【Good Bye 2014B】【Floyd or 并查集】New Year Permutation 全排列有位置交换序列 使得字典序尽可能小

New Year Permutation

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

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.

#include<stdio.h>#include<iostream>#include<string.h>#include<string>#include<ctype.h>#include<math.h>#include<set>#include<map>#include<vector>#include<queue>#include<bitset>#include<algorithm>#include<time.h>using namespace std;void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }#define MS(x,y) memset(x,y,sizeof(x))#define MC(x,y) memcpy(x,y,sizeof(x))#define MP(x,y) make_pair(x,y)#define ls o<<1#define rs o<<1|1typedef long long LL;typedef unsigned long long UL;typedef unsigned int UI;template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }const int N = 303, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;int a[N], p[N];char s[N][N];int n;bool use[N];int ans[N];int main(){while (~scanf("%d", &n)){for (int i = 1; i <= n; ++i){scanf("%d", &a[i]);p[a[i]] = i;}for (int i = 1; i <= n; ++i){scanf("%s", s[i] + 1);s[i][i] = '1';}for (int k = 1; k <= n; ++k){for (int i = 1; i <= n; ++i){for (int j = 1; j <= n; ++j){if (s[i][k] == '1'&&s[k][j] == '1')s[i][j] = '1';}}}MS(use, 0);for (int i = 1; i <= n; ++i){for (int j = 1; j <= n; ++j)if(!use[j]&&s[i][p[j]]=='1'){ans[i] = j;use[j] = 1;p[j] = i;p[a[i]] = p[j];break;}}for (int i = 1; i <= n; ++i)printf("%d ", ans[i]);puts("");}return 0;}/*【题意】给你一个长度为n（300）的全排列。同时给你一个n*n的01矩阵s[][]我们可以交换两个数，当且仅当它们所在的的下标i,j对应着的s[i][j]为1。问你我们可以得到的最小字典序的排列【类型】floyd【分析】这题可以用floyd做。我们对矩阵做闭包传递的floyd。(或者——并查集)那么，任意两个位置的数是否可以交换我们就知道了。这个类似于冒泡排序的过程，只要有一个交换链存在，任意一个数都可以位移到它想要的位置。于是这种做法是可行的。就可以AC啦。【时间复杂度&&优化】O(n^3)【数据】*/