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 Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

N=gets.to_i
dist_f=N.times.map{gets.split("").map{|e|e.to_i==1 ? 1 : Float::INFINITY}}
N.times do |i|
    dist_f[i][i]=0
end
dist=Marshal.load(Marshal.dump(dist_f))
cnt=0
N.times do |k|
    dist_next=Array.new(N){Array.new(N,Float::INFINITY)}
    N.times do |i|
        N.times do |j|
            a=dist[i][j]
            b=dist[i][k]+dist[k][j]
            bet=a<b ? a : b
            dist_next[i][j]=bet
            cnt=cnt < bet ? bet : cnt if bet!=Float::INFINITY
        end
    end
    dist=dist_next
end

color=Array.new(N,false)
stack=Array.new
stack << 0

until stack.empty?
    i=stack.pop
    next if color[i]
    
    z,o=false,false
    dist_f[i].each_with_index do |e,j|
        if e==1
            stack << j
            z=true if color[j]==0
            o=true if color[j]==1
        end   
    end
    if z && o
        puts -1
        exit
    elsif z && !o
        color[i]=1
    else
        color[i]=0
    end
    
end
puts cnt+1