#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
const int N=205;
int n;
int a[N][N];
int col[N];
void dfs(int u,int color)
{
	if(col[u]!=-1)
	{
		if(col[u]!=color)
		{
			printf("-1");
			exit(0);
		}
		return;
	}
	col[u]=color;
	for(int v=1;v<=n;v++)
		if(a[u][v]) dfs(v,color^1);
	return;
}
int f[N][N];
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			scanf("%1d",&a[i][j]);
	memset(col,-1,sizeof(col));
	dfs(1,0);
	memset(f,63,sizeof(f));
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			if(a[i][j]) f[i][j]=a[i][j];
	for(int i=1;i<=n;i++)
		f[i][i]=0;
	for(int k=1;k<=n;k++)
		for(int i=1;i<=n;i++)
			for(int j=1;j<=n;j++)
				f[i][j]=min(f[i][j],f[i][k]+f[k][j]);
	int ans=0;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			ans=max(ans,f[i][j]+1);
	printf("%d",ans);
	return 0;
}
### 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<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
const int N=205;
int n;
int a[N][N];
int col[N];
void dfs(int u,int color)
{
	if(col[u]!=-1)
	{
		if(col[u]!=color)
		{
			printf("-1");
			exit(0);
		}
		return;
	}
	col[u]=color;
	for(int v=1;v<=n;v++)
		if(a[u][v]) dfs(v,color^1);
	return;
}
int f[N][N];
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			scanf("%1d",&a[i][j]);
	memset(col,-1,sizeof(col));
	dfs(1,0);
	memset(f,63,sizeof(f));
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			if(a[i][j]) f[i][j]=a[i][j];
	for(int i=1;i<=n;i++)
		f[i][i]=0;
	for(int k=1;k<=n;k++)
		for(int i=1;i<=n;i++)
			for(int j=1;j<=n;j++)
				f[i][j]=min(f[i][j],f[i][k]+f[k][j]);
	int ans=0;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			ans=max(ans,f[i][j]+1);
	printf("%d",ans);
	return 0;
}
```