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 JAVA):

Solution:
import java.util.Arrays;
import java.util.Scanner;


public class C {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt(), m = sc.nextInt();
		for(int i=0; i<1001; i++) Arrays.fill(dp[i], -1);
		double res = 0;
		
		
		double[][] dp = new double[1001][1001];
		
		dp[0][0] = 1;
		
		for(int i=0; i<=n; i++) {
			for(int j=0; j<=m; j++) {
				if(i==0 && j==0) continue;
				
				if (i > 0) 
					dp[i][j] += dp[i-1][j]  * (1.0*m*n-i+1-m+j)/(1.0*m*n-i+1);
				
				if (i > 0 && j > 0)
					dp[i][j] += dp[i-1][j-1]  * (m-j+1.0) / (1.0*m*n-i+1);
					
				
				//dp[i][j] = (i>0 ? dp[i-1][j] : 0) * (1.0*m*n-i+1-m+j)/(1.0*m*n-i+1) + ((i>0 && j>0) ? dp[i-1][j-1] : 0) * (m-j+1.0) / (1.0*m*n-i+1);
			}
		}
		
		for(int cnt=1; cnt<=m; cnt++) {
	//		double f = (find(cnt, n, m, n));
			double f = dp[n][cnt];
			res += n * f * Math.pow((cnt*1.0)/n, 2);
		}
		System.out.println(res);
		
	//	System.out.println(dp[2][0]);
	//	System.out.println(dp[2][1]);
	//	System.out.println(dp[2][2]);
		
		
	}
	
	
	static double find(int cnt, int pos, int m, int n) {
		if(cnt==0 && pos==0) return 1;
		if(pos<0 || cnt<0) return 0;
		if(cnt > pos) return 0;
		if(dp[cnt][pos] > -0.5) return dp[cnt][pos];
		// i = pos, j = cnt
		return dp[cnt][pos] = find(cnt, pos-1, m, n) * (1.0*m*n+cnt-m-pos) / (1.0*m*n-pos) + 
				find(cnt-1, pos-1, m, n) * (1.0*m-cnt)/(1.0*m*n-pos); 
	}
	
	
	static double[][] dp = new double[1001][1001];
	
}