#include <bits/stdc++.h>
using namespace std;
char con[205][205];
vector<int>adj[205];
bool vis[205], col[205];
int dist[205][205];
void dfs(int x){
	//cout<<x<<endl;
	vis[x]=true;
	for(int y:adj[x]){
		if(vis[y]) continue;
		col[y]=1^col[x];
		dfs(y);
	}
}
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	int n;
	cin>>n;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			cin>>con[i][j];
			if(con[i][j]=='1'){
				adj[i].push_back(j);
				dist[i][j]=1;
			}
			else if(i==j) dist[i][j]=0;
			else dist[i][j]=1000;
		}
	}
	col[1]=0;
	dfs(1);
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			if(con[i][j]=='1' && col[i]==col[j]){
				cout<<-1;
				return 0;
			}
		}
	}
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			for(int k=1;k<=n;k++){
				dist[j][k]=min(dist[j][k], dist[j][i]+dist[k][i]);
			}
		}
	}
	int ans=0;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			ans=max(ans, dist[i][j]);
		}
	}
	cout<<ans+1;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
char con[205][205];
vector<int>adj[205];
bool vis[205], col[205];
int dist[205][205];
void dfs(int x){
	//cout<<x<<endl;
	vis[x]=true;
	for(int y:adj[x]){
		if(vis[y]) continue;
		col[y]=1^col[x];
		dfs(y);
	}
}
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	int n;
	cin>>n;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			cin>>con[i][j];
			if(con[i][j]=='1'){
				adj[i].push_back(j);
				dist[i][j]=1;
			}
			else if(i==j) dist[i][j]=0;
			else dist[i][j]=1000;
		}
	}
	col[1]=0;
	dfs(1);
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			if(con[i][j]=='1' && col[i]==col[j]){
				cout<<-1;
				return 0;
			}
		}
	}
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			for(int k=1;k<=n;k++){
				dist[j][k]=min(dist[j][k], dist[j][i]+dist[k][i]);
			}
		}
	}
	int ans=0;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			ans=max(ans, dist[i][j]);
		}
	}
	cout<<ans+1;
}

```