#include<bits/stdc++.h>
#define int long long
using namespace std;
signed main(){
  int N;
  cin>>N;
  vector<vector<int>> A(N);
  vector<vector<int>> C(N,vector<int>(N,1000000000000000000));
  for(int i=0;i<N;i++)
    for(int j=0;j<N;j++){
      char c;
      cin>>c;
      if(i<j)continue;
      if(i==j){
        C[i][j]=0;
        continue;
      }
      if(c=='1'){
        C[i][j]=1;C[j][i]=1;
        A[i].push_back(j);
        A[j].push_back(i);
      }
    }
  vector<int> B(N,-1);
  B[0]=0;
  queue<int> Q;
  Q.push(0);
  while(!Q.empty()){
    int I=Q.front();
    Q.pop();
    for(int i:A[I]){
      if(B[i]==-1){
        B[i]=B[I]^1;
        Q.push(i);
      }else if(B[i]==B[I]){
        puts("-1");
        return 0;
      }
    }
  }
  for(int k=0;k<N;k++)
    for(int i=0;i<N;i++)
      for(int j=0;j<N;j++)C[i][j]=min(C[i][j],C[i][k]+C[k][j]);
  int mx=0;
  for(int i=0;i<N;i++)
    for(int j=0;j<N;j++)mx=max(mx,C[i][j]);
  cout<<mx+1<<endl;
}
### Prompt

Please create a solution in cpp 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>
#define int long long
using namespace std;
signed main(){
  int N;
  cin>>N;
  vector<vector<int>> A(N);
  vector<vector<int>> C(N,vector<int>(N,1000000000000000000));
  for(int i=0;i<N;i++)
    for(int j=0;j<N;j++){
      char c;
      cin>>c;
      if(i<j)continue;
      if(i==j){
        C[i][j]=0;
        continue;
      }
      if(c=='1'){
        C[i][j]=1;C[j][i]=1;
        A[i].push_back(j);
        A[j].push_back(i);
      }
    }
  vector<int> B(N,-1);
  B[0]=0;
  queue<int> Q;
  Q.push(0);
  while(!Q.empty()){
    int I=Q.front();
    Q.pop();
    for(int i:A[I]){
      if(B[i]==-1){
        B[i]=B[I]^1;
        Q.push(i);
      }else if(B[i]==B[I]){
        puts("-1");
        return 0;
      }
    }
  }
  for(int k=0;k<N;k++)
    for(int i=0;i<N;i++)
      for(int j=0;j<N;j++)C[i][j]=min(C[i][j],C[i][k]+C[k][j]);
  int mx=0;
  for(int i=0;i<N;i++)
    for(int j=0;j<N;j++)mx=max(mx,C[i][j]);
  cout<<mx+1<<endl;
}
```