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 Python (3.4.3) (judge verdict: Wrong Answer)

N = int(input())
S = []
for i in range(N):
    S.append(list(input()))

memo1 = [50000]*N
nows = set([0])
t = -1
visited = set([])

while True:
    t += 1
    next = set([])
    for now in nows:
        if t < memo1[now]:
            memo1[now]=t
            visited.add(now)
        temp = []
        for j in range(N):
            if (S[now][j]=='1')and(j not in visited):
                temp.append(j)
        next |= set(temp)
    nows = next
    if len(nows)==0:
        break

far1 = memo1.index(max(memo1))

memo2 = [50000]*N
nows = set([far1])
t = -1
visited = set([])

while True:
    t += 1
    next = set([])
    for now in nows:
        if t < memo2[now]:
            memo2[now]=t
            visited.add(now)
        temp = []
        for j in range(N):
            if (S[now][j]=='1')and(j not in visited):
                temp.append(j)
        next |= set(temp)
    nows = next
    if len(nows)==0:
        break
far2 = memo2.index(max(memo2))
output = max(memo2)
if (output == 1)and(N!=2):
    print(-1)
else:
    print(output+1)