#include<bits/stdc++.h>
using namespace std;
int n;
int dis[1001];
int e[1001][1001];
bool bfs(int s)
{
	memset(dis,0,sizeof(dis));
	queue<int>q;
	q.push(s);
	dis[s]=1;
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		for(int i=1;i<=n;i++)
			if(e[u][i])
			{
				if(!dis[i])
				{
					dis[i]=dis[u]+1;
					q.push(i);
				}
				else if(abs(dis[i]-dis[u])!=1)
					return false;
			}
	}
	return true;
}
signed main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
		{
			scanf("%1d",&e[i][j]);
		}
	int ans=-1;
	for(int i=1;i<=n;i++)
	{
		if(bfs(i))
			for(int j=1;j<=n;j++)
				ans=max(ans,dis[j]);
	}
	printf("%d",ans);
	return 0;
}
/*
aqsqqsqqwww
10
*/
### Prompt

Please create a solution in Cpp 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;
int n;
int dis[1001];
int e[1001][1001];
bool bfs(int s)
{
	memset(dis,0,sizeof(dis));
	queue<int>q;
	q.push(s);
	dis[s]=1;
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		for(int i=1;i<=n;i++)
			if(e[u][i])
			{
				if(!dis[i])
				{
					dis[i]=dis[u]+1;
					q.push(i);
				}
				else if(abs(dis[i]-dis[u])!=1)
					return false;
			}
	}
	return true;
}
signed main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
		{
			scanf("%1d",&e[i][j]);
		}
	int ans=-1;
	for(int i=1;i<=n;i++)
	{
		if(bfs(i))
			for(int j=1;j<=n;j++)
				ans=max(ans,dis[j]);
	}
	printf("%d",ans);
	return 0;
}
/*
aqsqqsqqwww
10
*/
```