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 i, j, n, m;
double p, a[1001], ans, first, second;
int main() {
  cin.tie(0);
  cout.precision(7);
  ios_base::sync_with_stdio(0);
  cin >> n >> m;
  if (n == 1) {
    cout << fixed << 1;
    return 0;
  }
  a[0] = 1;
  for (i = 1; i < n; i++)
    for (j = min(i, m); j >= 0; j--) {
      p = (m - j - 1.) / (n * m - i - 0.);
      a[j + 1] += a[j] * p;
      a[j] *= (1 - p);
    }
  for (i = 0; i <= m; i++) ans += a[i] * (i + 1.) / n;
  cout << fixed << ans;
}