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 PyPy3 (7.3.0) (judge verdict: Accepted)

from collections import deque

N=int(input())
E=[[] for _ in range(N+1)]
P=[]

for a in range(1,N+1):
    S=input()
    for b in range(1,N+1):
        if S[b-1]=="1":
            E[a].append(b)

            if a<b:
                P.append((a,b))

M=-1
for a in range(1,N+1):
    Q=deque([a])
    T=[0 for i in range(N+1)]
    T[a]=1
    while Q:
        v=Q.popleft()
        for w in E[v]:
            if T[w]==0:
                T[w]=T[v]+1
                Q.append(w)

    F=True
    for (u,v) in P:
        F&=(abs(T[u]-T[v])==1)

    if F:
        M=max(M,max(T))

print(M)