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 <iostream>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <algorithm>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <cmath>
using namespace std;

using ll = long long;
using P = pair<int, int>;

const int INF = 1e9;
const int MOD = 1e9 + 7;

// 4近傍、8近傍
int dx[] = {1, 0, -1, 0, 1, -1, -1, 1};
int dy[] = {0, 1, 0, -1, 1, 1, -1, -1};

int main() {
  // 入力
  int r1, r2;
  cin >> r1 >> r2;

  // 解答
  printf("%.8f\n", (double)(r1 * r2) / (r1 + r2));

  return 0;
}