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;
#pragma comment(linker, "/STACK:64000000")
int l, r, h, g, j, i, y, q1, q2, sum, m, cur, p, n;
double c[1005][1005], ans;
inline void solve() {
  scanf("%d%d", &n, &m);
  c[1][1] = 1.0;
  for (i = 2; i <= n; i++) {
    c[i][1] = c[i - 1][1] * (double)(n * m - i + 1 - (m - 1)) /
              (double)(n * m - i + 1);
    for (j = 2; j <= i; j++) {
      if (j > m) {
        break;
      }
      c[i][j] =
          c[i - 1][j - 1] * (double)(m - j + 1) / (double)(n * m - i + 1) +
          c[i - 1][j] * (double)(n * m - (i - 1) - (m - j)) /
              (double)(n * m - (i - 1));
    }
  }
  for (j = 1; j <= min(n, m); j++) {
    ans += c[n][j] * (double)(j) / (double)(n);
  }
  printf("%.10lf\n", ans);
}
inline void init() {}
inline void answer() {}
int main() {
  init();
  solve();
  answer();
  return 0;
}