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

Permutation a1, a2, ..., an is prettier than permutation b1, b2, ..., bn, if and only if there exists an integer k (1 ≤ k ≤ n) where a

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

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, ..., pn — 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 Ai, 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, Ai, j = Aj, i holds. Also, for all integers i where 1 ≤ i ≤ n, Ai, i = 0 holds.

Output

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

Examples

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

題意:給n個數，要求這n個數字小的儘量放到前面，求一個最小的。

給一個矩陣a[i][j]==1，表示位置 i 的數字可以和 位置 j 的數字交換。

題解：Floyd 確定下哪些位置可換，資料並不大，選擇排序搞一下即可

#include<bits/stdc++.h>
using namespace std;
const int N=300+10;
int limit[N][N];
int n,a[N];
int main()
{
	cin>>n;
	for(int i=1;i<=n;i++)cin>>a[i];
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			scanf("%1d",&limit[i][j]);
	for(int k=1;k<=n;k++)
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++)
	if(limit[i][k]&&limit[k][j])
		limit[i][j]=1;
	
	for(int i=1;i<n;i++)
	{
		int k=i;
		for(int j=i+1;j<=n;j++)
		{
			if(a[k]>a[j]&&limit[i][j])
				k=j;
		}	
		swap(a[i],a[k]);
	} 
	for(int i=1;i<=n;i++)printf("%d%c",a[i]," \n"[i==n]);
	
	return 0;
}