#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
const int MAXN=205;
int n;
int e[MAXN][MAXN];
int grp[MAXN];
int main()
{
	cin>>n;
	bool flag=false; 
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=n;j++)
		{
			char k;
			cin>>k;
			e[i][j]=k-'0'; 
		}
	}
	int ans=0;
	for(int i=1;i<=n;i++) 
	{
		queue<int> xtq; 
		memset(grp,0,sizeof(grp));
		int cnt=1;
		grp[i]=1;
		xtq.push(i);
		bool tmp=true;
		while(!xtq.empty())
		{
			int u=xtq.front();
			xtq.pop();
			for(int j=1;j<=n;j++)
			{
				if(e[u][j])
				{
					if((grp[j]==grp[u]-1||grp[j]==grp[u]+1)&&grp[j]!=0)
						continue;
					if(grp[j]!=0)
						tmp=false;
					else
					{
						grp[j]=grp[u]+1;
						xtq.push(j);
						cnt=max(cnt,grp[j]);
					}
				}
			}
		}
		ans=max(ans,cnt);
		if(tmp==true)
			flag=true;
	}
	if(!flag)
		cout<<-1<<endl;
	else 
		cout<<ans<<endl;
	return 0;
}

### 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 <cstring>
#include <queue>
using namespace std;
const int MAXN=205;
int n;
int e[MAXN][MAXN];
int grp[MAXN];
int main()
{
	cin>>n;
	bool flag=false; 
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=n;j++)
		{
			char k;
			cin>>k;
			e[i][j]=k-'0'; 
		}
	}
	int ans=0;
	for(int i=1;i<=n;i++) 
	{
		queue<int> xtq; 
		memset(grp,0,sizeof(grp));
		int cnt=1;
		grp[i]=1;
		xtq.push(i);
		bool tmp=true;
		while(!xtq.empty())
		{
			int u=xtq.front();
			xtq.pop();
			for(int j=1;j<=n;j++)
			{
				if(e[u][j])
				{
					if((grp[j]==grp[u]-1||grp[j]==grp[u]+1)&&grp[j]!=0)
						continue;
					if(grp[j]!=0)
						tmp=false;
					else
					{
						grp[j]=grp[u]+1;
						xtq.push(j);
						cnt=max(cnt,grp[j]);
					}
				}
			}
		}
		ans=max(ans,cnt);
		if(tmp==true)
			flag=true;
	}
	if(!flag)
		cout<<-1<<endl;
	else 
		cout<<ans<<endl;
	return 0;
}

```