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: Wrong Answer)

N = gets.to_i
S = N.times.map{
	gets.strip.chars.map{|_| _=='1' }
}

Ds = lambda{|dbase|
	ds, d, q = [nil]*N, 1, {dbase=>true}
	ds[dbase] = d
	until q.empty?
		r = {}
		q.keys.each{|i|
			(0...N).each{|j|
				next unless S[i][j]
				if ds[j].nil?
					ds[j] = d+1
					r[j] = true
				elsif ds[j] == d+1
				elsif ds[j] == d-1
				else
					raise
				end
			}
		}
		q, d = r, d+1
	end
	return ds
}

Heuristics = lambda{
	return 0...N if N < 100
	ds0 = Ds[0]
	mx0 = ds0.each_with_index.max
	dsA = Ds[mx0[1]]
	mxA = dsA.each_with_index.max
	dsB = mxA[1] == 0 ? ds0 : Ds[mxA[1]]
	siA, siB = [dsA, dsB].map{|ds| (0...N).sort_by{|i| ds[i] } }
	return siA[-N/5, N/5]|siB[-N/5, N/5]|(siA[N/2-N/10, N/5]&siB[N/2-N/10, N/5])
}

begin
	p Heuristics.call.map{|i| Ds[i].max }.max
rescue RuntimeError
	p(-1)
end