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 double dp[1024][1024];
long double solve(int n, int m) {
  memset(dp, 0, sizeof(dp));
  int i, k;
  dp[1][1] = 1.0;
  for (i = 2; i <= n; i++) {
    for (k = 0; k <= n && k <= m && k <= n * m; k++) {
      dp[i][k] = 0.0;
      if (k > 0)
        dp[i][k] += dp[i - 1][k - 1] *
                    ((long double)(m - k + 1) / ((long double)n * m - i + 1));
      dp[i][k] += dp[i - 1][k] * ((long double)(m * n - i + 1 - (m - k)) /
                                  ((long double)n * m - i + 1));
    }
  }
  long double res = 0.0;
  for (i = 1; i <= n && i <= m; i++) {
    res += dp[n][i] * ((long double)i / (long double)n);
  }
  return res;
}
int main() {
  int n, m;
  cin >> n >> m;
  cout << solve(n, m) << endl;
  return 0;
}