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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int Nmax = 1005;
double Dp[Nmax][Nmax];
int main() {
  int N, M;
  cin >> N >> M;
  Dp[0][0] = 1;
  for (int i = 1; i <= N; i++) {
    for (int j = 0; j <= M; j++) {
      double p1 = (double)(M - j) / ((double)(N * M - (i - 1)));
      double p2 = (double)(M - j + 1) / ((double)(N * M - (i - 1)));
      Dp[i][j] = Dp[i - 1][j] * (1 - p1) + (j > 0 ? Dp[i - 1][j - 1] * p2 : 0);
    }
  }
  double Sol = 0;
  for (int i = 1; i <= M; i++) Sol = Sol + Dp[N][i] * i * i / ((double)N);
  printf("%.14lf\n", Sol);
}