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;
typedef long double ld;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> pi;

vvi G;
vi V;

bool dfs1(int x) {
  for (auto y : G[x]) {
    if (V[y] == 0) {
      V[y] = -V[x];
      if (!dfs1(y)) return false;
    }
    else if (V[y] == V[x]) return false;
  }
  return true;
}

int bfs(int s) {
  queue<int> Q;
  int res = 0;
  V[s] = 0;
  Q.push(s);
  while (!Q.empty()) {
    int x = Q.front();
    Q.pop();
    for (auto y : G[x]) {
      if (V[y] == -1) {
        V[y] = V[x] + 1;
        res = max(V[y], res);
        Q.push(y);
      }
    }
  }
  return res;
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  G = vvi(n);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      char x;
      cin >> x;
      if (x == '1') G[i].push_back(j);
    }
  }
  V = vi(n, 0);
  V[0] = 1;
  if (!dfs1(0)) {
    cout << -1 << '\n';
    return 0;
  }
  int res = 0;
  for (int i = 0; i < n; ++i) {
    V = vi(n,-1);
    res = max(res, bfs(i)+1);
  }
  cout << res <<'\n';
}