#include<bits/stdc++.h>
#define rep(i,n) for(int i = 0; i < n; i++)
#define pb push_back
using namespace std;
typedef long long ll;

int main(){
  int n;
  cin>>n;
  vector<string> c(n); 
  vector<vector<int>> g(n);
  rep(i,n) cin>>c[i];
  rep(i,n)rep(j,n) if(c[i][j]=='1') g[i].pb(j);
  int ans=0;
  
  rep(i,n){
    queue<int> que;
    vector<int> dist(n,-1);
 
    que.push(i);
    dist[i]=1;
    while(!que.empty()){
      int x=que.front();
      que.pop();
      for(auto nx:g[x]){
        if(dist[nx]!=-1) continue;
        que.push(nx);
        dist[nx]=dist[x]+1;
      }
    }
    vector<vector<int>> l(n);
    
    rep(i,n){
      l[dist[i]-1].pb(i);
      ans=max(ans,dist[i]);
    }
    rep(i,n){
      rep(j,(int)l[i].size()) rep(k,(int)l[i].size()){
        if(c[l[i][j]][l[i][k]]=='1'){
          cout<<"-1"<<endl;
          return 0;
        }
      }
    }
  }
  
  cout<<ans<<endl;
}
### Prompt

Develop a solution in cpp to the problem described below:

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>
#define rep(i,n) for(int i = 0; i < n; i++)
#define pb push_back
using namespace std;
typedef long long ll;

int main(){
  int n;
  cin>>n;
  vector<string> c(n); 
  vector<vector<int>> g(n);
  rep(i,n) cin>>c[i];
  rep(i,n)rep(j,n) if(c[i][j]=='1') g[i].pb(j);
  int ans=0;
  
  rep(i,n){
    queue<int> que;
    vector<int> dist(n,-1);
 
    que.push(i);
    dist[i]=1;
    while(!que.empty()){
      int x=que.front();
      que.pop();
      for(auto nx:g[x]){
        if(dist[nx]!=-1) continue;
        que.push(nx);
        dist[nx]=dist[x]+1;
      }
    }
    vector<vector<int>> l(n);
    
    rep(i,n){
      l[dist[i]-1].pb(i);
      ans=max(ans,dist[i]);
    }
    rep(i,n){
      rep(j,(int)l[i].size()) rep(k,(int)l[i].size()){
        if(c[l[i][j]][l[i][k]]=='1'){
          cout<<"-1"<<endl;
          return 0;
        }
      }
    }
  }
  
  cout<<ans<<endl;
}
```