User ainta has a permutation *p*<sub class="lower-index">1</sub>, *p*<sub class="lower-index">2</sub>, ..., *p*<sub class="lower-index">*n*</sub>. As the New Year is coming, he wants to make his permutation as pretty as possible.

Permutation *a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub> is prettier than permutation *b*<sub class="lower-index">1</sub>, *b*<sub class="lower-index">2</sub>, ..., *b*<sub class="lower-index">*n*</sub>, if and only if there exists an integer *k* (1 ≤ *k* ≤ *n*) where *a*<sub class="lower-index">1</sub> = *b*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub> = *b*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*k* - 1</sub> = *b*<sub class="lower-index">*k* - 1</sub> and *a*<sub class="lower-index">*k*</sub> &lt; *b*<sub class="lower-index">*k*</sub> 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*<sub class="lower-index">*i*</sub> and *p*<sub class="lower-index">*j*</sub> (1 ≤ *i*, *j* ≤ *n*, *i* ≠ *j*) if and only if *A*<sub class="lower-index">*i*, *j*</sub> = 1.

Given the permutation *p* and the matrix *A*, user ainta wants to know the prettiest permutation that he can obtain.
The first line contains an integer *n* (1 ≤ *n* ≤ 300) — the size of the permutation *p*.

The second line contains *n* space-separated integers *p*<sub class="lower-index">1</sub>, *p*<sub class="lower-index">2</sub>, ..., *p*<sub class="lower-index">*n*</sub> — 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 *A*<sub class="lower-index">*i*, *j*</sub> 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* &lt; *j* ≤ *n*, *A*<sub class="lower-index">*i*, *j*</sub> = *A*<sub class="lower-index">*j*, *i*</sub> holds. Also, for all integers *i* where 1 ≤ *i* ≤ *n*, *A*<sub class="lower-index">*i*, *i*</sub> = 0 holds.
import sys
input = lambda: sys.stdin.readline().rstrip()
from collections import deque,defaultdict,Counter
from itertools import permutations,combinations
from bisect import *
from heapq import *
from math import ceil,gcd,lcm,floor,comb
alph = 'abcdefghijklmnopqrstuvwxyz'
#pow(x,mod-2,mod)

def dfs(i):
    global N,A,P
    l = [i]
    ans = []
    index = []
    while l:
        x = l.pop()
        if vis[x]==1:continue
        index.append(x)
        ans.append(A[x])
        vis[x] = 1
        for y in P[x]:
            if vis[y]==1:continue
            l.append(y)

    ans.sort()
    num = 0
    for i in index:
        A[i] = ans[num]
        num+=1


N = int(input())
A = list(map(int,input().split()))
l = [input() for _ in range(N)]
P = [[] for _ in range(N)]

for x in range(N):
    for y in range(N):
        if l[y][x]=='1':
            P[x].append(y)
            P[y].append(x)

vis = [0]*N
for i in range(N):
    if vis[i]==1:continue
    dfs(i)
print(*A)

