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;
double p[1005][1005];
int main() {
  int n, i, j, m, t;
  cin >> n >> m;
  t = n * m;
  p[1][1] = 1.0;
  for (i = 2; i <= n; i++)
    for (j = 0; j <= n; j++) {
      if (j > i)
        p[i][j] = 0.0;
      else if (j == 0)
        p[i][j] = 0.0;
      else {
        p[i][j] =
            p[i - 1][j] * ((t - i + 1 - m + j)) + p[i - 1][j - 1] * (m - j + 1);
        p[i][j] /= (t - i + 1);
      }
    }
  double prob = 0.0;
  for (i = 0; i <= n; i++) prob += p[n][i] * ((i) / (n * 1.0));
  printf("%.9lf\n", prob);
  return 0;
}