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.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException; 
import java.io.InputStreamReader; 
import java.util.Scanner; 
import java.util.StringTokenizer; 
import java.util.*;
import java.io.*;
public class codeforces 
{ 
   static class Student{
        double e;
       String s;
        Student(String s,double e){
            this.s=s;
            this.e=e;
        }
    }
    static int sieveOfEratosthenes(int n) 
    { 
        // Create a boolean array "prime[0..n]" and initialize 
        // all entries it as true. A value in prime[i] will 
        // finally be false if i is Not a prime, else true. 
        int pos=0;
        boolean prime[] = new boolean[n+1]; 
        for(int i=0;i<=n;i++) 
            prime[i] = true; 
          
        for(int p = 2; p*p <=n; p++) 
        { 
            // If prime[p] is not changed, then it is a prime 
            if(prime[p] == true) 
            { 
                // Update all multiples of p 
                for(int i = p*p; i <= n; i += p) 
                    prime[i] = false; 
            } 
        } 
          
        // Print all prime numbers 
        for(int i = 2; i <= n; i++) 
        { 
            if(prime[i] == true) 
                ++pos;
        }
        return pos;
    }
static    class Sortbyroll implements Comparator<Student> 
{ 
    // Used for sorting in ascending order of 
    // roll number 
    public int compare(Student c, Student b) 
    { 
        String s1,s2;
        s1=c.s.toLowerCase();
        s2=b.s.toLowerCase();
    if(c.e>b.e)
    return -1;
    if(c.e<b.e)
    return 1;
    return s1.compareTo(s2);
    } 
} 
    static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    }
    static class Edge{
        int a,b;
        Edge(int a,int b){
            this.a=a;
            this.b=b;
        }
    }
    public static void main(String[] args){
        //long sum=0;
        int t,i,j,k,l,pos,n;
        String r,p;
        FastReader sc=new FastReader();
        n=sc.nextInt();
        int a[]=new int[n];
        for(i=0;i<n;i++)
            a[i]=sc.nextInt()-1;
        String s[]=new String[n];
        for(i=0;i<n;i++){
            s[i]=sc.nextLine();
        }
        int b[]=new int[n];
        Queue<Integer>q=new LinkedList<>();
        ArrayList<Integer>v;
        ArrayList<Integer>val;
        int ans[]=new int[n];
        for(i=0;i<n;i++){
            if(b[i]==0){
                b[i]=1;
                v=new ArrayList<>();
                val=new ArrayList<>();
                v.add(i);
                val.add(a[i]);
                q.add(i);
                while(!q.isEmpty()){
                    for(j=0;j<n;j++){
                        if(b[j]==0&&s[q.peek()].charAt(j)=='1'){
                            b[j]=1;
                            v.add(j);
                            val.add(a[j]);
                            q.add(j);
                        }
                    }
                    q.poll();
                }
                Collections.sort(v);
                Collections.sort(val);
                      /*  for(i=0;i<v.size();i++)
            System.out.print(val.get(i)+" ");
                break;*/
                for(j=0;j<v.size();j++){
                    ans[v.get(j)]=val.get(j)+1;
                }
            }
        }
        for(i=0;i<n;i++)
            System.out.print(ans[i]+" ");
    }
 static long power(long x, long y, long p) 
{ 
    long res = 1;      // Initialize result 
  
    x = x % p;  // Update x if it is more than or  
                // equal to p 
  
    while (y > 0) 
    { 
        // If y is odd, multiply x with result 
        if ((y & (long)1)%2!=0) 
            res = (res*x) % p; 
  
        // y must be even now 
        y = y>>1; // y = y/2 
        x = (x*x) % p;   
    } 
    return res%p; 
} 
   static int find(int x,int parent[]) 
    { 
        // Finds the representative of the set 
        // that x is an element of 
        while(parent[x]!=x) 
        { 
            // if x is not the parent of itself 
            // Then x is not the representative of 
            // his set,
            x=parent[x];
  
            // so we recursively call Find on its parent 
            // and move i's node directly under the 
            // representative of this set 
        } 
  
        return x; 
    }
    static void union(int x, int y,int rank[],int parent[]) 
    { 
        // Find representatives of two sets 
        int xRoot = find(x,parent), yRoot = find(y,parent); 
  
        // Elements are in the same set, no need 
        // to unite anything. 
        if (xRoot == yRoot) 
            return; 
  
         // If x's rank is less than y's rank 
        if (rank[xRoot] < rank[yRoot]) 
  
            // Then move x under y  so that depth 
            // of tree remains less 
            parent[xRoot] = yRoot; 
  
        // Else if y's rank is less than x's rank 
        else if (rank[yRoot] < rank[xRoot]) 
  
            // Then move y under x so that depth of 
            // tree remains less 
            parent[yRoot] = xRoot; 
  
        else // if ranks are the same 
        { 
            // Then move y under x (doesn't matter 
            // which one goes where) 
            parent[yRoot] = xRoot; 
  
            // And increment the the result tree's 
            // rank by 1 
            rank[xRoot] = rank[xRoot] + 1; 
        } 
    } 
    static long gcd(long a, long b) 
    { 
        if (a == 0) 
            return b; 
        return gcd(b % a, a); 
    }
}