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 mpow(long long a, long long n) {
  long long ret = 1;
  long long b = a;
  while (n) {
    if (n & 1) ret = (ret * b) % 1000000007;
    b = (b * b) % 1000000007;
    n >>= 1;
  }
  return (long long)ret;
}
double arr[1009][1009];
void pre() {
  int i, j;
  for (i = 0; i < 1008; i++) {
    for (j = 0; j <= i; j++) {
      if (i == j || j == 0)
        arr[i][j] = 1;
      else
        arr[i][j] = (arr[i - 1][j] + arr[i - 1][j - 1]);
    }
  }
}
int main() {
  pre();
  int n, m, t, j, k, l, i;
  double newval;
  double x;
  double p;
  cin >> n >> m;
  int r = (n * m);
  double myval = 0.0;
  for (i = 1; i <= n; i++) {
    x = (double)(i) / (double)(n);
    p = (x * x);
    k = m, l = r;
    for (j = 1; j <= i; j++, k--, l--) {
      newval = (1.0 * k) / (1.0 * l);
      p *= newval;
    }
    p *= arr[n][i];
    k = r - m;
    for (j = i + 1; j <= n; j++, k--, l--) {
      newval = (1.0 * k) / (1.0 * l);
      p *= newval;
    }
    myval += p;
  }
  myval = myval * n;
  printf("%0.7lf\n", myval);
  return 0;
}