#include<bits/stdc++.h>
using namespace std;
int a[222][222];
int col[222];
int n;
bool dfs(int x,int c)
{
	if(col[x]!=-1)
	{
		if(col[x]!=c){
			return false;
		}
		return true;
	}
	col[x] = c;
	for(int i=1;i<=n;++i)
	{
		if(a[x][i]==1){
			if(!dfs(i,!c)){
				return false;
			}
		}
	}
	return true;
}
int main() {
	cin>>n;
	memset(a,0x3f,sizeof a);
	memset(col,-1,sizeof col);
	
	for(int i=1;i<=n;++i) 
	{
		a[i][i] = 0;
		for(int j=1;j<=n;++j)
		{
			char c;
			cin>>c;
			if(c!='0')
			{
				a[i][j] = 1;
			}
		}
	}
	if(!dfs(1,0)){
		cout<<-1<<endl;
		return 0;
	}
	for(int k=1;k<=n;++k)
	{
		for(int i=1;i<=n;++i)
		{
			for(int j=1;j<=n;++j)
			{
				a[i][j] = min(a[i][j],a[i][k] + a[k][j]);
			}
		}
	}
	int ans = 0;
	for(int i=1;i<=n;++i)
	{
		for(int j=1;j<=n;++j)
		{
			//cout<<i<<" "<<j<<": "<<a[i][j]<<endl;
			ans=max(ans,a[i][j]);
		}
	}
	cout<<ans+1<<endl;
} 
### Prompt

Your challenge is to write 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<bits/stdc++.h>
using namespace std;
int a[222][222];
int col[222];
int n;
bool dfs(int x,int c)
{
	if(col[x]!=-1)
	{
		if(col[x]!=c){
			return false;
		}
		return true;
	}
	col[x] = c;
	for(int i=1;i<=n;++i)
	{
		if(a[x][i]==1){
			if(!dfs(i,!c)){
				return false;
			}
		}
	}
	return true;
}
int main() {
	cin>>n;
	memset(a,0x3f,sizeof a);
	memset(col,-1,sizeof col);
	
	for(int i=1;i<=n;++i) 
	{
		a[i][i] = 0;
		for(int j=1;j<=n;++j)
		{
			char c;
			cin>>c;
			if(c!='0')
			{
				a[i][j] = 1;
			}
		}
	}
	if(!dfs(1,0)){
		cout<<-1<<endl;
		return 0;
	}
	for(int k=1;k<=n;++k)
	{
		for(int i=1;i<=n;++i)
		{
			for(int j=1;j<=n;++j)
			{
				a[i][j] = min(a[i][j],a[i][k] + a[k][j]);
			}
		}
	}
	int ans = 0;
	for(int i=1;i<=n;++i)
	{
		for(int j=1;j<=n;++j)
		{
			//cout<<i<<" "<<j<<": "<<a[i][j]<<endl;
			ans=max(ans,a[i][j]);
		}
	}
	cout<<ans+1<<endl;
} 
```