New Year Permutation



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

Permutation a₁, a₂, ..., a_n is prettier than permutation b₁, b₂, ..., b_n, if and only if there exists an integer k (1 ≤ k ≤ n) wherea₁ = b₁, a₂ = b₂, ..., a_k - 1 = b_k - 1 and a_k < b_k 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_i and p_j (1 ≤ i, j ≤ n, i ≠ j) if and only ifA_i, 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, ..., p_n — the permutation p that user ainta has. Each integer between 1and 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 thei-th line A_i, 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 where1 ≤ i < j ≤ n, A_i, j = A_j, i holds. Also, for all integers i where 1 ≤ i ≤ n, A_i, i = 0 holds.

Output

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

Sample test(s)

input

75 2 4 3 6 7 10001001000000000000101000001000000000100001001000

output

1 2 4 3 6 7 5

input

54 2 1 5 30010000011100100110101010

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).

A permutation p is a sequence of integers p₁, p₂, ..., p_n, consisting of n distinct positive integers, each of them doesn't exceed n. Thei-th element of the permutation p is denoted as p_i. The size of the permutation p is denoted as n.

一个数列，如果前k个数相等，a[k+1]<b[k+1]那么 a比b好，给出一个矩阵，如果aij=1，那么i和j位置的数是可以交换的，求给出矩阵状态下最好的数列。其实这个比较法就是尽量把小的数放在前面，但是加了一个交换的位置条件。自己想法就是把每个位置遍历，然后把连通分支整个找出来，再内部排序，赋值，接着找下一个连通分支，重复上述动作，循环n次，虽然有点重复，但是只有300个点，并不会超时。话说1A真开心。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<queue>using namespace std;int b[303][303];int a[303], leap[303], dp[303];queue<int>p;int main(){int i, j, m, n, ans, temp;char c;cin >> n;for (i = 1; i <= n; i++)cin >> a[i];getchar();for (i = 1; i <= n; i++){for (j = 1; j <= n; j++){c = getchar();b[i][j] = c - '0';}getchar();}for (i = 1; i <= n; i++){p.push(i);memset(leap, 0, sizeof(leap));leap[i] = 1;while (p.size()){temp = p.front();for (j = 1; j <= n; j++){if (b[temp][j] == 1 && !leap[j]){p.push(j);leap[j] = 1;}}p.pop();}ans = 0;for (j = 1; j <= n; j++){if (leap[j]){ans++;dp[ans] = a[j];}}sort(dp + 1, dp + 1 + ans);ans = 0;for (j = 1; j <= n; j++){if (leap[j]){ans++;a[j] = dp[ans];}}}for (i = 1; i <= n; i++)cout << a[i] << " ";cout << endl;return 0;}

这是我当时第一次做cf的b题，当时的a题还是50分钟这样做出来的，b题就更不敢奢望了，现在感觉自己有提升，但是 作为一个初学者，想要获得提升是很容易的，还要继续走下去。