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;
int n, m;
long double p(int k) {
  vector<long double> A, B;
  for (int i = m - k + 1; i < m + 1; i++)
    if (i > 0)
      A.push_back(i);
    else if (i < 0)
      return 0;
  for (int i = n - k + 1; i < n + 1; i++)
    if (i > 0)
      A.push_back(i);
    else if (i < 0)
      return 0;
  for (int i = m * n - m - n + k + 1; i < m * n - m + 1; i++)
    if (i > 0)
      A.push_back(i);
    else if (i < 0)
      return 0;
  for (int i = 0; i < k + 1; i++)
    if (i > 0)
      B.push_back(i);
    else if (i < 0)
      return 0;
  for (int i = m * n - n + 1; i < m * n + 1; i++)
    if (i > 0)
      B.push_back(i);
    else if (i < 0)
      return 0;
  sort((A).rbegin(), (A).rend());
  sort((B).rbegin(), (B).rend());
  long double res = 1;
  while (B.size() < A.size()) B.push_back(1);
  while (A.size() < B.size()) A.push_back(1);
  for (int i = 0; i < A.size(); i++) res *= A[i] / B[i];
  return res;
}
int main() {
  cin >> n >> m;
  long double res = 0;
  for (int k = 1; k < min(n, m) + 1; k++) {
    res += p(k) * k * k / n;
  }
  printf("%.09f\n", (double)res);
}