#include<bits/stdc++.h> 
#define inf 1e18
using namespace std;
typedef long long int ll;
int n;
char w[205][205];
vector<int>g[205];
queue<int>q;
int cl[205];
int ans=0;
int main()
{
	scanf("%d",&n);getchar();
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=n;j++)
		{
			w[i][j]=getchar();
			if(w[i][j]=='1')g[i].push_back(j);
		}
		getchar();
	}
	for(int i=1;i<=n;i++)
	{
		memset(cl,0,sizeof(cl));
		cl[i]=1;
		q.push(i);
		int flag=1,tmpans=0;;
		while(!q.empty())
		{
			int t=q.front();q.pop();
			tmpans=max(tmpans,cl[t]);
			for(int j=0;j<g[t].size();j++)
			{
				int v=g[t][j];
				if(cl[v]&&abs(cl[v]-cl[t])!=1)
				{
					flag=0;
					break;
				}
				else if(!cl[v])cl[v]=cl[t]+1,q.push(v);
			}
			if(!flag)break;
		}
		if(flag)ans=max(ans,tmpans);
	}
	if(!ans)printf("-1\n");
	else printf("%d\n",ans);
}
### 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> 
#define inf 1e18
using namespace std;
typedef long long int ll;
int n;
char w[205][205];
vector<int>g[205];
queue<int>q;
int cl[205];
int ans=0;
int main()
{
	scanf("%d",&n);getchar();
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=n;j++)
		{
			w[i][j]=getchar();
			if(w[i][j]=='1')g[i].push_back(j);
		}
		getchar();
	}
	for(int i=1;i<=n;i++)
	{
		memset(cl,0,sizeof(cl));
		cl[i]=1;
		q.push(i);
		int flag=1,tmpans=0;;
		while(!q.empty())
		{
			int t=q.front();q.pop();
			tmpans=max(tmpans,cl[t]);
			for(int j=0;j<g[t].size();j++)
			{
				int v=g[t][j];
				if(cl[v]&&abs(cl[v]-cl[t])!=1)
				{
					flag=0;
					break;
				}
				else if(!cl[v])cl[v]=cl[t]+1,q.push(v);
			}
			if(!flag)break;
		}
		if(flag)ans=max(ans,tmpans);
	}
	if(!ans)printf("-1\n");
	else printf("%d\n",ans);
}
```