#include<bits/stdc++.h>
using namespace std;
using pi=pair<int,int>;
#define mp make_pair
int main(){
  int n;
  string t;
  cin>>n;
  vector<vector<bool>> s(n,vector<bool>(n,false));
  for(int i=0;i<n;++i){
    cin>>t;
    for(int j=0;j<n;++j)if(t[j]=='1')s[i][j]=true;
  }
  int res=-1;
  for(int src=0;src<n;++src){
    bool cont=true;
    vector<int> p(n,-1);
    queue<pi> que,dumq;
    que.push(mp(src,0));
    while(!que.empty()){
      pi q=que.front();que.pop();
      if(p[q.first]<0)p[q.first]=q.second;
      else continue;
      for(int i=0;i<n;++i){
        if(s[i][q.first])que.push(mp(i,q.second+1));
      }
    }
    int maxp=0;
    for(int i=0;cont&&i<n;++i){
      for(int j=0;cont&&j<n;++j){
        if(s[i][j]&&abs(p[i]-p[j])!=1)cont=false;
        else maxp=max(maxp,max(p[i],p[j]));
      }
    }
    if(cont)res=max(res,maxp+1);
  }
  cout<<res;
}
### Prompt

Develop a solution in Cpp 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;
using pi=pair<int,int>;
#define mp make_pair
int main(){
  int n;
  string t;
  cin>>n;
  vector<vector<bool>> s(n,vector<bool>(n,false));
  for(int i=0;i<n;++i){
    cin>>t;
    for(int j=0;j<n;++j)if(t[j]=='1')s[i][j]=true;
  }
  int res=-1;
  for(int src=0;src<n;++src){
    bool cont=true;
    vector<int> p(n,-1);
    queue<pi> que,dumq;
    que.push(mp(src,0));
    while(!que.empty()){
      pi q=que.front();que.pop();
      if(p[q.first]<0)p[q.first]=q.second;
      else continue;
      for(int i=0;i<n;++i){
        if(s[i][q.first])que.push(mp(i,q.second+1));
      }
    }
    int maxp=0;
    for(int i=0;cont&&i<n;++i){
      for(int j=0;cont&&j<n;++j){
        if(s[i][j]&&abs(p[i]-p[j])!=1)cont=false;
        else maxp=max(maxp,max(p[i],p[j]));
      }
    }
    if(cont)res=max(res,maxp+1);
  }
  cout<<res;
}
```