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;
int n, m;
bool solve() {
  if (scanf("%d%d", &n, &m) == EOF) return false;
  double ans = 0.0;
  for (int k = 1; k <= min(n, m); ++k) {
    long double x = 1.0;
    for (int i = 0; i < n; ++i) {
      x *= n * m - i;
      x /= n - i;
    }
    long double y = 1.0;
    for (int i = 0; i < n - k; ++i) {
      y *= n * m - m - i;
      y /= n - k - i;
    }
    long double z = 1.0;
    for (int i = 0; i < k; ++i) {
      z *= m - i;
      z /= k - i;
    }
    long double prob = y * z / x;
    ans += prob * k / n * k / n;
  }
  printf("%.12f\n", ans * n);
  return true;
}
int main() {
  while (solve())
    ;
  return 0;
}