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 maxn = 1100 * 1100;
double a[maxn];
void init() {
  a[1] = log(1.0);
  for (int i = 2; i <= 1000000; i++) {
    a[i] = a[i - 1] + log(i + 0.0);
  }
}
int main() {
  init();
  int n, m;
  double ans;
  while (scanf("%d%d", &n, &m) != EOF) {
    ans = 0;
    for (int i = 1; i <= min(n, m); i++) {
      ans += 1.0 * i * i / n *
             (exp(a[m] + a[n * m - m] + a[n] + a[n * m - n] - a[i] - a[m - i] -
                  a[n - i] - a[n * m - m - n + i] - a[n * m]));
    }
    printf("%.8f\n", ans);
  }
  return 0;
}