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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
template <class T>
inline bool read(T &n) {
  T x = 0, tmp = 1;
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();
  if (c == EOF) return false;
  if (c == '-') c = getchar(), tmp = -1;
  while (c >= '0' && c <= '9') x *= 10, x += (c - '0'), c = getchar();
  n = x * tmp;
  return true;
}
int main() {
  int n, m;
  read(n), read(m);
  if (n == 1 && m == 1)
    printf("1.0000000\n");
  else
    printf("%.7f\n", (1.0 / (double)n + (double)((m - 1) * (n - 1)) /
                                            (double)(n * m - 1) / (double)n));
  return 0;
}