#include<bits/stdc++.h>
#define N 205
using namespace std;

int a[N][N],ans=-1,n,dis[N];

inline int bfs(int x)
{
	memset(dis,0x3f,sizeof dis);
	queue<int>q;
	q.push(x);
	dis[x]=1;
	int cur=1;
	while(!q.empty())
	{
		int x=q.front();
		q.pop();cur=max(cur,dis[x]);
		for(int i=1;i<=n;++i)
			if(a[x][i])
			{
				if(dis[i]<0x3f3f3f3f)
				{
					if(dis[i]!=dis[x]-1&&dis[i]!=dis[x]+1) return -1;
				}
				else dis[i]=dis[x]+1,q.push(i);
			}
	}
	return cur;
}

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;++i)
		for(int j=1;j<=n;++j)
			scanf("%1d",&a[i][j]);
	for(int i=1;i<=n;++i) ans=max(ans,bfs(i));
	printf("%d",ans);
}
### 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>
#define N 205
using namespace std;

int a[N][N],ans=-1,n,dis[N];

inline int bfs(int x)
{
	memset(dis,0x3f,sizeof dis);
	queue<int>q;
	q.push(x);
	dis[x]=1;
	int cur=1;
	while(!q.empty())
	{
		int x=q.front();
		q.pop();cur=max(cur,dis[x]);
		for(int i=1;i<=n;++i)
			if(a[x][i])
			{
				if(dis[i]<0x3f3f3f3f)
				{
					if(dis[i]!=dis[x]-1&&dis[i]!=dis[x]+1) return -1;
				}
				else dis[i]=dis[x]+1,q.push(i);
			}
	}
	return cur;
}

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;++i)
		for(int j=1;j<=n;++j)
			scanf("%1d",&a[i][j]);
	for(int i=1;i<=n;++i) ans=max(ans,bfs(i));
	printf("%d",ans);
}
```