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 a1 = b1, a2 = b2, ..., ak - 1 = bk - 1 and ak < bk 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 pi and pj (1 ≤ i, j ≤ n, i ≠ j) if and only if Ai, 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, ..., 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

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). 

<image>

A permutation p is a sequence of integers p1, p2, ..., pn, consisting of n distinct positive integers, each of them doesn't exceed n. The i-th element of the permutation p is denoted as pi. The size of the permutation p is denoted as n.
import java.util.*;
import java.io.*;
 
 
public class gfg {

    //static int n =0;
    public static void main(String[] args) throws IOException{
        Reader.init(System.in);
        //PrintWriter out = new PrintWriter(System.out);

        int t = 1;//Reader.nextInt();
        while(t-->0){
           solve();
        }
        //out.close();
    }


    static TreeSet<Integer>[] graph;
    static int n=0;
    static int m=0;
    static boolean[] vis;
    static TreeSet<Integer> set= new TreeSet<>();
    static int cnt=0;
    static long mod =1000000007;
    static int[] arr;
    static PriorityQueue<Integer> ind = new PriorityQueue<>();
    static PriorityQueue<Integer> val = new PriorityQueue<>();

    static void solve()throws IOException{
        n = Reader.nextInt();
        graph = new TreeSet[n];
        arr  = new int[n];
        for(int i=0;i<n;i++){
            graph[i] = new TreeSet<Integer>();
        }
        for(int i=0;i<n;i++){
            arr[i] = Reader.nextInt();
        }
        vis = new boolean[n];

        for(int i=0;i<n;i++){
            String s = Reader.next();
            char[] c = s.toCharArray();
            for(int j=0;j<n;j++){
                if(c[j]=='1')
                {
                    graph[i].add(j);
                    graph[j].add(j);
                }     
            }
        }
        for(int i=0;i<n;i++){
            if(!vis[i]){
                dfs(i);
                while(ind.size()>0){
                    int a = ind.poll();
                    int v = val.poll();
                    arr[a]=v;
                }
            }
            
        }
        for(int i=0;i<n;i++){
            System.out.print(arr[i]+" ");
        }
        return;

    }

    static void dfs(int a){
        vis[a]=true;
        ind.add(a);
        val.add(arr[a]);
        for(int x:graph[a]){
            if(!vis[x])
                dfs(x);
        }
    }
    
    
    // ggratest common divisor
    static int gcd(int a , int b){
        if(b==0)
            return a;
        else
            return gcd(b,a%b);
    }
 
    // least common multiple
    static int lcm(int a, int b){
        return (a*b)/gcd(a,b);
    }
    
    // a^b
    static long fastpow(long a, long b){
        long res = 1;a=a%mod;
        while(b>0){
            if(b%2==1)
                res = (res*a)%mod;
            a = (a*a)%mod;
            b = b>>1;
        }
        return res;
    }
 
    
}
 
 
class Reader {
    static BufferedReader reader;
    static StringTokenizer tokenizer;
 
    /** call this method to initialize reader for InputStream */
    static void init(InputStream input) {
        reader = new BufferedReader(
                     new InputStreamReader(input) );
        tokenizer = new StringTokenizer("");
    }
 
    /** get next word */
    static String next() throws IOException {
        while ( ! tokenizer.hasMoreTokens() ) {
            //TODO add check for eof if necessary
            tokenizer = new StringTokenizer(
                   reader.readLine() );
        }
        return tokenizer.nextToken();
    }
 
    static int nextInt() throws IOException {
        return Integer.parseInt( next() );
    }
    static long nextLong() throws IOException {
        return Long.parseLong( next() );
    }
    
    static double nextDouble() throws IOException {
        return Double.parseDouble( next() );
    }
}
class newcomparator implements Comparator<Integer>{
    //@Override
    public int compare(Integer a, Integer  b){
        return a<=b?1:-1;
    }
 }
class node {
    int a;
    long b;// cost to rreach\
    node(int s,long va){
        a=s;
        b=va;
    }
}
class mergesort{
    static void sort(int l, int h, int[] arr){
        if(l<h){
            int mid = (l+h)/2;
            
            sort(l,mid,arr);
            sort(mid+1,h,arr);
            merge(l,mid,h,arr);
        }
    }
    static void merge(int l, int m , int h , int [] arr){
 
        int[] left = new int[m-l+1];
        int[] right = new int[h-m];
        for(int i= 0 ;i< m-l+1;i++){
            left[i] = arr[l+i];
        }
        for(int i=0;i<h-m;i++){
            right[i] = arr[m+1+i];
        }
        //now left and right arrays are assumed to be sorted and we have tp merge them together 
        // int the original aaray.
        int i=l;
        int lindex = 0;
        int rindex =  0;
        
        while(lindex<m-l+1 && rindex<h-m){
            if(left[lindex]<=right[rindex]){
                arr[i]=left[lindex];
                lindex++;
                i++;
            }
            else{
                arr[i]=right[rindex];
                rindex++;
                i++;
            }
        }
        while(lindex<m-l+1){
            arr[i]=left[lindex];
            lindex++;
            i++;
        }
        while(rindex<h-m){
            arr[i]=right[rindex];
            rindex++;
            i++;
        }
    }
    
}



