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;
double DP[1010][1010];
int N, M, UP;
int main() {
  cin >> N >> M;
  UP = min(N, M);
  DP[1][1] = 1.0;
  for (int i = 2; i <= N; ++i)
    for (int j = 0; j <= UP; ++j) {
      DP[i][j] += DP[i - 1][j] * (1.0 - (double(M - j)) / (N * M - (i - 1)));
      if (j > 0)
        DP[i][j] +=
            DP[i - 1][j - 1] * ((double(M - (j - 1))) / (N * M - (i - 1)));
    }
  double ans = 0;
  for (int i = 1; i <= UP; ++i) ans += DP[N][i] * (double(i) / N);
  printf("%.9lf\n", ans);
  return 0;
}