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 c[1000005];
void init(void) {
  int i;
  c[0] = 0;
  for (i = 1; i <= 1000000; i++) {
    c[i] = c[i - 1] + log(i * 1.0);
  }
}
int main() {
  int n, m, i;
  double s, f;
  init();
  while (cin >> n >> m) {
    s = 0;
    for (i = 0; i <= min(m, n); i++) {
      f = 0;
      f += c[m] - c[i] - c[m - i];
      f += c[m * (n - 1)] - c[n - i] - c[m * (n - 1) - (n - i)];
      f -= log(n * 1.0);
      f += 2 * log(i * 1.0);
      f -= (c[n * m] - c[n] - c[n * m - n]);
      s += exp(f);
    }
    printf("%.9lf\n", s);
  }
  return 0;
}