#include<bits/stdc++.h>
using namespace std;

int edge[200][200];
string s[300];
int l[200];

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n;
	cin>>n;
	for(int i=0;i<n;i++){
		cin>>s[i];
		for(int j=0;j<n;j++)
			edge[i][j]=int(s[i][j]-'0');
	}
	queue<int> q;
	int ans=-1;
	for(int i=0;i<n;i++){
		memset(l,-1,sizeof l);
		l[i]=0;
		q.push(i);
		bool ok=true;
		while(!q.empty()){
			int now=q.front();q.pop();
			for(int j=0;j<n;j++)
				if(edge[now][j]){
					
					if(l[j]==-1){
						l[j]=l[now]+1;
						q.push(j);
					}
					else{
						if(abs(l[j]-l[now])!=1){
							ok=false;
							break;
						}
					}
				}
			if(!ok){
				while(!q.empty())	q.pop();
				break;
			}
		}
		if(!ok)	continue;
		for(int j=0;j<n;j++)
			ans=max(ans,l[j]+1);
	}
	cout<<ans<<endl;
}

### Prompt

Develop a solution in CPP 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>
using namespace std;

int edge[200][200];
string s[300];
int l[200];

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n;
	cin>>n;
	for(int i=0;i<n;i++){
		cin>>s[i];
		for(int j=0;j<n;j++)
			edge[i][j]=int(s[i][j]-'0');
	}
	queue<int> q;
	int ans=-1;
	for(int i=0;i<n;i++){
		memset(l,-1,sizeof l);
		l[i]=0;
		q.push(i);
		bool ok=true;
		while(!q.empty()){
			int now=q.front();q.pop();
			for(int j=0;j<n;j++)
				if(edge[now][j]){
					
					if(l[j]==-1){
						l[j]=l[now]+1;
						q.push(j);
					}
					else{
						if(abs(l[j]-l[now])!=1){
							ok=false;
							break;
						}
					}
				}
			if(!ok){
				while(!q.empty())	q.pop();
				break;
			}
		}
		if(!ok)	continue;
		for(int j=0;j<n;j++)
			ans=max(ans,l[j]+1);
	}
	cout<<ans<<endl;
}

```