Description

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.

Input

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

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.

Sample Input

Input

2
0 0.5
0.5 0

Output

0.500000 0.500000 

Input

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

Output

1.000000 0.000000 0.000000 0.000000 0.000000 

此題可以輕易地想到狀態壓縮動態規劃。

首先  狀態  100010010011  中  1表示fish還活著   0反之。

接下來  我們把11111111111   賦值為1，開始運算。

每次都讓一條fish被吃掉，也就是說找2條還活著的魚。

注意這裡魚從0開始計數到n-1。

第j條fish 活著  可以用  i&(1<<j)==1  判斷。

  接下來再找一條k。新狀態為i^(1<<k) 。

至於概率的計算可以看程式理解一下。

程式碼如下

#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;

const int Ms=1<<20;
const int Mn=20;

double f[Ms],a[Mn][Mn];
int n,all;

int getfish(int s)
{
    int fish=0;
    while (s!=0)
    {
        fish+=s&1;
        s>>=1;
    }
    return fish;
}

int main()
{
    scanf("%d",&n);
    for (int i=0;i<n;i++)
    for (int j=0;j<n;j++)
        scanf("%lf",&a[i][j]);
    all=(1<<n)-1;
    f[all]=1;
    for (int i=all;i>=1;i--)
    {
        int fish=getfish(i);
        if (fish==1) continue;
        double c=2*f[i]/fish/(fish-1);
        for (int j=0;j<n;j++)
        if (i&(1<<j))
        {
            for (int k=0;k<n;k++)
            if (i&(1<<k))
            {
                f[i^(1<<k)]+=c*a[j][k];
            }
        }
    }
    for (int i=0;i<n;i++)
        printf("%.6lf ",f[1<<i]);
    return 0;
}