#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
  int n;
  cin >> n;
  vector<vector<ll>> s(n,vector<ll>(n));
  for(int i=0;i<n;i++){
    string u;cin >> u;
    for(int j=0;j<n;j++){
      if(u[j]=='1'){
        s[i][j]=1;
      }
      else{
        s[i][j]=1e9;
      }
    }
    s[i][i]=0;
  }
  for(int i=0;i<n;i++){
    for(int j=0;j<n;j++){
      for(int k=0;k<n;k++){
        s[j][k]=min(s[j][k],s[j][i]+s[i][k]);
      }
    }
  }
  ll ans=0;
  int q;
  for(int i=0;i<n;i++){
    for(int j=0;j<n;j++){
      if(ans<s[i][j]){
        q=i;
        ans=max(ans,s[i][j]);
      }
    }
  }
  map<ll,ll> mp;
  for(int i=0;i<n;i++){
    mp[i]=s[q][i];
  }
  for(int i=0;i<n;i++){
    for(int j=0;j<n;j++){
      if(mp[i]==mp[j]){
        if(s[i][j]==1){
          cout << -1 << endl;
          return 0;
        }
      }
    }
  }
  cout << ans+1 << 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>
using namespace std;
typedef long long ll;
int main(){
  int n;
  cin >> n;
  vector<vector<ll>> s(n,vector<ll>(n));
  for(int i=0;i<n;i++){
    string u;cin >> u;
    for(int j=0;j<n;j++){
      if(u[j]=='1'){
        s[i][j]=1;
      }
      else{
        s[i][j]=1e9;
      }
    }
    s[i][i]=0;
  }
  for(int i=0;i<n;i++){
    for(int j=0;j<n;j++){
      for(int k=0;k<n;k++){
        s[j][k]=min(s[j][k],s[j][i]+s[i][k]);
      }
    }
  }
  ll ans=0;
  int q;
  for(int i=0;i<n;i++){
    for(int j=0;j<n;j++){
      if(ans<s[i][j]){
        q=i;
        ans=max(ans,s[i][j]);
      }
    }
  }
  map<ll,ll> mp;
  for(int i=0;i<n;i++){
    mp[i]=s[q][i];
  }
  for(int i=0;i<n;i++){
    for(int j=0;j<n;j++){
      if(mp[i]==mp[j]){
        if(s[i][j]==1){
          cout << -1 << endl;
          return 0;
        }
      }
    }
  }
  cout << ans+1 << endl;
}
```