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;
int main() {
  scanf("%d", &(N)), scanf("%d", &(M));
  double dp[N + 1];
  for (int i = (0); i < (N + 1); i++) dp[i] = 0;
  dp[0] = 1;
  double ndp[N + 1];
  for (int COUNTER = (0); COUNTER < (N); COUNTER++) {
    for (int i = (0); i < (N + 1); i++) ndp[i] = 0;
    for (int i = (0); i < (N); i++) {
      double c = (double)(M - i) / (N * M - COUNTER);
      ndp[i + 1] += dp[i] * c;
      ndp[i] += dp[i] * (1 - c);
    }
    for (int i = (0); i < (N + 1); i++) dp[i] = ndp[i];
  }
  double sum = 0;
  for (int i = (0); i < (N + 1); i++) {
    dp[i] *= i;
    sum += dp[i];
  }
  for (int i = (0); i < (N + 2); i++) dp[i] /= sum;
  double ans = 0;
  for (int i = (0); i < (N + 1); i++) {
    ans += dp[i] * ((double)i / N);
  }
  printf("%.12lf\n", ans);
}