#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
const int inf=1e9;
int n,sl,fh,ans,col[210],dis[210][210];
int rd()
{
	sl=0;fh=1;
	char ch=getchar();
	while(ch<'0'||'9'<ch) {if(ch=='-') fh=-1; ch=getchar();}
	while('0'<=ch&&ch<='9') sl=sl*10+ch-'0',ch=getchar();
	return sl*fh;
}
void dfs(int u,int c)
{
	col[u]=c;
	for(int i=1;i<=n;++i)
		if(dis[u][i]<inf)
		{
			if(!~col[i]) dfs(i,c^1);
			else if(col[i]==col[u]) {puts("-1"); exit(0);}
		}
}
int main()
{
	n=rd();
	for(int i=1;i<=n;++i) col[i]=-1;
	for(int i=1;i<=n;++i)
		for(int j=1;j<=n;++j)
		{
			scanf("%1d",&dis[i][j]);
			if(!dis[i][j]) dis[i][j]=inf;
		}
	dfs(1,0);
	for(int k=1;k<=n;++k) 
		for(int i=1;i<=n;++i)
			if(i!=k)
				for(int j=1;j<=n;++j)
					if(i!=j&&j!=k)
						dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);
	for(int i=1;i<=n;++i)
		for(int j=1;j<=n;++j)
			if(dis[i][j]<inf)
				ans=max(ans,dis[i][j]);
	printf("%d\n",ans+1);
	return 0;
}
### Prompt

Please formulate 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<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
const int inf=1e9;
int n,sl,fh,ans,col[210],dis[210][210];
int rd()
{
	sl=0;fh=1;
	char ch=getchar();
	while(ch<'0'||'9'<ch) {if(ch=='-') fh=-1; ch=getchar();}
	while('0'<=ch&&ch<='9') sl=sl*10+ch-'0',ch=getchar();
	return sl*fh;
}
void dfs(int u,int c)
{
	col[u]=c;
	for(int i=1;i<=n;++i)
		if(dis[u][i]<inf)
		{
			if(!~col[i]) dfs(i,c^1);
			else if(col[i]==col[u]) {puts("-1"); exit(0);}
		}
}
int main()
{
	n=rd();
	for(int i=1;i<=n;++i) col[i]=-1;
	for(int i=1;i<=n;++i)
		for(int j=1;j<=n;++j)
		{
			scanf("%1d",&dis[i][j]);
			if(!dis[i][j]) dis[i][j]=inf;
		}
	dfs(1,0);
	for(int k=1;k<=n;++k) 
		for(int i=1;i<=n;++i)
			if(i!=k)
				for(int j=1;j<=n;++j)
					if(i!=j&&j!=k)
						dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);
	for(int i=1;i<=n;++i)
		for(int j=1;j<=n;++j)
			if(dis[i][j]<inf)
				ans=max(ans,dis[i][j]);
	printf("%d\n",ans+1);
	return 0;
}
```