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 fs first
#define sc second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define ALL(A) A.begin(),A.end()
#define RALL(A) A.rbegin(),A.rend()
typedef long long LL;
typedef pair<LL,LL> P;
template<class T> inline bool chmax(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; }
const LL mod=1000000007;
const LL LINF=1LL<<62;
const int INF=1<<30;
int dx[]={1,0,-1,0,1,-1,1,-1};
int dy[]={0,1,0,-1,1,-1,-1,1};


int main(){
    int a,b;cin >> a >> b;
    cout << setprecision(30) << fixed << 1. * a * b / (a + b) << endl;
    return 0;
}