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 inf = (int)1e9, maxn = (int)1e5 + 1;
const double eps = (double)1e-8;
const int mod = (int)1000000009;
double d[1001][1001], s, o;
int n, m, k;
int main() {
  cout << fixed << setprecision(10);
  cin >> n >> m;
  k = n * m;
  memset(d, double(0), sizeof d);
  d[1][1] = 1;
  for (int i = (2); i <= (n); i++) {
    s = double(0);
    d[i][1] = 1;
    for (int j = (2); j <= (min(i, m)); j++) {
      d[i][j] =
          d[i - 1][j] * (double((k - i + 1) - (m - j)) / double(k - i + 1)) +
          d[i - 1][j - 1] * (double(m - j + 1) / double(k - i + 1));
      s += double(d[i][j]);
    }
    d[i][1] = double(1) - s;
  }
  o = double(0);
  for (int i = (1); i <= (min(n, m)); i++) {
    o += (double(d[n][i])) * (double(i) / double(n));
  }
  cout << o;
  return 0;
}