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;
long double c(int n, int p) {
  if (p > n - p) p = n - p;
  long double res = 1;
  for (int i = (1); i < (p + 1); ++i) {
    res *= (n - i + 1);
    res /= i;
  }
  return res;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  long double res = 0;
  int k = ((n < m) ? (n) : (m)) + 1;
  for (int p = (1); p < (k); ++p) {
    long double val = c(m * n - m, n - p);
    val /= c(m * n, n);
    val *= c(m, p);
    val *= (p / (long double)n);
    val *= (p / (long double)n);
    res += val;
  }
  res *= n;
  printf("%.15lf\n", (double)res);
  return 0;
}