E

题目链接：http://codeforces.com/problemset/problem/16/E

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.

Examples

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 
题目大意：

有一堆鱼，n条，每两只相遇都会有其中一只吃掉对方，现在给出每条鱼吃掉其他鱼的概率，问最后每条鱼存活的概率。

 思路：状压dp的模板题

看代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<stack>
#include<map>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<stack>
#include<map>
#include<queue>
#include<cmath>
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
#define sc1(a) scanf("%lld",&a)
#define pf1(a) printf("%lld
",a)
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
const LL INF=1e18;
const ull base=2333;
const int maxn=1e1+50;
const int maxm=1e3+50;
const int maxv=1e6+5;
const int mod=1e9+7;
const int ba=3e5;
double a[maxn][maxn];
double dp[1<<20];
int main()
{
    LL N;sc1(N);
    for(int i=0;i<N;i++)
    {
        for(int j=0;j<N;j++) scanf("%lf",&a[i][j]);
    }
    dp[(1<<N)-1]=1;//全部鱼存在的情况下 结果为1
    for(int s=(1<<N)-1;s>=0;s--)
    {
        LL sum=0;
        for(int i=0;i<N;i++)  if(s&(1<<i)) sum++;
        for(int i=0;i<N;i++)//枚举i吃掉 j
        {
            if(s&(1<<i)) //该位还活着
            {
                for(int j=0;j<N;j++)
                {
                    if(i==j) continue;//自己不会吃自己
                    if(s&(1<<j))
                    {
                        dp[s^(1<<j)]+=dp[s]*2.0/(sum*(sum-1))*a[i][j];
                    }
                }
            }
        }
    }
    for(int i=0;i<N;i++) printf("%lf ",dp[(1<<i)]);
    return 0;
}
/**

*/