#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
using namespace std;
int s[210][210];
int que[210],dep[210];
int n,k;

inline int read()
{
	register char ch=getchar();
	register int sum(0);
	while(ch<'0'||ch>'9') ch=getchar();
	while(ch>='0'&&ch<='9') sum=sum*10+(ch^48),ch=getchar();
	return sum;
}

int work(int v)
{
	memset(dep,0,sizeof(dep));
	memset(que,0,sizeof(que));
	int head(0),tail(1);
	dep[v]=1,que[1]=v;
	while(head<tail)
	{
		++head;
		int u=que[head];
		for(int i=1;i<=n;++i)
		if(s[u][i])
			{
				if(!dep[i]) que[++tail]=i,dep[i]=dep[u]+1;
				else if((dep[i]-1!=dep[u])&&(dep[i]+1!=dep[u])) return 0;
			}
	}
	for(int i=1;i<=n;++i) k=max(k,dep[i]);
	return 1;
}

int main()
{
	n=read();
	for(int i=1;i<=n;++i)
	{
		char ch=getchar();
		while(ch!='0'&&ch!='1') ch=getchar();
		for(int j=1;j<=n;++j)
			s[i][j]=ch-'0',ch=getchar();
	}
	int flag(0);
	for(int i=1;i<=n;++i) 
	flag|=work(i);
	if(flag) cout<<k;
	else cout<<-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<iostream>
#include<cstdio>
#include<cstring>
#include<string>
using namespace std;
int s[210][210];
int que[210],dep[210];
int n,k;

inline int read()
{
	register char ch=getchar();
	register int sum(0);
	while(ch<'0'||ch>'9') ch=getchar();
	while(ch>='0'&&ch<='9') sum=sum*10+(ch^48),ch=getchar();
	return sum;
}

int work(int v)
{
	memset(dep,0,sizeof(dep));
	memset(que,0,sizeof(que));
	int head(0),tail(1);
	dep[v]=1,que[1]=v;
	while(head<tail)
	{
		++head;
		int u=que[head];
		for(int i=1;i<=n;++i)
		if(s[u][i])
			{
				if(!dep[i]) que[++tail]=i,dep[i]=dep[u]+1;
				else if((dep[i]-1!=dep[u])&&(dep[i]+1!=dep[u])) return 0;
			}
	}
	for(int i=1;i<=n;++i) k=max(k,dep[i]);
	return 1;
}

int main()
{
	n=read();
	for(int i=1;i<=n;++i)
	{
		char ch=getchar();
		while(ch!='0'&&ch!='1') ch=getchar();
		for(int j=1;j<=n;++j)
			s[i][j]=ch-'0',ch=getchar();
	}
	int flag(0);
	for(int i=1;i<=n;++i) 
	flag|=work(i);
	if(flag) cout<<k;
	else cout<<-1;
	return 0;
}
```