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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <utility>
#include <map>
#include <vector>
#include <algorithm>
#include <queue>
#include <math.h>
#include <stack>
#include <set>
#include <deque>
#include <cmath>
#include <cassert>

using namespace std;
typedef long long ll;
ll mod = 1e9+7;
#define rep(i,n) for(int i = 0; i < (n); ++i)

vector<int> bfs(int node, vector<string> &a) {
    int n = a.size();
    vector<int> d(n, n + 1);
    vector<int> color(n, -1);
    d[node] = 0;
    color[node] = 0;

    queue<int> q;
    q.push(node);

    while(!q.empty()) {
        int node = q.front();
        q.pop();
        rep(i, n) {
            if (a[node][i] == '1') {
                if (d[i] >= n) {
                    d[i] = d[node] + 1;
                    color[i] = 1 - color[node];
                    q.push(i);
                }
                if (color[i] != 1 - color[node]) {
                    return vector<int> ();
                }
            }
        }
    }
    return d;
}

int main() {
    int n;
    cin >> n;
    vector<string> mat(n);
    rep(i, n) {
        cin >> mat[i];
    }
    int ans = -1;
    rep(i, n) {
        auto d = bfs(i, mat);
        if (d.empty()) {
            cout << -1 << endl;
            return 0;
        }
        ans = max(ans, *max_element(d.begin(), d.end()));
    }
    cout << ans + 1 << endl;
    return 0;
}