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;
double dp[1001], f, sum, pr;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    dp[i] = (1.0 - sum) / (m * n - i + 1);
    sum += dp[i];
  }
  f = sum * m / n;
  sum = 0.0;
  pr = 1.0;
  for (int i = 1; i < n; i++) {
    double takeNow = pr / (m * n - i + 1);
    double pr2 = 1.0;
    for (int j = i + 1; j <= n; j++) {
      double k = pr2 / (m * n - j + 1);
      sum += takeNow * k;
      pr2 *= (m * n * 1.0 - j) / (m * n * 1.0 - j + 1);
    }
    pr *= (m * n * 1.0 - i - 1) / (m * n * 1.0 - i + 1);
  }
  sum *= 2.0;
  f += sum * m * (m - 1) / n;
  printf("%.10lf", f);
  return 0;
}