#include<bits/stdc++.h>
#define inf 0x3f3f3f3f/3
using namespace std;
const int N=209;

int flag,ans;
int n,dis[N][N];

void floyd()
{
	for(int k=1;k<=n;k++){
		for(int i=1;i<=n;i++){
			for(int j=1;j<=n;j++){
				if(dis[i][j]<inf&&dis[i][k]<inf&&dis[k][j]<inf){
					if(dis[i][j]!=dis[i][k]+dis[k][j]){
						int tmp=dis[i][j]+dis[i][k]+dis[k][j];
						if(tmp%2)flag=1;
					}
				}
				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]);
		}
	}
}

int main()
{
	ios::sync_with_stdio(false);
	cin>>n;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			if(i!=j)dis[i][j]=inf;
		}
	}
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			char c;cin>>c;
			if(c=='1')dis[i][j]=1;
		}
	}
	floyd();
	if(flag==1){
		cout<<"-1"<<endl;
	} else {	
		cout<<ans+1<<endl;
	}
	return 0;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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>
#define inf 0x3f3f3f3f/3
using namespace std;
const int N=209;

int flag,ans;
int n,dis[N][N];

void floyd()
{
	for(int k=1;k<=n;k++){
		for(int i=1;i<=n;i++){
			for(int j=1;j<=n;j++){
				if(dis[i][j]<inf&&dis[i][k]<inf&&dis[k][j]<inf){
					if(dis[i][j]!=dis[i][k]+dis[k][j]){
						int tmp=dis[i][j]+dis[i][k]+dis[k][j];
						if(tmp%2)flag=1;
					}
				}
				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]);
		}
	}
}

int main()
{
	ios::sync_with_stdio(false);
	cin>>n;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			if(i!=j)dis[i][j]=inf;
		}
	}
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			char c;cin>>c;
			if(c=='1')dis[i][j]=1;
		}
	}
	floyd();
	if(flag==1){
		cout<<"-1"<<endl;
	} else {	
		cout<<ans+1<<endl;
	}
	return 0;
}
```