500_B. New Year Permutation
User ainta has a permutation p1, p2, ..., pn. As the New Year is coming, he wants to make his permutation as pretty as possible.

Permutation a1, a2, ..., an is prettier than permutation b1, b2, ..., bn, if and only if there exists an integer k (1 ≤ k ≤ n) where a1 = b1, a2 = b2, ..., ak - 1 = bk - 1 and ak < bk 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 pi and pj (1 ≤ i, j ≤ n, i ≠ j) if and only if Ai, j = 1.

Given the permutation p and the matrix A, user ainta wants to know the prettiest permutation that he can obtain.

Input

The first line contains an integer n (1 ≤ n ≤ 300) — the size of the permutation p.

The second line contains n space-separated integers p1, p2, ..., pn — 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 Ai, j 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 < j ≤ n, Ai, j = Aj, i holds. Also, for all integers i where 1 ≤ i ≤ n, Ai, i = 0 holds.

Output

In the first and only line, print n space-separated integers, describing the prettiest permutation that can be obtained.

Examples

Input

7
5 2 4 3 6 7 1
0001001
0000000
0000010
1000001
0000000
0010000
1001000


Output

1 2 4 3 6 7 5


Input

5
4 2 1 5 3
00100
00011
10010
01101
01010


Output

1 2 3 4 5

Note

In the first sample, the swap needed to obtain the prettiest permutation is: (p1, p7).

In the second sample, the swaps needed to obtain the prettiest permutation is (p1, p3), (p4, p5), (p3, p4). 

<image>

A permutation p is a sequence of integers p1, p2, ..., pn, consisting of n distinct positive integers, each of them doesn't exceed n. The i-th element of the permutation p is denoted as pi. The size of the permutation p is denoted as n.
```python
import sys

n=int(input())
a=list(map(lambda x:int(x),sys.stdin.readline().split()))
matrix=[]
for _ in range(n):
    row=input()
    row=list(row)
    row=list(map(lambda x:int(x),row))
    matrix.append(row)

class DisjSet:
    def __init__(self, n):
        # Constructor to create and
        # initialize sets of n items
        self.rank = [1] * n
        self.parent = [i for i in range(n)]
        self.extra = []

    # Finds set of given item x
    def find(self, x):

        # Finds the representative of the set
        # that x is an element of
        if (self.parent[x] != x):
            # if x is not the parent of itself
            # Then x is not the representative of
            # its set,
            self.parent[x] = self.find(self.parent[x])

        # so we recursively call Find on its parent
        # and move i's node directly under the
        # representative of this set

        return self.parent[x]

    # Do union of two sets represented
    # by x and y.
    def Union(self, x, y):

        # Find current sets of x and y
        xset = self.find(x)
        yset = self.find(y)

        # If they are already in same set
        if xset == yset:
            self.extra.append((x, y))
            return

        # Put smaller ranked item under
        # bigger ranked item if ranks are
        # different
        if self.rank[xset] < self.rank[yset]:
            self.parent[xset] = yset

        elif self.rank[xset] > self.rank[yset]:
            self.parent[yset] = xset

        # If ranks are same, then move y under
        # x (doesn't matter which one goes where)
        # and increment rank of x's tree
        else:
            self.parent[yset] = xset
            self.rank[xset] = self.rank[xset] + 1
obj=DisjSet(n)
for i in range(n):
    for j in range(n):
        if matrix[i][j]==1:
            obj.Union(i,j)
hashmap={}
for i in range(n):
    temp=obj.find(i)
    if temp not in hashmap:
        hashmap[temp]=[]
    hashmap[temp].append(i)
final=[0 for _ in range(n)]
for k in hashmap:
    temp=[a[i] for i in hashmap[k]]
    temp.sort()
    for i in range(0,len(temp)):
        final[hashmap[k][i]]=temp[i]
print(*final)


```