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;
const int N = 1e6 + 34;
const long double eps = 1e-10;
const int C = 3;
int n, m, t;
int k1, kol, k2, c[1111][1111];
long double cur, p;
vector<int> ch;
vector<int> zn;
void go() {
  if (ch.size() == 0 && zn.size() == 0) return;
  if (ch.size() == 0 || (zn.size() != 0 && cur > 2)) {
    cur /= zn.back();
    zn.pop_back();
    return;
  }
  cur *= ch.back();
  ch.pop_back();
}
int main() {
  cin >> n >> m;
  kol = m * n;
  c[0][0] = 1;
  for (int i = 1; i < 1100; i++) {
    c[i][0] = c[i][i] = 1;
    for (int j = 1; j < i; j++) c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
  }
  for (int i = 1; i <= min(n, m); i++) {
    k1 = m - 1;
    k2 = kol - 1 - k1;
    cur = 1;
    ch.clear();
    zn.clear();
    t = min(i - 1, (n - 1) - (i - 1));
    for (int j = 1; j <= t; j++) {
      ch.push_back(n - t + j - 1);
      zn.push_back(j);
    }
    for (int j = kol - 1; j > kol - i; j--) {
      cur *= (0.0 + k1--) / (j + 0.0);
      go();
    }
    for (int j = kol - i; j > kol - n; j--) {
      cur *= (0.0 + k2--) / (j + 0.0);
      go();
    }
    while (ch.size() != 0 || zn.size() != 0) go();
    p += cur * (i + 0.0) / (n + 0.0);
  }
  cout.precision(15);
  cout << fixed << p << "\n";
  return 0;
}