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;
template <class T>
inline T sqr(T x) {
  return x * x;
}
const double PI = acos(-1.0);
using namespace std;
double dp[1005][1005];
int main(void) {
  int n, m;
  scanf("%d %d", &n, &m);
  if (n == 1) {
    puts("1.0000000");
    return 0;
  }
  dp[0][0] = 1;
  for (int i = 1; i <= n - 1; i++) {
    dp[i][0] = dp[i - 1][0] * (1 - (m - 1.0) / (n * m - i));
    for (int j = 1; j <= i && j <= m; j++) {
      dp[i][j] = dp[i - 1][j - 1] * (m - j + 0.0) / (n * m - i) +
                 dp[i - 1][j] * (1 - (m - j - 1.0) / (n * m - i));
    }
  }
  double ans = 0;
  for (int i = 0; i <= n - 1; i++) {
    ans += dp[n - 1][i] * (i + 1.0) / n;
  }
  printf("%.10lf\n", ans);
  return 0;
}