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)

// In the name of God

#include <bits/stdc++.h>

using namespace std;

const int N = 205;

vector<int> adj[N];
int col[N], last;
vector<int> components[N];
bool vis[N];

void dfs(int u) {
	vis[u] = 1;
	components[last].push_back(u);
	for(auto v : adj[u]) {
		if(vis[v]) continue;
		dfs(v);
	}
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	int n;
	cin >> n;
	for(int i=0;i<n;i++) {
		for(int j=0;j<n;j++) {
			char k;
			cin >> k;
			if(k == '1') adj[i].push_back(j);
		}
	}
	for(int i=0;i<n;i++) {
		if(vis[i]) continue;
		dfs(i);
		last++;
	}
	int ret = 0;
	for(int i=0;i<last;i++) {
		vector<int> &m = components[i];
		int mx = -1;
		for(int start=0;start<m.size();start++) {
			bool fail = false;
			memset(col, 0, sizeof col);
			int now = m[start];
			queue<int> q;
			q.push(now);
			col[now] = 1;
			while(!q.empty()) {
				if(fail) break;
				int u = q.front();
				q.pop();
				for(auto v : adj[u]) {
					if(col[v] && abs(col[v]-col[u])!=1) {
						fail = true;
						break;
					}
					else if(col[v]) continue;
					col[v] = col[u] + 1;
					q.push(v);
				}
			}
			if(fail) continue;
			for(int g=0;g<m.size();g++) {
				mx = max(mx, col[m[g]]);
			}
		}
		if(mx==-1) {
			cout << -1;
			exit(0);
		}
		ret += mx-1;
	}
	cout << ret + 1;
	return 0;
}