New Year Permutation（Floyd+并查集）

Description

User ainta has a permutation p₁, p₂, ..., p_n. As the New Year is coming, he wants to make his permutation as pretty as possible.

Permutation a₁, a₂, ..., a_n is prettier than permutation b₁, b₂, ..., b_n, if and only if there exists an integer k (1 ≤ k ≤ n) where a₁ = b₁, a₂ = b₂, ..., a_k - 1 = b_k - 1 and a_k < b_k all holds.

As known, permutation p is so sensitive that it could be only modified by swapping two distinct elements. But swapping two elements is harder than you think. Given an n × n binary matrix A, user ainta can swap the values of p_i and p_j (1 ≤ i, j ≤ n, i ≠ j) if and only if A_i, j = 1.

Given the permutation p and the matrix A, user ainta wants to know the prettiest permutation that he can obtain.

Input

The first line contains an integer n (1 ≤ n ≤ 300) — the size of the permutation p.

The second line contains n space-separated integers p₁, p₂, ..., p_n — the permutation p that user ainta has. Each integer between 1 and n occurs exactly once in the given permutation.

Next n lines describe the matrix A. The i-th line contains n characters '0' or '1' and describes the i-th row of A. The j-th character of the i-th line A_i, j is the element on the intersection of the i-th row and the j-th column of A. It is guaranteed that, for all integers i, j where 1 ≤ i < j ≤ n, A_i, j = A_j, i holds. Also, for all integers i where 1 ≤ i ≤ n, A_i, i = 0 holds.

Output

In the first and only line, print n space-separated integers, describing the prettiest permutation that can be obtained.

Sample Input

Input

7
5 2 4 3 6 7 1
0001001
0000000
0000010
1000001
0000000
0010000
1001000

Output

1 2 4 3 6 7 5

Input

5
4 2 1 5 3
00100
00011
10010
01101
01010

Output

1 2 3 4 5

思路：
用floyd先将左右潜在的可以连通的边在邻接矩阵中标记出来，然后像冒泡一样的给他们排序一下就可以了
要注意两点：
一是刚开始开数组的时候，要从0开始开，不然用scanf输入会很麻烦
二是floyd的三层for循环，注意k的顺序要在最前面

这题也是对并查集数据结构的一种很好的理解，不过忽略了并查集分组的性质，而只关注了并查集的连通性
即通过邻接矩阵得到扩展边与边的连通关系

---

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

int n;
int p[];
char f[][];

int main()
{
    while(~scanf("%d",&n))
    {
        for(int i = ;i < n;i++)
            scanf("%d",&p[i]);
        for(int i = ;i < n;i++)
        {
            scanf("%s",f[i]);
            for(int j = ;j < n;j++)
                if(f[i][j] == '')
                    f[i][j] = '';
        }
        for(int k = ;k < n;k++)
            for(int i = ;i < n;i++)
                for(int j = ;j < n;j++)
                    if(f[i][k]=='' && f[k][j]=='')
                        f[i][j] = '';

        for(int i = ;i < n;i++)
            for(int j = i+;j < n;j++)
                if(f[i][j]=='' && p[j] < p[i])
                    swap(p[i],p[j]);

        for(int i = ;i < n-;i++)
            printf("%d ",p[i]);
        printf("%d\n",p[n-]);
    }
    return ;
}