New Year Permutation

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) wherea₁ = 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 ann × n binary matrix A, user ainta can swap the values of p_i andp_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 permutationp.

The second line contains n space-separated integersp₁, p₂, ..., p_n — the permutation p that user ainta has. Each integer between1 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 thei-th row of A. Thej-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.

Example

Input

75 2 4 3 6 7 10001001000000000000101000001000000000100001001000

Output

1 2 4 3 6 7 5

Input

54 2 1 5 30010000011100100110101010

Output

1 2 3 4 5

Note

In the first sample, the swap needed to obtain the prettiest permutation is: (p₁, p₇).

In the second sample, the swaps needed to obtain the prettiest permutation is (p₁, p₃), (p₄, p₅), (p₃, p₄).

A permutation p is a sequence of integersp₁, p₂, ..., p_n, consisting of n distinct positive integers, each of them doesn't exceedn. The i-th element of the permutationp is denoted as p_i. The size of the permutationp is denoted as n.

题意：给n个数a[]，然后给n*n的关系矩阵g[][]，如果关系矩阵中a[i][j] == 1，则代表a[i]和a[j]等价，可以交换。问经过交换能得到的字典序最小的序列为多少。

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<stdio.h>#include<iostream>#include<string.h>#include<algorithm>using namespace std;int a[400];char b[400][400];int c[1000][1000];int main(){    int n;    while(scanf("%d",&n)!=-1)    {        for(int i=0;i<n;i++)        {            scanf("%d",&a[i]);        }        memset(c,0,sizeof(c));        for(int i=0;i<n;i++)        {            scanf("%s",b[i]);            for(int j=0;j<n;j++)            {                c[i][j]=b[i][j]-'0';            }        }        for(int i=0;i<n;i++)        {            for(int j=0;j<n;j++)            {                for(int k=0;k<n;k++)                {                    if(c[j][i]==1&&c[i][k]==1)                    {                        c[j][k]=1;                    }                }            }        }        int t;        for(int i=0;i<n;i++)        {            for(int j=i+1;j<n;j++)            {                if(a[i]>a[j]&&c[i][j]==1)                {                    t=a[i];                    a[i]=a[j];                    a[j]=t;                }            }        }        for(int i=0;i<n;i++)        {            printf("%d ",a[i]);        }        printf("\n");    }}