寒假训练——搜索 K - Cycle

A tournament is a directed graph without self-loops in which every pair of vertexes is connected by exactly one directed edge. That is, for any two vertexes u and v (u ≠ v) exists either an edge going from u to v, or an edge from v to u.

You are given a tournament consisting of n vertexes. Your task is to find there a cycle of length three.

Input

The first line contains an integer n (1 ≤ n ≤ 5000). Next n lines contain the adjacency matrix A of the graph (without spaces). A_i, j = 1 if the graph has an edge going from vertex i to vertex j, otherwise A_i, j = 0. A_i, j stands for the j-th character in the i-th line.

It is guaranteed that the given graph is a tournament, that is, A_i, i = 0, A_i, j ≠ A_j, i (1 ≤ i, j ≤ n, i ≠ j).

Output

Print three distinct vertexes of the graph a₁, a₂, a₃ (1 ≤ a_i ≤ n), such that A_a1, a2 = A_a2, a3 = A_a3, a1 = 1, or "-1", if a cycle whose length equals three does not exist.

If there are several solutions, print any of them.

Examples

Input

5
00100
10000
01001
11101
11000

Output

1 3 2 

Input

5
01111
00000
01000
01100
01110

Output

-1
 

cycle
思路：
就是用搜索，以一排展开，dfs含有两个参数，这两个参数代表这个位置为1，所有，再去搜索，按照没有搜索的一排展开，
找到这个值。

AC之后，之前这个题目思路有一些混乱，再整理一下
在一个dfs要考虑到三个数，dfs里面的两个参数就是其中两个数，然后找到这两个参数中任意一个对应得第三个数，
再判断第三个数和另一个参数有没有相互对应，有就返回了，没有继续判断这一行有没有被标记，标记了就不用，没有标记就
继续进行搜索。值得确定得是，第一确实每一个都搜到了，没有遗漏，第二，就是搜索得递归，可以再仔细想想

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
using namespace std;
const int maxn=5050;
char ana[maxn][maxn];
bool bna[maxn][maxn],vis[maxn];
int n,a,b,c;
 
int dfs(int x,int y)
{
    vis[x]=1;
    for(int i=1;i<=n;i++)
    {
        if(bna[x][i])
        {
            if(y&&bna[i][y])
            {
                a=y;
                b=x;
                c=i;
                return 1;
            }
            if(!vis[i]) if(dfs(i,x)) return 1;
        }
    }
    return 0;
}
 
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%s",ana[i]+1);
        for(int j=1;j<=n;j++)
        {
            bna[i][j]=ana[i][j]-'0';
        }
    }
    for(int i=1;i<=n;i++)
    {
        if(!vis[i])
        {
            if(dfs(i,0))
            {
                printf("%d %d %d\n",a,b,c);
                return 0;
            }
        }
    }
    printf("-1\n");
    return 0;
}