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:200000000")
const double EPS = 1E-9;
const int INF = 1000000000;
const long long INF64 = (long long)1E18;
const double PI = 3.1415926535897932384626433832795;
int n, m;
bool u[1100][1100];
double z[1100][1100];
double rec(int pos, int taken) {
  if (taken == n || pos == m) return double(taken) / n;
  if (u[pos][taken]) return z[pos][taken];
  double p = (n - taken) / double(n * m - pos);
  double res = rec(pos + 1, taken + 1) * p + rec(pos + 1, taken) * (1 - p);
  u[pos][taken] = true;
  return z[pos][taken] = res;
}
int main() {
  cin >> n >> m;
  double res = rec(1, 1);
  printf("%.10lf\n", res);
  return 0;
}