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 <iostream>
#include <vector>
#include <limits.h>
#include <algorithm>
#include <string>
#include <math.h>
#include <limits.h>
#include <queue>
#include <map>
#include <set>
#include <iomanip>
#include <bitset>
#include <cassert>
#include <random>
#include <functional>
#include <stack>
#include <iomanip>
#include <cassert>
using namespace std;
long long MOD = 1e9 + 7;

template< typename T >
struct edge {
	int src, to;
	T cost;

	edge(int to, T cost) : src(-1), to(to), cost(cost) {}

	edge(int src, int to, T cost) : src(src), to(to), cost(cost) {}

	edge& operator=(const int& x) {
		to = x;
		return *this;
	}

	operator int() const { return to; }
};

template< typename T >
using Edges = vector< edge< T > >;
template< typename T >
using WeightedGraph = vector< Edges< T > >;
using UnWeightedGraph = vector< vector< int > >;
template< typename T >
using Matrix = vector< vector< T > >;

template< typename T >
vector< T > dijkstra(WeightedGraph< T >& g, int s) {
	const auto INF = numeric_limits< T >::max();
	vector< T > dist(g.size(), INF);

	using Pi = pair< T, int >;
	priority_queue< Pi, vector< Pi >, greater< Pi > > que;
	dist[s] = 0;
	que.emplace(dist[s], s);
	while (!que.empty()) {
		T cost;
		int idx;
		tie(cost, idx) = que.top();
		que.pop();
		if (dist[idx] < cost) continue;
		for (auto& e : g[idx]) {
			auto next_cost = cost + e.cost;
			if (dist[e.to] <= next_cost) continue;
			dist[e.to] = next_cost;
			que.emplace(dist[e.to], e.to);
		}
	}
	return dist;
}

long long N;
WeightedGraph<long long> G;

int main() {
	cin >> N;
	G.resize(N);
	for (int t = 0; t < N; t++) {
		string S;
		cin >> S;
		for (int s = 0; s < N; s++) {
			if (S[s] == '1')G[s].push_back({ s,t,1 });
		}
	}
	long long ans = 0;
	for (int s = 0; s < N; s++) {
		vector<long long> dijkans = dijkstra(G, s);
		for (int t = 0; t < N; t++) {
			ans = max(ans, dijkans[t] + 1);
		}
	}
	if (N > 2 && ans == 2)cout << -1 << endl;
	else cout << ans << endl;
	return 0;
}