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 inv = 1000000000;
const int minv = -inv;
const int max_n = 1010;
int n, m;
double f[max_n];
int main() {
  scanf("%d", &n);
  scanf("%d", &m);
  if (n == 1) {
    printf("1.00000000\n");
    return 0;
  }
  if (m == 1) {
    printf("%.6f\n", 1.0 / (double)n);
    return 0;
  }
  f[0] = 1.0;
  int r = (n - 1) * m;
  int t = n * m;
  for (int i = 0; i < n; i++) f[0] *= (double)(r - i) / (double)(t - i);
  double res = 0.0;
  for (int s = 1; s <= min(n, m); s++) {
    f[s] = f[s - 1] * (double)((n - s + 1) * (m - s + 1)) /
           (double)(s * (r - n + s));
    res += f[s] * (double)(s * s) / (double)n;
  }
  printf("%.6f\n", res);
  return 0;
}