#include <bits/stdc++.h>
using namespace std;


int main(){
  int64_t N,m=0;
  cin >> N;
  vector<string> S(N);
  vector<int64_t> U(N,0);
  queue<int64_t> Q;
  for(int64_t i=0;i<N;i++){
    cin >> S.at(i);
  }
  U.at(0)=1;
  Q.push(0);
  while(!Q.empty()){
    int64_t q=Q.front();
    Q.pop();
    for(int64_t i=0;i<N;i++){
      if(S.at(q).at(i)=='1'&&U.at(i)==0){
        U.at(i)=3-U.at(q);
        Q.push(i);
      }else if(S.at(q).at(i)=='1'&&U.at(i)!=3-U.at(q)){
        cout << -1 << endl;
        return 0;
      }
    }
  }
  for(int64_t i=0;i<N;i++){
    for(int64_t j=0;j<N;j++){
      U.at(j)=0;
    }
    U.at(i)=1;
    Q.push(i);
    while(!Q.empty()){
      int64_t q=Q.front();
      Q.pop();
      for(int64_t j=0;j<N;j++){
        if(S.at(q).at(j)=='1'&&U.at(j)==0){
          U.at(j)=U.at(q)+1;
          Q.push(j);
          m=max(U.at(j),m);
        }
      }
    }
  }
  cout << m << endl;   
}
### Prompt

Create a solution in cpp for 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;


int main(){
  int64_t N,m=0;
  cin >> N;
  vector<string> S(N);
  vector<int64_t> U(N,0);
  queue<int64_t> Q;
  for(int64_t i=0;i<N;i++){
    cin >> S.at(i);
  }
  U.at(0)=1;
  Q.push(0);
  while(!Q.empty()){
    int64_t q=Q.front();
    Q.pop();
    for(int64_t i=0;i<N;i++){
      if(S.at(q).at(i)=='1'&&U.at(i)==0){
        U.at(i)=3-U.at(q);
        Q.push(i);
      }else if(S.at(q).at(i)=='1'&&U.at(i)!=3-U.at(q)){
        cout << -1 << endl;
        return 0;
      }
    }
  }
  for(int64_t i=0;i<N;i++){
    for(int64_t j=0;j<N;j++){
      U.at(j)=0;
    }
    U.at(i)=1;
    Q.push(i);
    while(!Q.empty()){
      int64_t q=Q.front();
      Q.pop();
      for(int64_t j=0;j<N;j++){
        if(S.at(q).at(j)=='1'&&U.at(j)==0){
          U.at(j)=U.at(q)+1;
          Q.push(j);
          m=max(U.at(j),m);
        }
      }
    }
  }
  cout << m << endl;   
}
```