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 Scala (2.11.7) (judge verdict: Accepted)

import java.util.Scanner

import scala.collection.{immutable, mutable}
import scala.collection.mutable.ArrayBuffer

object Main {
  def main(args: Array[String]) {
    val result = solve(new Scanner(System.in))
    if (result.length > 0) println(result)
  }

  def solve(in: String): String = solve(new Scanner(in))

  def solve(start: Int, N: Int, S: Array[Seq[Int]]): Int = {
    val Q = mutable.Queue[(Int, Int)]()
    val T = ArrayBuffer.fill(N)(0)

    T(start) = 1
    Q += Tuple2(start, 1)

    while (Q.nonEmpty) {
      val (i, t) = Q.dequeue()
      for (next <- S(i)) {
        if (T(next) == 0) {
          T(next) = t + 1
          Q += Tuple2(next, t + 1)
        } else if (T(next) != t - 1 && T(next) != t + 1) {
          return -1
        }
      }
    }

    T.max
  }

  def solve(sc: Scanner): String = {
    val N = sc.nextInt
    val S: Array[Seq[Int]] = Array.fill(N)(sc.next.zipWithIndex.filter(_._1 == '1').map(_._2))

    val answers = for (i <- 0 until N) yield solve(i, N, S)

    if (answers.isEmpty) {
      "-1"
    } else {
      answers.max.toString
    }
  }
}