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>
int n, m;
bool card[1000010];
double ct = 0.0;
void recur(int last, int level) {
  if (level == n) {
    for (int k = 0; k < m; ++k) {
      int ct2 = 0;
      for (int j = 0; j < n; ++j) {
        if (card[k * n + j]) {
          ++ct2;
        }
      }
      ct += ct2 * ct2;
    }
  } else {
    for (int k = last + 1; k < m * n; ++k) {
      card[k] = true;
      recur(k, level + 1);
      card[k] = false;
    }
  }
}
inline long long binom(int n, int k) {
  long long ans = 1;
  for (int i = 0; i < k; ++i) {
    ans *= n - i;
  }
  for (int i = 0; i < k; ++i) {
    ans /= i + 1;
  }
  return ans;
}
int main() {
  std::cin >> n >> m;
  if (n == 1 && m == 1) {
    std::cout << 1 << std::endl;
    return 0;
  }
  std::cout << (1 + ((double)(n - 1) * (m - 1)) / (n * m - 1)) / n << std::endl;
  return 0;
}