Problem:
Score : 500 points

Problem Statement

Given is a connected undirected graph with N vertices and M edges. The vertices are numbered 1 to N, and the edges are described by a grid of characters S.
If S_{i,j} is 1, there is an edge connecting Vertex i and j; otherwise, there is no such edge.

Determine whether it is possible to divide the vertices into non-empty sets V_1, \dots, V_k such that the following condition is satisfied. If the answer is yes, find the maximum possible number of sets, k, in such a division.

Every edge connects two vertices belonging to two "adjacent" sets. More formally, for every edge (i,j), there exists 1\leq t\leq k-1 such that i\in V_t,j\in V_{t+1} or i\in V_{t+1},j\in V_t holds.

Constraints

2 \leq N \leq 200

S_{i,j} is 0 or 1.

S_{i,i} is 0.

S_{i,j}=S_{j,i}

The given graph is connected.

N is an integer.

Input

Input is given from Standard Input in the following format:

N
S_{1,1}...S_{1,N}
:
S_{N,1}...S_{N,N}

Output

If it is impossible to divide the vertices into sets so that the condition is satisfied, print -1.
Otherwise, print the maximum possible number of sets, k, in a division that satisfies the condition.

Sample Input 1

2
01
10

Sample Output 1

2

We can put Vertex 1 in V_1 and Vertex 2 in V_2.

Sample Input 2

3
011
101
110

Sample Output 2

-1

Sample Input 3

6
010110
101001
010100
101000
100000
010000

Sample Output 3

4

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
int main()
{
    short a[201][201];
    short n,c=0;
    scanf("%hd",&n);
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            scanf("%hd",&a[i][j]);
        }
    }
    for(int i=2;i<=n;i++)
    {
        for(int j=1;j<i;j++)
        {
            if(a[i][j]==1)
            {
                for(int k=1;k<=n;k++)
                {
                    if(a[i][k]&a[j][k]==1)
                    {
                        c++;
                        break;
                    }
                }
            }
            if(c>0)
            {
                printf("-1\n");
                break;
            }
        }
        if(c>0)
            break;
    }
    if(c==0)
    {
        short b=0;
        int sum=0;
        for(short i=1;i<=n;i++)
        {
            b=0;
            for(short j=1;j<=n;j++)
            {
                b=b^a[j][i];
            }
            if(b==1)
                sum++;
        }
        printf("%d\n",sum);
    }
    return 0;
}