#include <iostream>
#include <queue>
using namespace std;
typedef pair<int,int> Paint;
int main(){
  int n;cin>>n;
  char mp[n][n];
  for(int i = 0; n > i; i++){
    for(int j = 0; n > j; j++){
      cin>>mp[i][j];
    }
  }
  int ans = -1;
  for(int i = 0; n > i; i++){
    int tmp[n];
    bool lock[n];
    for(int j = 0; n > j; j++)tmp[j] = -1,lock[j]=false;
    queue<Paint> z;
    z.push(make_pair(i,0));
    tmp[i]=0;lock[i]=true;
    while(!z.empty()){
      for(int j = 0; n > j; j++){
        if(mp[z.front().first][j] == '1'){
          if(!lock[j]){
            if(!(z.front().second+1==tmp[j] || tmp[j]==-1)){
              cout << -1 << endl;
              return 0;
            }else if(tmp[j]==-1){
              z.push(make_pair(j,z.front().second+1));
            }
          }
          tmp[j] = z.front().second+1;
        }
      }
      if(z.size()==1)ans = max(ans,z.front().second);
      lock[z.front().first]=true;
      z.pop();
    }
  }
  cout << ans+1 << endl;
}

### Prompt

Please provide a cpp coded solution 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 <iostream>
#include <queue>
using namespace std;
typedef pair<int,int> Paint;
int main(){
  int n;cin>>n;
  char mp[n][n];
  for(int i = 0; n > i; i++){
    for(int j = 0; n > j; j++){
      cin>>mp[i][j];
    }
  }
  int ans = -1;
  for(int i = 0; n > i; i++){
    int tmp[n];
    bool lock[n];
    for(int j = 0; n > j; j++)tmp[j] = -1,lock[j]=false;
    queue<Paint> z;
    z.push(make_pair(i,0));
    tmp[i]=0;lock[i]=true;
    while(!z.empty()){
      for(int j = 0; n > j; j++){
        if(mp[z.front().first][j] == '1'){
          if(!lock[j]){
            if(!(z.front().second+1==tmp[j] || tmp[j]==-1)){
              cout << -1 << endl;
              return 0;
            }else if(tmp[j]==-1){
              z.push(make_pair(j,z.front().second+1));
            }
          }
          tmp[j] = z.front().second+1;
        }
      }
      if(z.size()==1)ans = max(ans,z.front().second);
      lock[z.front().first]=true;
      z.pop();
    }
  }
  cout << ans+1 << endl;
}

```