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.
class DSU:

	def __init__(self, n):
		self.dsu = [i for i in range(n)]
		self.weights = [0] * n 

	def find(self, u):
		if u == self.dsu[u]:
			return u
		self.dsu[u] = self.find(self.dsu[u])
		return self.dsu[u]

	def union(self, u, v):
		u = self.find(u)
		v = self.find(v)
		if u!=v:
			if self.weights[u] < self.weights[v]:
				u, v = v, u
			self.dsu[v] = u 
			if self.weights[u] == self.weights[v]:
				self.weights[u] += 1

n = int(input())
p = [int(x) for x in input().split()]
dsu = DSU(n)

for i in range(n):
	line = input()
	for j in range(n):
		if line[j] == '1':
			dsu.union(i, j)

components = {}
for i in range(n):
	par = dsu.find(i)
	if par not in components:
		components[par] = []
	components[par].append(i)

for comp in components:
	arr = [p[a] for a in components[comp]]
	arr = sorted(arr)
	for idx, a in enumerate(components[comp]):
		p[a] = arr[idx]

print(" ".join([str(x) for x in p])) 