Fish

有编号从 $1$ 到 $n$ 的 $n$ 条鱼住在湖里。每天恰好有两条鱼相遇，任意两条鱼彼此相遇的概率是一样的。如果两条标号为 $i$ 和 $j$ 的鱼见面，第一条吃掉第二条的概率为 $a_{i,j}$，第二条吃掉第一条的概率为 $a_{j,i}=1-a_{i,j}$。这样的过程一直进行到湖里只剩下一条鱼，请你算出每只鱼最后存活在湖里的可能性。

$ n $ fish, numbered from $ 1 $ to $ n $ , live in a lake. Every day right one pair of fish meet, and the probability of each other pair meeting is the same. If two fish with indexes i and j meet, the first will eat up the second with the probability $ a_{ij} $ , and the second will eat up the first with the probability $ a_{ji}=1-a_{ij} $ . The described process goes on until there are at least two fish in the lake. For each fish find out the probability that it will survive to be the last in the lake.

The first line contains integer $ n $ ( $ 1<=n<=18 $ ) — the amount of fish in the lake. Then there follow $ n $ lines with $ n $ real numbers each — matrix $ a $ . $ a_{ij} $ ( $ 0<=a_{ij}<=1 $ ) — the probability that fish with index $ i $ eats up fish with index $ j $ . It's guaranteed that the main diagonal contains zeros only, and for other elements the following is true: $ a_{ij}=1-a_{ji} $ . All real numbers are given with not more than 6 characters after the decimal point.

Output $ n $ space-separated real numbers accurate to not less than 6 decimal places. Number with index $ i $ should be equal to the probability that fish with index $ i $ will survive to be the last in the lake.

2
0 0.5
0.5 0

0.500000 0.500000 

5
0 1 1 1 1
0 0 0.5 0.5 0.5
0 0.5 0 0.5 0.5
0 0.5 0.5 0 0.5
0 0.5 0.5 0.5 0

1.000000 0.000000 0.000000 0.000000 0.000000