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 (3.8.2) (judge verdict: Time Limit Exceeded)

n = int(input())
s = [list(map(int, list(input()))) for _ in range(n)]

colors = [0 for i in range(n)]

#頂点vをcolor(1 or -1)で塗り、再帰的に矛盾がないか調べる。矛盾があればFalse
def dfs(v,color):
    #今いる点を着色
    colors[v] = color
    #今の頂点から行けるところをチェック
    for i in range(n):
        if s[v][i] == 1:
            #同じ色が隣接してしまったらFalse
            if colors[i] == color:
                return False
            #未着色の頂点があったら反転した色を指定し、再帰的に調べる
            if colors[i] == 0 and not dfs(i, -color):
                return False
    #調べ終わったら矛盾がないのでTrue
    return True

#2部グラフならTrue, そうでなければFalse
def is_bipartite():
    return dfs(0,1) # 頂点0を黒(1)で塗ってDFS開始

##############################

def warshall_floyd(d):
    #d[i][j]: iからjへの最短距離
    for k in range(n):
        for i in range(n):
            for j in range(n):
                d[i][j] = min(d[i][j], d[i][k]+d[k][j])
    return d

##############################
d = [[float("inf")]*n for i in range(n)]
#d[u][v] : 辺uvのコスト(存在しないときはinf)
for i in range(n):
    for j in range(n):
        if s[i][j] != 0:
            d[i][j] = s[i][j]
for i in range(n):
    d[i][i] = 0 #自身のところに行くコストは０
    
if is_bipartite():
    print(max(max(warshall_floyd(d)))+1)
else:
    print(-1)