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

vector<int> g[222];
int c[222] = {};
bool used[222] = {};

void dfs(int now, int color){
  if(used[now]) return;
  used[now] = true;
  c[now] = color;
  for(auto p: g[now]){
    if(c[p] == color){
      cout << -1;
      exit(0);
    }
    dfs(p, 3 - color);
  }
}

int main(){
    int n;
    string s;
    cin >> n;
    if(n==2){cout<<2;return 0;}
    int dist[222][222]={};
    for(int i=0;i<n;i++)for(int j=0;j<n;j++)dist[i][j]=100000;
    for(int i=0;i<n;i++){
      cin >> s;
      for(int j = 0; j < n; j++){
        if(s[j] == '1'){
        	 g[i].push_back(j);
             dist[i][j]=dist[j][i]=1;
        }
      }
    }
    dfs(0, 1);
    for(int i=0;i<n;i++)for(int j=0;j<n;j++)for(int k=0;k<n;k++)dist[j][k]=min(dist[j][k],dist[j][i]+dist[i][k]);
    int res=0;
    for(int i=0;i<n;i++)for(int j=0;j<n;j++)res=max(res,dist[i][j]);
    cout<<res+1<<endl;
}

### Prompt

Generate a Cpp solution to 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;

vector<int> g[222];
int c[222] = {};
bool used[222] = {};

void dfs(int now, int color){
  if(used[now]) return;
  used[now] = true;
  c[now] = color;
  for(auto p: g[now]){
    if(c[p] == color){
      cout << -1;
      exit(0);
    }
    dfs(p, 3 - color);
  }
}

int main(){
    int n;
    string s;
    cin >> n;
    if(n==2){cout<<2;return 0;}
    int dist[222][222]={};
    for(int i=0;i<n;i++)for(int j=0;j<n;j++)dist[i][j]=100000;
    for(int i=0;i<n;i++){
      cin >> s;
      for(int j = 0; j < n; j++){
        if(s[j] == '1'){
        	 g[i].push_back(j);
             dist[i][j]=dist[j][i]=1;
        }
      }
    }
    dfs(0, 1);
    for(int i=0;i<n;i++)for(int j=0;j<n;j++)for(int k=0;k<n;k++)dist[j][k]=min(dist[j][k],dist[j][i]+dist[i][k]);
    int res=0;
    for(int i=0;i<n;i++)for(int j=0;j<n;j++)res=max(res,dist[i][j]);
    cout<<res+1<<endl;
}

```