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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include "bits/stdc++.h"

using namespace std;

using i64 = int_fast64_t;

template <class T>
using akari = vector<T>;
template <class T>
using yukari = vector<akari<T>>;
using seika = string;

void solve()
{
    int n;
    cin >> n;
    yukari<int> g(n);
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
        {
            char x;
            cin >> x;
            if (x == '1')
            {
                g[i].push_back(j);
            }
        }
    akari<int> dis(n, -1);
    dis[0] = 0;
    auto dfs = [&](auto dfs, int v = 0, int p = -1) -> void {
        for (int nv : g[v])
        {
            if (nv != p)
            {
                //cout << v << ' ' << nv << ' ' << dis[nv] << endl;
                if (dis[nv] != -1)
                {
                    if (dis[nv] % 2 == dis[v] % 2)
                    {
                        cout << -1 << endl;
                        exit(0);
                    }
                    else
                    {
                        continue;
                    }
                }
                dis[nv] = dis[v] + 1;
                dfs(dfs, nv, v);
            }
        }
    };
    dfs(dfs);
    int s = distance(dis.begin(), max_element(dis.begin(), dis.end()));
    //cout << s << endl;
    dis.assign(n, -1);
    dis[s] = 0;
    dfs(dfs, s, -1);
    cout << *max_element(dis.begin(), dis.end()) + 1 << endl;
}

int main()
{
    solve();
    return 0;
}