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 MAX = 1000010;
bool is_prime[MAX];
vector<int> p;
int prv[MAX], pos[MAX];
int son[MAX], mom[MAX];
set<int> used;
vector<int> vec1, vec2;
void put(int n, int k, int *a) {
  for (int i = 0; i < k; i++) {
    int x = n - i;
    while (x != 1) {
      a[pos[prv[x]]]++;
      used.insert(pos[prv[x]]);
      x /= prv[x];
    }
  }
}
int main() {
  fill(is_prime, is_prime + MAX, true);
  is_prime[1] = false;
  for (int i = 2; i < MAX; i++) {
    if (!is_prime[i]) continue;
    pos[i] = p.size();
    p.push_back(i);
    prv[i] = i;
    for (long long j = 1LL * i * i; j < MAX; j += i) {
      is_prime[j] = false;
      prv[j] = i;
    }
  }
  int n, m;
  scanf("%d%d", &n, &m);
  if (n == 1) {
    printf("%f\n", 1.0);
    return 0;
  }
  fill(son, son + p.size(), 0);
  fill(mom, mom + p.size(), 0);
  double ans = 0;
  for (int i = 1; i <= min(n, m); i++) {
    put(n, n, son);
    put(n * m, n, mom);
    put(n, 1, mom);
    put(i * i, 1, son);
    put((n - 1) * m, n - i, son);
    put(n - i, n - i, mom);
    put(m, i, son);
    put(i, i, mom);
    for (auto it = used.begin(); it != used.end(); it++) {
      int tmp = min(son[*it], mom[*it]);
      son[*it] -= tmp;
      mom[*it] -= tmp;
    }
    for (auto it = used.begin(); it != used.end(); it++) {
      for (int j = 0; j < son[*it]; j++) vec1.push_back(p[*it]);
      for (int j = 0; j < mom[*it]; j++) vec2.push_back(p[*it]);
      son[*it] = 0;
      mom[*it] = 0;
    }
    long double sum = 1;
    int k = min(vec1.size(), vec2.size());
    for (int j = 0; j < k; j++) {
      sum *= vec1[j];
      sum /= vec2[j];
      double tmp = sum;
    }
    for (int j = k; j < vec1.size(); j++) sum *= vec1[j];
    for (int j = k; j < vec2.size(); j++) sum /= vec2[j];
    double tmp = sum;
    ans += sum;
    used.clear();
    vec1.clear();
    vec2.clear();
  }
  printf("%.10f\n", ans);
  return 0;
}