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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("O3")
using namespace std;
const double eps = 1e-9;
const double pi = acos(-1.0);
const int inf = 2e5 + 10;
int mod = 1e9 + 7;
const int MAXN = 1e5 + 7;
long double C(int n, int k) {
  long double res = 1.;
  if (2 * k > n) k = n - k;
  for (int i = 1; i <= k; ++i) {
    res *= n - k + i;
    res /= i;
  }
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cout << fixed << setprecision(10);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  long double res = 0;
  for (int k = 1; k <= min(n, m); ++k) {
    res += C(m, k) * C(m * (n - 1), n - k) * k * k;
  }
  res /= n;
  res /= C(n * m, n);
  cout << res;
  return 0;
}