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# (Mono 4.6.2.0) (judge verdict: Time Limit Exceeded)

using System;
using System.Collections;
using System.Collections.Generic;
using System.IO;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace AtCorder
{
	public class Program
	{
		static void Main(string[] args)
		{
			Scanner cin = new Scanner();
			int n = cin.Int();

			int inf = 100000;
			int[,] d = new int[n, n];
			for (int i = 0; i < n; i++) {
				for (int j = 0; j < n; j++) {
					if (i != j) {
						d[i, j] = inf;
					}
				}
			}

			List<int>[] to = new List<int>[n];
			for (int i = 0; i < n; i++) {
				to[i] = new List<int>();
			}

			for (int i = 0; i < n; i++) {
				string t = cin.Next();
				for (int j = 0; j < n; j++) {
					if (t[j] == '1') {
						to[i].Add(j);
						d[i,j] = 1;
					}
				}
			}

			//わーシャルふろいど
			for (int i = 0; i < n; i++) {
				for (int j = 0; j < n; j++) {
					for (int k = 0; k < n; k++) {
						d[j, k] = Math.Min(d[j, k], d[j, i] + d[i, k]);
					}
				}
			}

			int start = 0;
			int max = 0;
			for (int i = 0; i < n; i++) {
				int tempMax = 0;
				for (int j = 0; j < n; j++) {
					if (d[i, j] < inf) {
						tempMax = Math.Max(d[i, j], tempMax);
					}
				}

				if (tempMax > max) {
					max = tempMax;
					start = i;
				}
			}

			var q = new Queue<Tuple<int, int>>();
			q.Enqueue(new Tuple<int, int>(start, 0));
			while(q.Count > 0){
				var target = q.Dequeue();
				foreach (int next in to[target.Item1]) {
					int dist = d[start, next];
					if (dist == target.Item2 + 1 || dist == target.Item2 - 1) {
						if (dist == target.Item2 + 1) {
							q.Enqueue(new Tuple<int, int>(next, dist));
						}
					} else {
						Console.WriteLine(-1);
						return;
					}
				}
			}

			Console.WriteLine(max + 1);
		}
	}

	public class Scanner
	{
		private readonly char[] delimiter_ = new char[] { ' ' };
		private string[] buf_;
		private int index_;

		public Scanner()
		{
			buf_ = new string[0];
			index_ = 0;
		}

		public string Next()
		{
			if (index_ < buf_.Length) {
				return buf_[index_++];
			}

			string st = Console.ReadLine();
			while (st == "") {
				st = Console.ReadLine();
			}

			buf_ = st.Split(delimiter_, StringSplitOptions.RemoveEmptyEntries);
			if (buf_.Length == 0) {
				return Next();
			}

			index_ = 0;
			return buf_[index_++];
		}

		public int Int() => int.Parse(Next());
		public long Long() => long.Parse(Next());
		public double Double() => double.Parse(Next());

		public int[] ArrayInt(int N, int add = 0)
		{
			int[] Array = new int[N];
			for (int i = 0; i < N; i++) {
				Array[i] = Int() + add;
			}
			return Array;
		}

		public long[] ArrayLong(int N, long add = 0)
		{
			long[] Array = new long[N];
			for (int i = 0; i < N; i++) {
				Array[i] = Long() + add;
			}
			return Array;
		}

		public double[] ArrayDouble(int N, double add = 0)
		{
			double[] Array = new double[N];
			for (int i = 0; i < N; i++) {
				Array[i] = Double() + add;
			}
			return Array;
		}
	}
}