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: Accepted)

#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<queue>

using namespace std;

#define rep(i, n) for(int i = 0; i < n; i++)

int main(){
    int n;
    cin >> n;
    string a[n];
    rep(i, n) cin >> a[i];
    int s[n][n];
    rep(i, n)rep(j, n) s[i][j] = a[i][j] - '0';

    int ans = 0;
    int r[n];

    rep(i, n){
        fill(r, r + n, 0);
        r[i] = 1;
        vector<int> v;
        rep(j, n){
            if (s[i][j]){
                v.push_back(j);
                r[j] = 2;
            }
        }
        int num = 2;
        bool flag = true;
        while(v.size() > 0){
            vector<int> v_n;
            while(v.size() > 0){
                int j = v.back();
                v.pop_back();
                rep(k, n){
                    if (s[j][k]){
                        if (r[k] == 0){
                            r[k] = num + 1;
                            v_n.push_back(k);
                        }else if (r[k] != num - 1 && r[k] != num + 1){
                            flag = false;
                            break;
                        }
                    }
                }
                if (!flag) break;
                
            }
            if (!flag) break;
            rep(j, v_n.size()){
                v.push_back(v_n[j]);
            }
            num++;
        }

        if (!flag) continue;

        // rep(i, n) cout << r[i] << " ";
        // cout << endl;

        ans = max(ans, num - 1);
    }

    if (ans == 0){
        cout << -1 << endl;
    }else{
        cout << ans << endl;
    }
    
    

}