codeforces500B

New Year Permutation

 CodeForces - 500B 

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) where a₁ = b₁, a₂ = b₂, ..., a_k - 1 = b_k - 1 and a_k < b_kall 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 if A_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 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_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 where 1 ≤ 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

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).

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. The i-th element of the permutation p is denoted as p_i. The size of the permutation p is denoted as n.

 

sol：十分显然的是a可以与b换，b可以与c换，a就可以和c换，于是把可以换的用并查集缩成一个个连通块，在每个联通块中做到最优，那么就是答案了

#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
    ll s=0;
    bool f=0;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<3)+(s<<1)+(ch^48); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<0)
    {
        putchar('-'); x=-x;
    }
    if(x<10)
    {
        putchar(x+'0');    return;
    }
    write(x/10);
    putchar((x%10)+'0');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int N=305;
int n,a[N],Pos[N];
bool Bo[N][N];
int Father[N]; 
inline int Get_Father(int x)
{
    return (Father[x]==x)?(x):(Father[x]=Get_Father(Father[x]));
}
struct Record
{
    int Shuz,Weiz;
};
inline bool cmp_Shuz(Record p,Record q)
{
    return p.Shuz<q.Shuz;
}
inline bool cmp_Weiz(Record p,Record q)
{
    return p.Weiz<q.Weiz;
}
vector<Record>Liantong[N];
vector<int>Xulie[N];
int main()
{
    int i,j;
    R(n);
    for(i=1;i<=n;i++)
    {
        R(a[i]); Father[i]=i;
    }
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=n;j++)
        {
            char ch=' ';
            while(!isdigit(ch)) ch=getchar();
            if(ch=='1')
            {
                Father[Get_Father(i)]=Get_Father(j);
            }
        }
    }
    for(i=1;i<=n;i++)
    {
        Liantong[Get_Father(i)].push_back((Record){a[i],i});
    }
    for(i=1;i<=n;i++) if(Liantong[i].size())
    {
        sort(Liantong[i].begin(),Liantong[i].end(),cmp_Shuz);
        for(j=0;j<Liantong[i].size();j++) Xulie[i].push_back(Liantong[i][j].Shuz);
    }
    for(i=1;i<=n;i++) if(Liantong[i].size())
    {
        sort(Liantong[i].begin(),Liantong[i].end(),cmp_Weiz);
        for(j=0;j<Liantong[i].size();j++) a[Liantong[i][j].Weiz]=Xulie[i][j];
    }
    for(i=1;i<n;i++) W(a[i]); Wl(a[n]);
    return 0;
}
/*
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
*/