#include <bits/stdc++.h>
#include <algorithm>
using namespace std;

int main(){
  int n;
  cin>>n;
  vector<string> edge(n);
  for (int i=0;i<n;i++){
    cin>>edge.at(i);
    int su=0;
  }
  if(n==2) cout<<2<<endl;
  else{
    int ans=0;
    for(int m=0;m<n;m++){
      vector<string> edges=edge;
      vector<int> v(n,0);
      queue<int> tmp;
      v.at(m)=1;
      tmp.push(m);
      while(!tmp.empty()){
        int x=tmp.front();
        int y=v.at(x);
        tmp.pop();
        for (int i=0;i<n;i++){
          if(edges.at(x).at(i)=='1'){
            if(v.at(i)==0) {v.at(i)=y+1; tmp.push(i);}
            else if(v.at(i)!=y+1){cout<<-1<<endl; return 0;}
          }
          edges.at(x).at(i)='0';
          edges.at(i).at(x)='0';
        }
      }
      int tmpans=0;
      for(int i=0;i<n;i++){
        if(tmpans<v.at(i)) tmpans=v.at(i);
      }
      ans=max(ans,tmpans);
    }
    cout<<ans<<endl;
  }
}
### Prompt

Construct a CPP code solution to the problem outlined:

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>
#include <algorithm>
using namespace std;

int main(){
  int n;
  cin>>n;
  vector<string> edge(n);
  for (int i=0;i<n;i++){
    cin>>edge.at(i);
    int su=0;
  }
  if(n==2) cout<<2<<endl;
  else{
    int ans=0;
    for(int m=0;m<n;m++){
      vector<string> edges=edge;
      vector<int> v(n,0);
      queue<int> tmp;
      v.at(m)=1;
      tmp.push(m);
      while(!tmp.empty()){
        int x=tmp.front();
        int y=v.at(x);
        tmp.pop();
        for (int i=0;i<n;i++){
          if(edges.at(x).at(i)=='1'){
            if(v.at(i)==0) {v.at(i)=y+1; tmp.push(i);}
            else if(v.at(i)!=y+1){cout<<-1<<endl; return 0;}
          }
          edges.at(x).at(i)='0';
          edges.at(i).at(x)='0';
        }
      }
      int tmpans=0;
      for(int i=0;i<n;i++){
        if(tmpans<v.at(i)) tmpans=v.at(i);
      }
      ans=max(ans,tmpans);
    }
    cout<<ans<<endl;
  }
}
```