#include<bits/stdc++.h>
using namespace std;
int n;
bool s[203][203];
vector<int>v,g[203];
int vis[203];
int main(){
	cin>>n;
	for(int i=0;i<n;i++){
		string s;cin>>s;
		for(int j=0;j<n;j++)
			if(s[j]=='1')g[i].push_back(j);
	}
	int ans=-1;
	for(int i=0;i<n;i++){
		queue<pair<int,int> >q;
		q.push(make_pair(i,1));
		int dis[203];
		memset(dis,0,sizeof(dis));
		dis[i]=1;
		while(!q.empty()){
			int x=q.front().first,len=q.front().second;
			q.pop();
			for(int j=0;j<g[x].size();j++)
				if(!dis[g[x][j]]){
					dis[g[x][j]]=len+1;
					q.push(make_pair(g[x][j],len+1));
				}else if(dis[g[x][j]]%2!=(len+1)%2){
					cout<<-1;
					return 0;
				}
		}
		int res=0;
		for(int j=0;j<n;j++)res=max(res,dis[j]);
		ans=max(ans,res);
	}
	cout<<ans;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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 n;
bool s[203][203];
vector<int>v,g[203];
int vis[203];
int main(){
	cin>>n;
	for(int i=0;i<n;i++){
		string s;cin>>s;
		for(int j=0;j<n;j++)
			if(s[j]=='1')g[i].push_back(j);
	}
	int ans=-1;
	for(int i=0;i<n;i++){
		queue<pair<int,int> >q;
		q.push(make_pair(i,1));
		int dis[203];
		memset(dis,0,sizeof(dis));
		dis[i]=1;
		while(!q.empty()){
			int x=q.front().first,len=q.front().second;
			q.pop();
			for(int j=0;j<g[x].size();j++)
				if(!dis[g[x][j]]){
					dis[g[x][j]]=len+1;
					q.push(make_pair(g[x][j],len+1));
				}else if(dis[g[x][j]]%2!=(len+1)%2){
					cout<<-1;
					return 0;
				}
		}
		int res=0;
		for(int j=0;j<n;j++)res=max(res,dis[j]);
		ans=max(ans,res);
	}
	cout<<ans;
}
```