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>
const int maxi = 2000000000;
const int maxq = 1000000000;
const double eps = 1e-10;
const double pi = 3.1415926535897932;
const double inf = 1e+18;
const int mo = 1000000007;
const int maxn = 500000;
using namespace std;
int n, m;
double ans, f[1111][1111];
double ver(int k) { return f[n][k] * k / n; }
int main() {
  cin >> n >> m;
  f[1][1] = 1.0;
  for (int i = 2; i <= n; i++)
    for (int j = 1; j <= n; j++)
      f[i][j] = f[i - 1][j - 1] * (m - (j - 1)) / (n * m - (i - 1)) +
                f[i - 1][j] * (n * m - m - ((i - 1) - j)) / (n * m - (i - 1));
  for (int k = 1; k <= min(n, m); k++) ans += ver(k);
  printf("%.9lf\n", ans);
  return 0;
}