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:134217728")
using namespace std;
double dp[1 << 10][1 << 10];
int n, m;
double p(int a, int b) {
  if (a + b == n - 1) return a + 1;
  double& res = dp[a][b];
  if (res != -1.0) return res;
  res = 0;
  int c = m - 1 - a;
  int d = n * m - m - b;
  int total = n * m - 1 - a - b;
  if (c > 0) res += (0.0 + c) / total * p(a + 1, b);
  if (d > 0) res += (0.0 + d) / total * p(a, b + 1);
  return res;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < 1 << 10; ++i)
    for (int j = 0; j < 1 << 10; ++j) dp[i][j] = -1.0;
  printf("%.9lf\n", p(0, 0) / n);
  return 0;
}