Problem: Alex enjoys performing magic tricks. He has a trick that requires a deck of n cards. He has m identical decks of n different cards each, which have been mixed together. When Alex wishes to perform the trick, he grabs n cards at random and performs the trick with those. The resulting deck looks like a normal deck, but may have duplicates of some cards.

The trick itself is performed as follows: first Alex allows you to choose a random card from the deck. You memorize the card and put it back in the deck. Then Alex shuffles the deck, and pulls out a card. If the card matches the one you memorized, the trick is successful.

You don't think Alex is a very good magician, and that he just pulls a card randomly from the deck. Determine the probability of the trick being successful if this is the case.

Input

First line of the input consists of two integers n and m (1 ≤ n, m ≤ 1000), separated by space — number of cards in each deck, and number of decks.

Output

On the only line of the output print one floating point number – probability of Alex successfully performing the trick. Relative or absolute error of your answer should not be higher than 10 - 6.

Examples

Input

2 2


Output

0.6666666666666666


Input

4 4


Output

0.4000000000000000


Input

1 2


Output

1.0000000000000000

Note

In the first sample, with probability <image> Alex will perform the trick with two cards with the same value from two different decks. In this case the trick is guaranteed to succeed.

With the remaining <image> probability he took two different cards, and the probability of pulling off the trick is <image>.

The resulting probability is <image>

Solution (in JAVA):

Solution:
import java.util.Scanner;

public class C {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int m = in.nextInt();
        double[][] dp = new double[n + 1][n + 1];
        dp[0][0] = 1;
        for (int i = 0; i <= n; i++) {
            for (int j = 0; j <= n; j++) {
                if (i + j == 0)
                    continue;
                int allI = m - 1;
                int allJ = m * (n - 1);
                double p0 = (allI - i + 1.0) / (allJ + allI - i - j + 1);
                double p1 = (allJ - j + 1.0) / (allJ + allI - i - j + 1);
                dp[i][j] = ((i > 0) ? dp[i - 1][j] * p0 : 0)
                        + (j > 0 ? dp[i][j - 1] * p1 : 0);
            }
        }
        double ans = 0;
        for (int i = 0; i < n && i < m; i++)
            ans += (i + 1.00) / n * dp[i][n - 1 - i];
        System.out.println(ans);
    }
}