#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long ll;
vector<string> s(210);
int dis[210][210];

int main(){
  int n;
  cin >> n;
  for(int i=0;i<n;i++) cin >> s[i];
  int ans=0;
  queue<int> q;
  vector<int> flag(n);
  for(int i=0;i<n;i++){
    q.push(i);
    flag[i]++;
    while(q.size()){
      int t=q.front(); q.pop();
      for(int j=0;j<n;j++){
        if(s[t][j]=='1'&&flag[j]==i){
          dis[i][j]=dis[i][t]+1;
          q.push(j);
          flag[j]++;
        }
      }
    }
  }
  vector<int> a(n,-1);
  a[0]=0;
  for(int i=1;i<n;i++){
    a[i]=dis[0][i]%2;
  }
  for(int i=0;i<n;i++){
    for(int j=0;j<n;j++){
      ans=max(ans,dis[i][j]);
      if(dis[i][j]%2!=(a[i]+a[j])%2){
        cout << -1 << endl;
        return 0;
      }
    }
  }
  cout << ans+1 << endl;
}
### Prompt

In cpp, your task is to solve 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 <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long ll;
vector<string> s(210);
int dis[210][210];

int main(){
  int n;
  cin >> n;
  for(int i=0;i<n;i++) cin >> s[i];
  int ans=0;
  queue<int> q;
  vector<int> flag(n);
  for(int i=0;i<n;i++){
    q.push(i);
    flag[i]++;
    while(q.size()){
      int t=q.front(); q.pop();
      for(int j=0;j<n;j++){
        if(s[t][j]=='1'&&flag[j]==i){
          dis[i][j]=dis[i][t]+1;
          q.push(j);
          flag[j]++;
        }
      }
    }
  }
  vector<int> a(n,-1);
  a[0]=0;
  for(int i=1;i<n;i++){
    a[i]=dis[0][i]%2;
  }
  for(int i=0;i<n;i++){
    for(int j=0;j<n;j++){
      ans=max(ans,dis[i][j]);
      if(dis[i][j]%2!=(a[i]+a[j])%2){
        cout << -1 << endl;
        return 0;
      }
    }
  }
  cout << ans+1 << endl;
}
```