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: Time Limit Exceeded)

from collections import defaultdict
import queue

INF = float("inf")

def getlist():
	return list(map(int, input().split()))

class Graph(object):
    def __init__(self):
        self.graph = defaultdict(list)

    def __len__(self):
        return len(self.graph)

    def add_edge(self, a, b):
        self.graph[a].append(b)

    def get_nodes(self):
        return self.graph.keys()

class BFS(object):
	def __init__(self, graph, s, N):
		self.judge = "Yes"
		self.g = graph.graph
		self.Q = queue.Queue()
		self.dist = [INF] * N
		self.visit = ["No"] * N
		self.visit[s] = "Yes"
		self.dist[s] = 0
		self.Q.put(s)
		while not self.Q.empty():
			v = self.Q.get()
			for i in self.g[v]:
				if self.visit[i] == "No":
					self.dist[i] = self.dist[v] + 1
					self.Q.put(i)
					self.visit[i] = "Yes"
				else:
					if (self.dist[i] - self.dist[v]) % 2 == 0:
						self.judge = "No"

	def ans(self):
		return max(self.dist)

#処理内容
def main():
	#初期化
	N = int(input())
	L = []
	for i in range(N):
		S = list(input())
		L.append(S)
	G = Graph()
	for i in range(N):
		for j in range(i + 1, N):
			if L[i][j] == "1":
				G.add_edge(i, j)
				G.add_edge(j, i)
	#BFS
	jud = "Yes"
	A = - INF
	for i in range(N):
		bfs = BFS(G, i, N)
		A = max(bfs.ans(), A)
		if bfs.judge == "No":
			jud = "No"
	if jud == "No":
		print(-1)
	else:
		print(A + 1)

if __name__ == '__main__':
	main()