Problem: The dragon and the princess are arguing about what to do on the New Year's Eve. The dragon suggests flying to the mountains to watch fairies dancing in the moonlight, while the princess thinks they should just go to bed early. They are desperate to come to an amicable agreement, so they decide to leave this up to chance.

They take turns drawing a mouse from a bag which initially contains w white and b black mice. The person who is the first to draw a white mouse wins. After each mouse drawn by the dragon the rest of mice in the bag panic, and one of them jumps out of the bag itself (the princess draws her mice carefully and doesn't scare other mice). Princess draws first. What is the probability of the princess winning?

If there are no more mice in the bag and nobody has drawn a white mouse, the dragon wins. Mice which jump out of the bag themselves are not considered to be drawn (do not define the winner). Once a mouse has left the bag, it never returns to it. Every mouse is drawn from the bag with the same probability as every other one, and every mouse jumps out of the bag with the same probability as every other one.

Input

The only line of input data contains two integers w and b (0 ≤ w, b ≤ 1000).

Output

Output the probability of the princess winning. The answer is considered to be correct if its absolute or relative error does not exceed 10 - 9.

Examples

Input

1 3


Output

0.500000000


Input

5 5


Output

0.658730159

Note

Let's go through the first sample. The probability of the princess drawing a white mouse on her first turn and winning right away is 1/4. The probability of the dragon drawing a black mouse and not winning on his first turn is 3/4 * 2/3 = 1/2. After this there are two mice left in the bag — one black and one white; one of them jumps out, and the other is drawn by the princess on her second turn. If the princess' mouse is white, she wins (probability is 1/2 * 1/2 = 1/4), otherwise nobody gets the white mouse, so according to the rule the dragon wins.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class C implements Runnable{
	public static void main (String[] args) {new Thread(null, new C(), "_cf", 1 << 28).start();}

	int w, b;
	double[][][] prob;

	public void run() {
		FastScanner fs = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);

		w = fs.nextInt();
		b = fs.nextInt();
		prob = new double[2][w+1][b+1];
		for(int i = 0; i < 2; i++)
			for(int j = 0; j < w+1; j++)
				for(int k = 0; k < b+1; k++)
					prob[i][j][k] = -1;

		System.out.printf("%.15f\n", solve(w, b, 0));

//		for(int t = 0; t <= 1; t++) {
//			System.out.printf("For %s\n", t == 0 ? "Princess" : "Dragon");
//			for(int i = 0; i <= w; i++) {
//				for(int j = 0; j <= b; j++) {
//					System.out.printf("Probability of winning with %d white and %d black is %G\n", i, j, prob[t][i][j]);
//				}
//			}
//			System.out.println();
//		}

		out.close();
	}

	double solve (int w, int b, int t) {
		if(prob[t][w][b] != -1) return prob[t][w][b];
//		System.out.printf("%s's turn white = %d black = %d\n", t==0?"Princess" : "Dragon", w, b);
		if(t == 0) {
			//base cases:
			if(w == 0) return prob[t][w][b] = 0; //0% chance of winning
			if(b == 0) return prob[t][w][b] = 1; //100% chance of winning

			//princess has a w/(w+b) chance of winning right now
			double res = (double)w / (w + b);

			//princess has a b/(w+b) chance of getting a black mouse
			//if she gets a black mouse, then her chances of winning are
			//1 - the chances of dragon winning with this new state *
			//the probability that she draws a black mouse here.

			res += (double)b/(w+b) * (1 - solve(w, b-1, t^1));

			return prob[t][w][b] = res;
		}
		else {
			//base cases:
			if(w + b == 0) return prob[t][w][b] = 1; //dragon wins if there's nothing left
			if(w == 0) return prob[t][w][b] = 0; //dragon has no chance of winning here
			if(b == 0) return prob[t][w][b] = 1; //dragon can only win here

			//probability that dragon draws a white mouse right now and wins is 
			//w / (w + b)

			double res = (double)w/(w+b);

			//if dragon draws a black mouse, there are two possibilities:
			//a white mouse jumps out
			//a black mouse jumps out
			//the probability of dragon winning is:

			//prob(BW) * (1 - solve(w-1, b-1, t^1) + 
			//prob(BB) * (1 - solve(w, b-2, t^1)

			if(w >= 1 && b >= 1) {
				double wb = ((double)b/(b+w)) * ((double)w/(b+w-1));			
				res += wb * (1 - solve(w-1, b-1, t^1));
			}

			if(b >= 2) {
				double bb = ((double)b/(b+w)) * ((double)(b-1)/(b+w-1));
				res += bb * (1 - solve(w, b-2, t^1));
			}

			return prob[t][w][b] = res;
		}
	}

	void sort (int[] a) {
		int n = a.length;
		for(int i = 0; i < 1000; i++) {
			Random r = new Random();
			int x = r.nextInt(n), y = r.nextInt(n);
			int temp = a[x];
			a[x] = a[y];
			a[y] = temp;
		}
		Arrays.sort(a);
	}

	class FastScanner {
		BufferedReader br;
		StringTokenizer st;
		public FastScanner() {
			try	{
				br = new BufferedReader(new InputStreamReader(System.in));
				//				br = new BufferedReader(new FileReader("cases.in"));
				st = new StringTokenizer("");
			} catch (Exception e){e.printStackTrace();}
		}
		public String next() {
			if (st.hasMoreTokens())	return st.nextToken();
			try {st = new StringTokenizer(br.readLine());}
			catch (Exception e) {e.printStackTrace();}
			return st.nextToken();
		}
		public int nextInt() {return Integer.parseInt(next());}
		public long nextLong() {return Long.parseLong(next());}
		public double nextDouble() {return Double.parseDouble(next());}

		public String nextLine() {
			String line = "";
			try {line = br.readLine();}
			catch (Exception e) {e.printStackTrace();}
			return line;
		}
		public Integer[] nextIntegerArray(int n) {
			Integer[] a = new Integer[n];
			for(int i = 0; i < n; i++) a[i] = nextInt();
			return a;
		}
		public int[] nextIntArray(int n) {
			int[] a = new int[n];
			for(int i = 0; i < n; i++) a[i] = nextInt();
			return a;
		}
		public char[] nextCharArray() {return nextLine().toCharArray();}
	}

}