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 ri() {
  int res;
  scanf("%d", &res);
  return res;
}
int numbits(int n) { return n ? 1 + numbits(n & (n - 1)) : 0; }
class timer {
 public:
  timer() : t_(clock()) {}
  void start() { t_ = clock(); }
  float elapsed() { return float(clock() - t_) / CLOCKS_PER_SEC; }

 private:
  clock_t t_;
};
double dp[1024][1024];
int main() {
  int n, m;
  cin >> n >> m;
  memset(dp, 0, sizeof(dp));
  dp[0][0] = 1.0;
  for (int i = 0; i <= m; ++i) {
    for (int j = 0; j < n; ++j) {
      if (dp[i][j] == 0.0) {
        continue;
      }
      int ones = m - i;
      int left = n * m - j;
      double p = 1.0 * ones / left;
      dp[i + 1][j + 1] += p * dp[i][j];
      dp[i][j + 1] += (1.0 - p) * dp[i][j];
    }
  }
  double res = 0.0;
  for (int i = 1; i <= min(n, m); ++i) {
    double p = 1.0 * i / n;
    double subres = dp[i][n] * p * p;
    res += subres;
  }
  printf("%.9f\n", res * n);
  return 0;
}