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' }
}

d = [nil]*N
(0..(N-2)).each{|i|
	((i+1)..(N-1)).each{|j|
		next unless S[i][j]
		d[i] = true if d[i].nil?
		d[j] = ! d[i] if d[j].nil?
		if d[i] == d[j]
			p(-1)
			exit
		end
	}
}

d = [1]*(N*(N-1)/2)
(0..(N-2)).each{|i|
	ij0 = (N-1)*N/2-(N-i-1)*(N-i)/2
	((i+1)..(N-1)).each{|j|
		next unless S[i][j]
		ij = ij0 + (j-i-1)
		dij = d[ij]
		dij = d[ij] = 2 if dij == 1
		jk0 = (N-1)*N/2-(N-j-1)*(N-j)/2
		((j+1)..(N-1)).each{|k|
			next unless S[j][k]
			next if S[i][k]
			jk = jk0 + (k-j-1)
			djk = d[jk]
			djk = d[jk] = dij+1 if djk <= dij
		}
	}
}
p d.max