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>
double fac[1111000];
double calc(int n, int k) { return fac[n] - fac[k] - fac[n - k]; }
int main() {
  fac[1] = 0.;
  for (int i = 2; i < 1001000; i++) fac[i] = fac[i - 1] + log(1. * i);
  int n, m;
  scanf("%d%d", &n, &m);
  double ans = 0;
  double c = calc(n * m, n);
  for (int k = 0; k <= n; k++) {
    if (0 <= k && k <= m) {
      if (0 <= k && n - k <= (n - 1) * m) {
        double tmp = calc(m, k);
        tmp += calc((n - 1) * m, n - k);
        tmp -= c;
        tmp = exp(tmp);
        ans += tmp * (1. * k * k) / (1. * n * n);
      }
    }
  }
  ans = ans * 1. * n;
  printf("%.10lf\n", ans);
  return 0;
}