#include<bits/stdc++.h>
using namespace std;
long N;
vector<string>S;
long solve(long s){
  queue<long>q;q.push(s);
  vector<long>D(N,-1);D[s]=0;
  while(q.size()){
    long p=q.front();q.pop();
    for(long i=0;i<N;i++)if(S[p][i]=='1'){
      if(D[i]!=-1&&abs(D[p]-D[i])!=1)return -1;
      if(D[i]==-1){D[i]=D[p]+1;q.push(i);}
    }
  }
  long ans=-1;
  for(long x:D)ans=max(ans,x);
  return ans+1;
}
int main(){
  cin>>N;S=vector<string>(N);
  for(long i=0;i<N;i++)cin>>S[i];
  long ans=-1;
  for(long i=0;i<N;i++)ans=max(ans,solve(i));
  cout<<ans;
}
### 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;
long N;
vector<string>S;
long solve(long s){
  queue<long>q;q.push(s);
  vector<long>D(N,-1);D[s]=0;
  while(q.size()){
    long p=q.front();q.pop();
    for(long i=0;i<N;i++)if(S[p][i]=='1'){
      if(D[i]!=-1&&abs(D[p]-D[i])!=1)return -1;
      if(D[i]==-1){D[i]=D[p]+1;q.push(i);}
    }
  }
  long ans=-1;
  for(long x:D)ans=max(ans,x);
  return ans+1;
}
int main(){
  cin>>N;S=vector<string>(N);
  for(long i=0;i<N;i++)cin>>S[i];
  long ans=-1;
  for(long i=0;i<N;i++)ans=max(ans,solve(i));
  cout<<ans;
}
```