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 <algorithm>
#include <vector>
#include <queue>
#include <iomanip>
#include <cmath>
#include <map>
#include <functional>
using namespace std;
#define REP(i,n) for (int i=0;i<(n);++i)
#define rep(i,a,b) for(int i=a;i<(b);++i)
constexpr long INF = 1L << 60;
constexpr int MOD = 1e9 + 7;

int n;
string s[200];
int deg[222], depth[222], used[222];
vector<vector<int>> g;
// ながさ奇数の閉路があったら無理

void dfs(int v, int p, int d) {
  if(used[v]) return;
  depth[v] = d;
  used[v] = 1;
  for(int nv : g[v]) {
    if(nv == p) continue;
    dfs(nv, v, d+1);
  }
}

int main() {
  cin.tie(0); ios_base::sync_with_stdio(false);
  int n; cin >> n;
  g.resize(n);
  REP(i,n) cin >> s[i];
  REP(i,n) REP(j,n) {
    if(s[i][j] == '1') {
      g[i].push_back(j);
    }
  }

  int ans = 2;
  REP(i,n) {
    REP(j,n) depth[j] = 0, used[j] = 0;
    dfs(i,-1,0);
    REP(j,n) ans = max(ans, depth[j]);
  }

  cout << ans << endl;
  return 0;
}