3 can be derived from the following formula:

* \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_3}



Given R_1 and R_2, find R_3.

Constraints

* 1 \leq R_1, R_2 \leq 100
* R_1 and R_2 are integers.

Input

The input is given from Standard Input in the following format:


R_1 R_2


Output

Print the value of R_3.

The output is considered correct if the absolute or relative error is at most 10^{-6}.

Examples

Input

2 3


Output

1.2000000000


Input

100 99


Output

49.7487437186

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

#define FOR(i,l,r) for(int i = (l);i < (r);i++)
#define ALL(x) (x).begin(),(x).end()
template<typename T> bool chmax(T& a,const T& b){return a < b ? (a = b,true) : false;}
template<typename T> bool chmin(T& a,const T& b){return b < a ? (a = b,true) : false;}
typedef long long ll;

double R1,R2;

double f(double m)
{
	return 1.0 / m;
}

int main()
{
	scanf("%lf%lf",&R1,&R2);

	double lo = 0.1,hi = 100.0,mid;
	FOR(i,0,100){
		mid = (lo + hi) / 2.0;
		if(f(mid) >= 1.0 / R1 + 1.0 / R2){
			lo = mid;
		}
		else{
			hi = mid;
		}
	}

	printf("%.10f\n",(lo + hi) / 2.0);

	return 0;
}