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;
long double f[1011][1011];
int m, n;
void init() {
  memset(f, 0, sizeof f);
  f[1][1] = 1;
  for (int i = (1), _b = (n - 1); i <= _b; i++) {
    for (int x = (0), _b = (min(i, m)); x <= _b; x++)
      if (f[i][x] > 1e-50) {
        f[i + 1][x] += f[i][x] * ((m * n - m) - (i - x)) / (double)(m * n - i);
        f[i + 1][x + 1] += f[i][x] * (m - x) / (double)(m * n - i);
      }
  }
}
int main() {
  ios ::sync_with_stdio(false);
  cout << (fixed) << setprecision(8);
  while (cin >> n >> m) {
    init();
    double res = 0.0;
    for (int k = (1), _b = (min(m, n)); k <= _b; k++) {
      res += f[n][k] * k / (double)n;
    }
    cout << res << endl;
  }
  return 0;
}