CF 148D D. Bag of mice （概率DP||数学期望）

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.

Example

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.

题意：

一对情侣开房玩抓老鼠游戏，老鼠有黑白两色，女的为先手，先抓到白老鼠胜。特别的，男的每抓一只老鼠后，也会随机放走一只老鼠。问女的赢的概率是多少。如果输了，后果会很严重，当天晚上只能睡沙发。

思路：

dp[i][j]为当前状态，有i只白老鼠，j只黑老鼠，女的赢的概率。那么dp[][] = 这一次赢 + 以后赢= i/(i+j) + 。。。具体如下。

#include<cstdio>

#include<cstdlib>

#include<iostream>

#include<algorithm>

#include<cstring>

using namespace std;

double dp[][];    int  w,b,i,j;

int main()

{

    while(~scanf("%d%d",&w,&b)){

        memset(dp,,sizeof(dp));

        for(i=;i<=w;i++)

         for(j=;j<=b;j++){

            if(i==) { dp[i][j]=;   continue;}

            if(j==) { dp[i][j]=1.0; continue;}

             dp[i][j]=1.0*i/(i+j);

            if(j>=) dp[i][j]+=(1.0*j/(i+j))*(j-)/(i+j-)*i/(i+j-)*dp[i-][j-];

            if(j>=) dp[i][j]+=(1.0*j/(i+j))*(j-)/(i+j-)*(j-)/(i+j-)*dp[i][j-];

        }

        printf("%.9lf\n",dp[w][b]);

    }return ;

}