#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int N;
string s[200];
int d[200][200];
vector<int>G[200];
int color[200];
bool dfs(int u,int col)
{
	color[u]=col;
	int nxt=3-col;
	for(int v:G[u])
	{
		if(color[v]!=0)
		{
			if(color[v]!=nxt)return false;
		}
		else if(!dfs(v,nxt))return false;
	}
	return true;
}
int main()
{
	cin>>N;
	for(int i=0;i<N;i++)
	{
		cin>>s[i];
		for(int j=0;j<N;j++)
		{
			d[i][j]=i==j?0:s[i][j]=='1'?1:1e9;
			if(s[i][j]=='1')G[i].push_back(j);
		}
	}
	if(!dfs(0,1))
	{
		cout<<-1<<endl;
		return 0;
	}
	for(int k=0;k<N;k++)for(int i=0;i<N;i++)for(int j=0;j<N;j++)
	{
		d[i][j]=min(d[i][j],d[i][k]+d[k][j]);
	}
	int ans=0;
	for(int i=0;i<N;i++)for(int j=0;j<N;j++)ans=max(ans,d[i][j]);
	cout<<ans+1<<endl;
}

### Prompt

Create a solution in Cpp for 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<vector>
#include<algorithm>
using namespace std;
int N;
string s[200];
int d[200][200];
vector<int>G[200];
int color[200];
bool dfs(int u,int col)
{
	color[u]=col;
	int nxt=3-col;
	for(int v:G[u])
	{
		if(color[v]!=0)
		{
			if(color[v]!=nxt)return false;
		}
		else if(!dfs(v,nxt))return false;
	}
	return true;
}
int main()
{
	cin>>N;
	for(int i=0;i<N;i++)
	{
		cin>>s[i];
		for(int j=0;j<N;j++)
		{
			d[i][j]=i==j?0:s[i][j]=='1'?1:1e9;
			if(s[i][j]=='1')G[i].push_back(j);
		}
	}
	if(!dfs(0,1))
	{
		cout<<-1<<endl;
		return 0;
	}
	for(int k=0;k<N;k++)for(int i=0;i<N;i++)for(int j=0;j<N;j++)
	{
		d[i][j]=min(d[i][j],d[i][k]+d[k][j]);
	}
	int ans=0;
	for(int i=0;i<N;i++)for(int j=0;j<N;j++)ans=max(ans,d[i][j]);
	cout<<ans+1<<endl;
}

```