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 mod = 1e9 + 7;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void solve() {
  long double n, m;
  cin >> n >> m;
  long double ans;
  if (n * m == 1)
    ans = 1;
  else
    ans = (1 + ((n - 1) * (m - 1)) / (n * m - 1)) / n;
  cout << setprecision(10) << fixed << ans << '\n';
}
signed main() {
  ios ::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  clock_t clk = clock();
  int t;
  t = 1;
  while (t--) {
    solve();
  }
  cerr << '\n'
       << setprecision(10) << fixed << (double)(clock() - clk) / CLOCKS_PER_SEC;
  return 0;
}