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.

题意：龙王和公主就晚上应该干什么产生了分歧。于是他们决定从袋子里抓取老鼠来决定该去干什么。袋子里有w只白鼠，b只黑鼠，如果谁抓到了白鼠谁就赢了，然后如果都抓完了还没有人赢的话，那就是龙王赢了。公主先抓，而且每次龙王抓一只老鼠之后，会有一只其他老鼠跑出去，这个跑出去的老鼠是随机的。问你公主能赢的概率是多少。

思路：我们设dp[i][j]代表了公主能赢的情况下，剩i只白鼠和j只黑鼠。显然没有黑鼠时dp[i][0]=1,而没有白鼠dp[0][i]=0

公主一次抓到白鼠胜利的概率：i/(i+j)

（前提黑鼠数量>=2）公主黑鼠，龙王黑鼠，跑出白鼠概率：j/(i+j)*(j-1)/(i+j-1)i/(i+j-2)dp[i-1][j-2]
（前提黑鼠数量>=3）公主黑鼠，龙王黑鼠，跑出黑鼠概率：j/(i+j)(j-1)/(i+j-1)(j-2)/(i+j-2)*dp[i][j-3]

AC代码：

#include <bits/stdc++.h>
typedef long long ll;
const int maxx=1010;
const int inf=0x3f3f3f3f;
using namespace std;
double dp[maxx][maxx];
int w,b;
void init()
{
    memset(dp,0,sizeof(dp));
    for(int i=1; i<=w; i++)
    {
        dp[i][0]=1;
    }
    for(int i=1; i<=b; i++)
    {
        dp[0][i]=0;
    }
}
int main()
{
    while(cin>>w>>b)
    {

        init();
        for(int i=1; i<=w; i++)
        {
            for(int j=1; j<=b; j++)
            {
                dp[i][j]+=(double)i/(i+j);
                if(j>=2)
                    dp[i][j]+=((double)j/(i+j)*(double)(j-1)/(i+j-1)*(double)i/(i+j-2)*dp[i-1][j-2]);
                if(j>=3)
                    dp[i][j]+=((double)j/(i+j)*(double)(j-1)/(i+j-1)*(double)(j-2)/(i+j-2)*dp[i][j-3]);
            }
        }
        cout<<fixed<<setprecision(9)<<dp[w][b]<<endl;
    }
    return 0;
}