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.chomp}

dp = Array.new(n) {Array.new(n, 10000)}
n.times do |i|
    dp[i][i] = 0
    n.times do |j|
        dp[i][j] = 1 if s[i][j] == ?1
    end
end

n.times do |k|
    n.times do |i|
        n.times do |j|
            x = dp[i][k] + dp[k][j]

            if dp[i][j] != 10000 && (dp[i][j] - x).odd?
                puts -1
                exit
            end

            dp[i][j] = x if dp[i][j] > x
        end
    end
end

#dp.each {|e| p e}
puts dp.map(&:max).max + 1