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;
typedef long long ll;
#define rep(i, n) for(ll i = 0, i##_len = (n); i < i##_len; i++)
#define reps(i, s, n) for(ll i = (s), i##_len = (n); i < i##_len; i++)
#define rrep(i, n) for(ll i = (n) - 1; i >= 0; i--)
#define rreps(i, e, n) for(ll i = (n) - 1; i >= (e); i--)
#define all(x) (x).begin(), (x).end()
#define sz(x) ((ll)(x).size())
#define len(x) ((ll)(x).length())

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    // ifstream in("input.txt");
    // cin.rdbuf(in.rdbuf());
    ll r1, r2;
    cin >> r1 >> r2;
    double a = 1.0 / r1;
    double b = 1.0 / r2;
    printf("%.10f\n", 1.0 / (a + b));
    return 0;
}