#include <bits/stdc++.h>
using namespace std;
#define int long long
const int MOD = 1000000007;

signed main() {
  int n;
  cin>>n;
  string s[n];
  for(int i=0;i<n;i++)cin>>s[i];
  int used[n];
  queue<int> k,k2;
  int ans=-1,ans2;
  for(int i=0;i<n;i++){
    for(int j=0;j<n;j++)used[j]=0;
    k.push(i);
    used[i]=1;
    ans2=0;
    while(!k.empty()){
    while(!k.empty()){
      for(int j=0;j<n;j++){
        if(s[k.front()][j]=='1'){
          if(used[j]==0){
             used[j]=-1*used[k.front()];
          k2.push(j);
          }else{
            if(used[j]==used[k.front()]){
              cout<<-1;
              return 0;
            }
          }
         
        }
      }
      k.pop();
   
    }
    swap(k,k2);
    ans2++;
    }
  //  cerr<<ans2<<' '<<k2.size()<<endl;
    ans=max(ans,ans2);
  }
  cout<<ans;

  


  return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

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.

Examples

Input

2
01
10


Output

2


Input

3
011
101
110


Output

-1


Input

6
010110
101001
010100
101000
100000
010000


Output

4

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int MOD = 1000000007;

signed main() {
  int n;
  cin>>n;
  string s[n];
  for(int i=0;i<n;i++)cin>>s[i];
  int used[n];
  queue<int> k,k2;
  int ans=-1,ans2;
  for(int i=0;i<n;i++){
    for(int j=0;j<n;j++)used[j]=0;
    k.push(i);
    used[i]=1;
    ans2=0;
    while(!k.empty()){
    while(!k.empty()){
      for(int j=0;j<n;j++){
        if(s[k.front()][j]=='1'){
          if(used[j]==0){
             used[j]=-1*used[k.front()];
          k2.push(j);
          }else{
            if(used[j]==used[k.front()]){
              cout<<-1;
              return 0;
            }
          }
         
        }
      }
      k.pop();
   
    }
    swap(k,k2);
    ans2++;
    }
  //  cerr<<ans2<<' '<<k2.size()<<endl;
    ans=max(ans,ans2);
  }
  cout<<ans;

  


  return 0;
}

```