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;
double lf[1000514];
double inf = 10e9;
double lc(int n, int m) { return lf[n] - lf[m] - lf[n - m]; }
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  lf[0] = 0;
  for (int i = 1; i <= 1000000; i++) {
    lf[i] = lf[i - 1] + log((double)i);
  }
  double ans = 0;
  for (int i = 0; i <= n; i++) {
    if (i <= m && n - i <= n * m - m) {
      ans += (double)i * (double)((double)i / n) *
             exp(lc(m, i) + lc(n * m - m, n - i) - lc(n * m, n));
    }
  }
  printf("%.10f\n", ans);
  return 0;
}