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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)


import java.util.Scanner;

public class Main {
    private static final int INITIAL_VALUE = Integer.MAX_VALUE / 2;

    public static void main(final String[] args) {
        final Scanner scanner = new Scanner(System.in);
        final int n = scanner.nextInt();
        final int[][] table = new int[n + 1][n + 1];

        for (int i = 1; i <= n; i++) {
            final String s = scanner.next();
            for (int j = 1; j <= n; j++) {
                final int value = s.charAt(j - 1) - '0';
                table[i][j] = value == 0 ? INITIAL_VALUE : 1;
            }
        }

        for (int k = 1; k <= n; k++) {
            for (int i = 1; i <= n; i++) {
                for (int j = 1; j <= n; j++) {
                    if (i == j) {
                        table[i][j] = 0;
                    }
                    if (table[i][j] < INITIAL_VALUE && table[i][k] + table[k][j] < INITIAL_VALUE
                            && table[i][j] % 2 != (table[i][k] + table[k][j]) % 2) {
                        System.out.println(-1);
                        return;
                    }
                    table[i][j] = Math.min(table[i][j], table[i][k] + table[k][j]);
                }
            }
        }

        int max = 0;
        for (final int[] array : table) {
            for (final int value : array) {
                max = Math.max(max, value + 1);
            }
        }

        System.out.println(max);
    }
}