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 <cstdio>
#include <iostream>
#include <cassert>
#include <algorithm>    
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <cstring>
#include <cmath>
#include <bitset>
#define mk make_pair
#define eb emplace_back
#define eps 1e-8
#define fi first
#define se second
#define all(x) (x).begin(),(x).end()
// #define int long long
using namespace std;

typedef long double ld;
typedef unsigned int ui;
typedef pair<int,int> pii;
typedef unsigned long long ull; 
typedef vector<int> vii;
typedef vector<long double> vd;
const int inf = 1e9;
const int M = 1e9 + 7;
//__int128

const int maxn = 202;
string s[maxn];
int vis[maxn];
int n;
int bfs(int u) {
	queue<int> q;
	q.push(u);
	int res = 0;
	vis[u] = 1; 
	while(!q.empty()) {
		int x = q.front(); q.pop();
		res = max(res, vis[x]);
		for(int i = 0; i < n; ++i) {
			if(s[x][i] == '1') {
				if(!vis[i]) {
					q.push(i);
					vis[i] = vis[x] + 1;
				} else if(vis[i] == vis[x]) {
					return -inf;
				}
			}
		}
	}
	return res;
}

signed main()
{
  ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	cin >> n;
	set<pii> d;
	for(int i = 0; i < n; ++i) {
		cin >> s[i];
		int x = 0;
		for(int j = 0; j < n; ++j) if(s[i][j] == '1') ++x;
		d.insert(pii(x, i));
	}
	int res = 0;
	while((int)d.size()) {
		int u = d.begin()->se;
		if(vis[u]) {
			d.erase(d.begin());
			continue;
		}
		res += bfs(u);
	}  
	cout << max(res, -1) << endl;
	return 0;
}