Edward is the headmaster of Marjar University. He is enthusiastic about chess and often plays chess with his friends. What's more, he bought a large decorative chessboard with N rows and M columns.

Every day after work, Edward will place a chess piece on a random empty cell. A few days later, he found the chessboard was dominated

"That's interesting!" Edward said. He wants to know the expectation number of days to make an empty chessboard of N × M dominated. Please write a program to help him.

Input

There are multiple test cases. The first line of input contains an integer Tindicating the number of test cases. For each test case:

There are only two integers N and M (1 <= N, M <= 50).

Output

For each test case, output the expectation number of days.

Any solution with a relative or absolute error of at most 10-8 will be accepted.

Sample Input

2
1 3
2 2

Sample Output

3.000000000000
2.666666666667

題解:要求次數的期望 我們先明確 期望 = 次數 * 該次數對應的概率  那明確這一點 動態轉移方程就很容易寫了

dp[k][i][j] 表示 放了k個 已經有i行 j列 滿足條件然後就可以分4種情況了  在放一個 加一行，加一列，都不變，都加1 詳見程式碼

#include<iostream>
#include<cstdio>
#include<cstring> 
using namespace std;
const int N=51;
int n,m;
double dp[N*N][N][N];
int main()
{
	int T;
	cin>>T;
	while(T--)
	{
		scanf("%d%d",&n,&m);
		memset(dp,0,sizeof(dp));
		dp[0][0][0]=1;
		for(int k=0;k<=n*m;k++)
			for(int i=0;i<=n;i++)
				for(int j=0;j<=m;j++)
				{
					if(i==n&&j==m) continue;//當然 已經滿足條件後就不要繼續求了 
					if(i*j>=k) dp[k+1][i][j]+=dp[k][i][j]*(i*j-k)/(n*m-k);
					if(i<n) dp[k+1][i+1][j]+=dp[k][i][j]*(n-i)*j/(n*m-k);
					if(j<m) dp[k+1][i][j+1]+=dp[k][i][j]*(m-j)*i/(n*m-k);
					if(i<n&&j<m) dp[k+1][i+1][j+1]+=dp[k][i][j]*(n-i)*(m-j)/(n*m-k);
				}
		double ans=0;
		for(int i=max(n,m);i<=n*m;i++)ans+=dp[i][n][m]*i;
		printf("%.12f\n",ans);
	}	
	return 0;
}