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)

#pragma GCC optimize ("Ofast")
#include <bits/stdc++.h>
#define endl '\n'
#define int long long
//#define inf 10000000000000001ll
//#define fi first
//#define se second
typedef long long ll;
using namespace std;
int a[201][201];
vector<int>g[201];
int used[201];
void dfs(int v){
    for (auto to:g[v]){
        if (used[to]==used[v]){
            cout<<-1;
            exit(0);
        }
        if (!used[to]){
            used[to]=used[v]%2+1;
            dfs(to);
        }
    }
}
int d[201][201];
int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n;cin>>n;
    for (int i=1;i<=n;i++){
        for (int j=1;j<=n;j++){
            char c;cin>>c;
            a[i][j]=(c=='1');
        }
    }

    for (int i=1;i<=n;i++) for (int j=1;j<=n;j++) d[i][j]=100000001;
    for (int i=1;i<=n;i++){
        for (int j=1;j<=n;j++){
            if (a[i][j]) {
                d[i][j]=1;
                g[i].push_back(j);
            }
        }
    }
    used[1]=1;
    dfs(1);
    for (int i=1;i<=n;i++) d[i][i]=0;


    for (int k=1;k<=n;k++){
        for (int i=1;i<=n;i++){
            for (int j=1;j<=n;j++){
                d[i][j]=min(d[i][j],d[i][k]+d[k][j]);
            }
        }
    }
    int maxx=0;
    for (int i=1;i<=n;i++){
        for (int j=1;j<=n;j++){
            maxx=max(maxx,d[i][j]);
        }
    }
    cout<<maxx+1;
    return 0;
}