The rules of Sith Tournament are well known to everyone. n Sith take part in the Tournament. The Tournament starts with the random choice of two Sith who will fight in the first battle. As one of them loses, his place is taken by the next randomly chosen Sith who didn't fight before. Does it need to be said that each battle in the Sith Tournament ends with a death of one of opponents? The Tournament ends when the only Sith remains alive.

Jedi Ivan accidentally appeared in the list of the participants in the Sith Tournament. However, his skills in the Light Side of the Force are so strong so he can influence the choice of participants either who start the Tournament or who take the loser's place after each battle. Of course, he won't miss his chance to take advantage of it. Help him to calculate the probability of his victory.

Input

The first line contains a single integer n (1 ≤ n ≤ 18) — the number of participants of the Sith Tournament.

Each of the next n lines contains n real numbers, which form a matrix pij (0 ≤ pij ≤ 1). Each its element pij is the probability that the i-th participant defeats the j-th in a duel.

The elements on the main diagonal pii are equal to zero. For all different i, j the equality pij + pji = 1 holds. All probabilities are given with no more than six decimal places.

Jedi Ivan is the number 1 in the list of the participants.

Output

Output a real number — the probability that Jedi Ivan will stay alive after the Tournament. Absolute or relative error of the answer must not exceed 10 - 6.

Examples

Input

3
0.0 0.5 0.8
0.5 0.0 0.4
0.2 0.6 0.0

Output

0.680000000000000

description
403机房最近决定举行一场锦标赛。锦标赛共有N个人参加，共进行N-1轮。第一轮随机挑选两名选手进行决斗，胜者进入下一轮的比赛，第二轮到第N-1轮再每轮随机挑选1名选手与上一轮胜利的选手决斗，最后只剩一轮选手。第i名选手与第j名选手决斗，第i名选手胜利的概率是a[i][j].
作为一号选手的富榄想知道如何安排每轮出场的选手可以使得他获胜的概率最大，并求出这个最大概率。
 

一看数据范围肯定是状压DP，不过虽然是概率DP，但是需要倒着推；我们如果正着推式子的话，初始状态是不确定的，因为并不知道一开始把哪个人放在擂台上最后主角获胜的概率最大。所以我们可以假设主角最后获胜的概率是1，然后倒着推。设dp[i][j]表示现在站在擂台上的是i号选手，状态是j，主角获胜的最大概率，其中状态j的k位置是1代表第k - 1个选手还没有被淘汰。所以dp[i][j] = max(dp[i][j ^ (1 << k)] * a[i][k] + dp[k][j * (1 << i)] * a[k][i]).代表的决策是：现在站在擂台上的人是i,如果k去挑战有2种情况：1，k获胜了，那么转移到dp[k][j * (1 << i)], 转移概率是a[k][j]，k失败了同理。那么在当前状态选择k去挑战擂台的获胜的总概率是两种可能获胜概率的总和。

 

#include<bits/stdc++.h>
using namespace std;
double dp[(1<<18)+5][20];
double p[20][20];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<n;j++)
        {
            scanf("%lf",&p[i][j]);
        }
    }
    memset(dp,0,sizeof(dp));
    dp[1][0]=1;
    for(int i=2;i<(1<<n);i++)
    {
        for(int j=0;j<n;j++)
        {
            for(int k=0;k<n;k++)
            {
                if((i&(1<<k)))
                {
                    dp[i][j]=max(dp[i][j],dp[i^(1<<k)][j]*p[j][k]+dp[i^(1<<j)][k]*p[k][j]);
                }
            }
        }
    }
    double ma=0;
    for(int i=0;i<n;i++)
    {
        ma=max(dp[(1<<n)-1][i],ma);
    }
    printf("%.10f\n",ma);
    return 0;
}