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.
def dfs(pos, visited, graph, p, prettiest):
    visited[pos] = True
    for i in graph[pos]:
        if not visited[i] and p[i] < prettiest:
            prettiest = p[i]
        if p[pos] < p[i] and p[i] < prettiest:
            prettiest = p[i]
        if graph[pos][i] and not visited[i]:
            graph[pos][i] = False
            graph[i][pos] = True
            p[pos], p[i] = p[i], p[pos]
            prettiest = dfs(i, visited, graph, p, prettiest)
            p[pos], p[i] = p[i], p[pos]
            graph[pos][i] = True
            graph[i][pos] = False
    return prettiest

n = int(input())
p = list(map(int, input().split()))
p_copy = p[:]
A = [[0] * n for _ in range(n)]
graph = [[False] * n for _ in range(n)]
for i in range(n):
    A[i] = list(map(int, input().split()))
for i in range(n):
    for j in range(n):
        if A[i][j]:
            graph[i][j] = True
prettiest = min(p)
visited = [False] * n
for i in range(n):
    if not visited[i]:
        prettiest = dfs(i, visited, graph, p_copy, prettiest)
print(' '.join(map(str, p_copy)))
