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 int INF = 1e5;
int main(){
  int N; scanf("%d", &N);
  vector<vector<int>> to(N,vector<int>(N,0));
  for(int i = 0; i < N; i++){
    for(int j = 0; j < N; j++){
      char c; cin >> c; int d = c-'0'; 
      if( d == 0 ){ d = INF;} 
      if( i == j ){ d = 0;}
      to[i][j] = d;}}
  
  vector<int> color(N,0); //0が未到達、1が白、２が黒
  
  queue<int> Q; Q.push(0); color[0] = 1;
  bool flag = true;
  while(!Q.empty()){
    int p = Q.front(); Q.pop();
    for(int i = 0; i < N; i++){ 
      if( to[p][i] == 1 ){ //隣接しているところに進む
        if( color[i] == 0){ 
          if(color[p] == 1){ color[i] = 2;}
          else{ color[i] = 1;}
          
          Q.push(i);
        }
        else{
          if( color[i] == color[p]){ //二部じゃない
            flag = false; break;}
        }
      }
    }
      if(!flag){ break;}
  }
    
  if(!flag){ printf("%d", -1); return 0;}
  
  for(int k = 0; k < N; k++){
    for(int i = 0; i < N; i++){
      for(int j = 0; j < N; j++){
        to[i][j] = min(to[i][j], to[i][k]+to[k][j]);
      }
    }
  }
  
  int ans = 0;
  
  for(int i = 0; i < N; i++){
    for(int j = 0; j < N; j++){
      ans = max(ans, to[i][j]);
    }
  }
  
  printf("%d\n", ans+1); return 0;
}