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 = sys.stdin.readline
from collections import defaultdict

n = int(input())
w = list(map(int, input().split()))
g = [input()[:-1] for _ in range(n)]
d = [[] for i in range(n)]
for i in range(n):
    for j in range(i+1, n):
        if g[i][j] == '1':
            d[i].append(j)
            d[j].append(i)
x = [0]*n
c = 1
for i in range(n):
    if x[i] == 0:
        q = [i]
        s = set()
        while q:
            a = q.pop()
            x[a] = c
            for j in d[a]:
                if x[j] == 0:
                    q.append(j)
    c += 1
q = defaultdict(list)
for i, j in enumerate(x):
    q[j].append((w[i], i))
x = [0] * n
for i in q:
    d1 = []
    d2 = []
    for j, l in q[i]:
        d1.append(j)
        d2.append(l)
    d1.sort()
    d2.sort()
    for j in range(len(d1)):
        x[d2[j]] = d1[j]
print(x)