cf 500B

B - New Year Permutation

Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Appoint description:

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 ifA_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 andn 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 thei-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

Output

1 2 3 4 5

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<queue>
#include<vector>
#include<set>
using namespace std;
int a[100010],n;
char s[410][410];
int main()
{
     scanf("%d",&n);
     for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
     for(int i=0;i<n;i++)
      scanf("%s",s[i]);
     for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
                  for(int k=0;k<n;k++)
                        if(s[j][i]=='1'&&s[i][k]=='1')
                              s[j][k]='1';
     for(int i=0;i<n;i++)
            for(int j=i+1;j<n;j++)
                  if(s[i][j]=='1'&&a[i]>a[j])
                        swap(a[i],a[j]);
     for(int i=0;i<n-1;i++)
            printf("%d ",a[i]);
     printf("%d
",a[n-1]);
     return 0;
}