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 long int ABS(long long int a) {
  if (a < 0) return (-a);
  return a;
}
void input() {}
int main() {
  input();
  int n, m;
  cin >> n >> m;
  double ans = 0.0;
  for (int i = (0); i < (n + 1); ++i) {
    if (i > m) continue;
    double res = (double)(i) / (double)(n);
    res = res * res;
    for (int j = (0); j < (i); ++j) {
      res *= (double)(m - j) / (double)(j + 1);
    }
    for (int j = (0); j < (n); ++j) {
      res *= (double)(j + 1) / (double)(n * m - j);
    }
    for (int j = (0); j < ((n - i)); ++j) {
      res *= (double)(n * m - m - j) / (double)(j + 1);
    }
    ans += res;
  }
  ans *= n;
  printf("%.15lf\n", ans);
  return 0;
}