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.
n = int(input())
p = list(map(int, input().split()))
A = [list(map(int, input())) for _ in range(n)]

parent = list(range(n))
rank = [0] * n

def find(x):
    if parent[x] != x:
        parent[x] = find(parent[x])
    return parent[x]

def union(x, y):
    x_root = find(x)
    y_root = find(y)
    if x_root == y_root:
        return
    if rank[x_root] < rank[y_root]:
        parent[x_root] = y_root
    elif rank[x_root] > rank[y_root]:
        parent[y_root] = x_root
    else:
        parent[y_root] = x_root
        rank[x_root] += 1

for i in range(n):
    for j in range(i + 1, n):
        if A[i][j] == 1:
            union(i, j)

groups = {}
for i in range(n):
    root = find(i)
    if root not in groups:
        groups[root] = []
    groups[root].append(p[i])

for group in groups.values():
    group.sort()

ans = [0] * n
used = [False] * n
for i in range(n):
    root = find(i)
    if not used[root]:
        used[root] = True
        j = 0
        for k in range(n):
            if find(k) == root:
                ans[k] = groups[root][j]
                j += 1

print(*ans)