#include<bits/stdc++.h>
using namespace std;
vector<int> v[210];
bool vis[210];
int dep[210];
inline void dfs(int np)
{
	vis[np]=1;
	for(int x:v[np])
	{
		if(vis[x])
		{
			if(dep[x]==dep[np])
			{
				puts("-1");
				exit(0);
			}
		}
		else
		{
			dep[x]=dep[np]^1;
			dfs(x);
		}
	}
}
int n,ans=0;
void bfs(int st)
{
	memset(vis,0,sizeof(vis));
	memset(dep,0,sizeof(dep));
	queue<int> q;
	vis[st]=1;
	q.push(st);
	dep[st]=1;
	while(!q.empty())
	{
		int fr=q.front();q.pop();
		for(int x:v[fr])
		{
			if(!vis[x])
			{
				dep[x]=dep[fr]+1;
				vis[x]=1;
				q.push(x);
			}
		}
	}
	for(int i=1;i<=n;i++)ans=max(ans,dep[i]);
}
char s[210][210];
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%s",s[i]+1);
		for(int j=1;j<=n;j++)
		{
			if(s[i][j]=='1')v[i].push_back(j);
		}
	}
	dfs(1);
	for(int i=1;i<=n;i++)bfs(i);
	printf("%d\n",ans);
	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<bits/stdc++.h>
using namespace std;
vector<int> v[210];
bool vis[210];
int dep[210];
inline void dfs(int np)
{
	vis[np]=1;
	for(int x:v[np])
	{
		if(vis[x])
		{
			if(dep[x]==dep[np])
			{
				puts("-1");
				exit(0);
			}
		}
		else
		{
			dep[x]=dep[np]^1;
			dfs(x);
		}
	}
}
int n,ans=0;
void bfs(int st)
{
	memset(vis,0,sizeof(vis));
	memset(dep,0,sizeof(dep));
	queue<int> q;
	vis[st]=1;
	q.push(st);
	dep[st]=1;
	while(!q.empty())
	{
		int fr=q.front();q.pop();
		for(int x:v[fr])
		{
			if(!vis[x])
			{
				dep[x]=dep[fr]+1;
				vis[x]=1;
				q.push(x);
			}
		}
	}
	for(int i=1;i<=n;i++)ans=max(ans,dep[i]);
}
char s[210][210];
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%s",s[i]+1);
		for(int j=1;j<=n;j++)
		{
			if(s[i][j]=='1')v[i].push_back(j);
		}
	}
	dfs(1);
	for(int i=1;i<=n;i++)bfs(i);
	printf("%d\n",ans);
	return 0;
}
```