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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

import java.lang.Math

fun main(args: Array<String>) {
    val INF = 1000000000
    val N = readInputLine().toInt()
    val S = Array(N) { readInputLine().toCharArray() }
    val d = Array(N) { IntArray(N) { INF } }

    for (i in 0 until N) {
        d[i][i] = 0
    }

    for (i in 0 until N) {
        for (j in 0 until N) {
            if (S[i][j] == '1') {
                d[i][j] = 1
            }
        }
    }

    for (k in 1 until N) {
        for (i in 1 until N) {
            for (j in 1 until N) {
                d[i][j] = Math.min(d[i][j], d[i][k] + d[k][j])
            }
        }
    }

    var ans = 0

    for (i in 0 until N) {
        for (j in 0 until N) {
            if (d[i][j] < INF) {
                ans = Math.max(d[i][j], ans)
            }
        }
    }


    if (N != 2 && ans == 1) {
        println(-1)
    } else {
        println(ans + 1)
    }
}

fun readInputLine(): String {
    return readLine()!!
}