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<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll INF = 1LL<<60;
#define all(x) (x).begin(),(x).end()
#define REP(i,n) for(int i=0;i<(n);++i)
#define FOR(i,k,n) for(int i=(k);i<(n);++i)

struct edge{int to; ll cost;};
typedef pair<ll,int> P;

void dijkstra(int s,vector<ll>&d,const vector<vector<edge>>&G){
    priority_queue<P,vector<P>,greater<P>> que;
    fill(d.begin(),d.end(),INF);
    d[s]=0;
    que.push(P(0,s));
    while(!que.empty()){
        P p=que.top();que.pop();
        int v=p.second;
        if(d[v]<p.first)continue;
        for(int i=0;i<G[v].size();i++){
            edge e=G[v][i];
            if(d[e.to]>d[v]+e.cost){
                d[e.to]=d[v]+e.cost;
                que.push(P(d[e.to],e.to));
            }
        }
    }
}

int main() {
    int N; cin >> N;
    vector<string> v(N);
    REP(i,N) cin >> v[i];
    vector<vector<edge>> G(N);
    vector<ll> d(N);
    REP(i,N)REP(j,N) {
        if(v[i][j]=='1') G[i].push_back((edge){j,1});
    }
    ll ans = 0;
    REP(i,N) {
        dijkstra(i,d,G);
        ll tmp = 0;
        REP(j,N) tmp = max(tmp, d[j]);
        bool check = true;
        REP(j,N)REP(k,N) {
            if(v[j][k]=='1'&&d[j]==d[k]) check = false;
        }
        if(check) ans = max(ans, tmp);
    }
    cout << (ans?ans+1:-1) << endl;
}