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;
int n, m;
long double a[1000010], b[1000010];
long double C(int n, int m) {
  if (n < m) return -1e100;
  return b[n] - b[m] - b[n - m];
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n * m; ++i) {
    a[i] = log(i);
    b[i] = b[i - 1] + a[i];
  }
  long double ans = 0;
  for (int k = 1; k <= n; ++k) {
    ans = ans +
          exp(2 * a[k] + C(m, k) + C((n - 1) * m, n - k) - a[n] - C(n * m, n));
  }
  printf("%.9f\n", (double)ans);
  return 0;
}