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

int main() {
  int N,Z=0;
  cin>>N;
  vector<vector<int>> G(N);
  for(int i=0;i<N;i++){
    for(int j=0;j<N;j++){
      char a;
      cin>>a;
      if(a=='1'){
        G.at(i).push_back(j);
      }
    }
  }
  vector<int> c(N,0);
  c.at(0)=1;
  for(int i=0;i<N;i++){
  queue<int> q;
  vector<int> r(N,-1);
  q.push(i);
  r.at(i)=0;
  while(!q.empty()){
    int b=q.front();
    q.pop();
    for(int x:G.at(b)){
      if(r.at(x)==-1){
        r.at(x)=r.at(b)+1;
        q.push(x);
      }
      if(i==0&&c.at(x)==0){
        c.at(x)=c.at(b)*(-1);
      }
      if(c.at(x)==c.at(b)){
        Z=-2;
        break;
      }
    }
    if(Z==-2){
      break;
    }
    else{
      Z=max(Z,r.at(b));
    }
  }
  }
    cout<<Z+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 <bits/stdc++.h>
using namespace std;

int main() {
  int N,Z=0;
  cin>>N;
  vector<vector<int>> G(N);
  for(int i=0;i<N;i++){
    for(int j=0;j<N;j++){
      char a;
      cin>>a;
      if(a=='1'){
        G.at(i).push_back(j);
      }
    }
  }
  vector<int> c(N,0);
  c.at(0)=1;
  for(int i=0;i<N;i++){
  queue<int> q;
  vector<int> r(N,-1);
  q.push(i);
  r.at(i)=0;
  while(!q.empty()){
    int b=q.front();
    q.pop();
    for(int x:G.at(b)){
      if(r.at(x)==-1){
        r.at(x)=r.at(b)+1;
        q.push(x);
      }
      if(i==0&&c.at(x)==0){
        c.at(x)=c.at(b)*(-1);
      }
      if(c.at(x)==c.at(b)){
        Z=-2;
        break;
      }
    }
    if(Z==-2){
      break;
    }
    else{
      Z=max(Z,r.at(b));
    }
  }
  }
    cout<<Z+1<<endl;
}



```