#include<bits/stdc++.h>
using namespace std;
using Graph=vector<vector<int>>;
int main(){
int N;
  cin>>N;
  Graph G(N);
  for(int i=0;i<N;i++){
    string s; cin>>s;
    for(int j=0;j<N;j++){
    if(s.at(j)=='1')
      G[i].push_back(j);
   }
}int ans=0;bool y=true;
  for(int i=0;i<N;i++){
    queue<int>que;
      que.push(i);
      vector<int>dist(N,-1);
    dist[i]=0;
    bool x=true;
    int a;
      while(!que.empty()){
      int v=que.front();
        que.pop();
        for(auto u:G[v]){
        if(dist[u]!=-1){
          if(dist[u]==dist[v]){x=false; break;}
        }else{
        dist[u]=dist[v]+1; if(a<dist[u])a=dist[u];
          que.push(u);
        
        }
        
        }  if(x){if(ans<a)ans=a;}else{y=false; break;}
      
      }  if(!y)break;
    
    }if(y)cout<<ans+1<<endl;
  else cout<<-1<<endl;
    return 0;
}

### Prompt

Please formulate 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;
using Graph=vector<vector<int>>;
int main(){
int N;
  cin>>N;
  Graph G(N);
  for(int i=0;i<N;i++){
    string s; cin>>s;
    for(int j=0;j<N;j++){
    if(s.at(j)=='1')
      G[i].push_back(j);
   }
}int ans=0;bool y=true;
  for(int i=0;i<N;i++){
    queue<int>que;
      que.push(i);
      vector<int>dist(N,-1);
    dist[i]=0;
    bool x=true;
    int a;
      while(!que.empty()){
      int v=que.front();
        que.pop();
        for(auto u:G[v]){
        if(dist[u]!=-1){
          if(dist[u]==dist[v]){x=false; break;}
        }else{
        dist[u]=dist[v]+1; if(a<dist[u])a=dist[u];
          que.push(u);
        
        }
        
        }  if(x){if(ans<a)ans=a;}else{y=false; break;}
      
      }  if(!y)break;
    
    }if(y)cout<<ans+1<<endl;
  else cout<<-1<<endl;
    return 0;
}

```