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 bblack 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.
PS：这个题如果想到是一个概率DP后，递推公式还是比较好推的，现在声明一个数组dp[i][j]表示公主在有i个白鼠，j个黑鼠获胜的概率，我们现在是算公主要赢的概率，这样可以分成三种情况。
1.公主直接抽到了白鼠，概率为：i/(i+j);
2.当公主抽到了黑鼠，龙也抽到了黑鼠，跑出去的也是黑鼠，（这种情况要黑鼠的数量不小于3个时候前提下），当这种情况下公主赢得概率等于又跳到dp[i][j-3]，所以在这种情况下的概率为：j/(i+j)*(j-1)/(i+j-1)*(j-2)/(i+j-2)*dp[i][j-3];
3.当公主抽到了黑鼠，龙也抽到了黑鼠，跑出去的是白鼠，（这种情况要黑鼠的数量不小于2个时候前提下），当这种情况下公主赢得概率等于又跳到dp[i-1][j-2]，所以在这种情况下的概率为：j/(i+j)*(j-1)/(i+j-1)*i/(i+j-2)*dp[i-1][j-2];
初始化需要注意一下dp[i][0],表示有白鼠，但是没得黑鼠，公主直接赢，反之dp[0][i]赋值为0.
细节看代码
#include <iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<map>
#include<queue>
#include<set>
#include<cmath>
#include<stack>
#include<string>
const int maxn=1e3+10;
const int mod=1e9+7;
const int inf=1e8;
#define me(a,b) memset(a,b,sizeof(a))
#define lowbit(x) x&(-x)
typedef long long ll;
using namespace std;
double dp[maxn][maxn];
int main()
{
int w,b;
cin>>w>>b;
for(int i=1;i<=w;i++)
dp[i][0]=1;//只有白鼠
for(int i=1;i<=b;i++)
dp[0][i]=0;//只有黑鼠
for(int i=1;i<=w;i++)
for(int j=1;j<=b;j++)
{
dp[i][j]+=(double)i/(i+j);//公主抽到白鼠
if(j>=2)
dp[i][j]+=(double)j/(i+j)*(j-1)/(i+j-1)*i/(i+j-2)*dp[i-1][j-2];//当公主抽到了黑鼠，龙也抽到了黑鼠，跑出去的是白鼠
if(j>=3)
dp[i][j]+=(double)j/(i+j)*(j-1)/(i+j-1)*(j-2)/(i+j-2)*dp[i][j-3];//当公主抽到了黑鼠，龙也抽到了黑鼠，跑出去的也是黑鼠
}
printf("%.9f\n",dp[w][b]);
return 0;
}