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

int N;
vector<string> S(200);
vector<int> bw(200,0);
int dist[200][200];

int BFS(){
	queue<int> que;
	que.push(0);
	bw[0]=1;
	while(!que.empty()){
		int f=que.front();
		que.pop();
		for(int j=0;j<N;j++){
			if(S[f][j]=='1'){
				if(bw[f]==bw[j])return 0;
				if(bw[j]==0){
					bw[j]=bw[f]*-1;
					que.push(j);
				}
			}
		}
	}
	return 1;
}

int main(){
	cin >> N;
	for(int i=0;i<N;i++)cin >> S[i];
	
	if(BFS()==0){
		cout << "-1" << endl;
		return 0;
	}
	
	for(int i=0;i<N;i++)for(int j=0;j<N;j++){
		if(i==j)dist[i][j]=0;
		else if(S[i][j]=='1')dist[i][j]=1;
		else dist[i][j]=9999999;
	}
	
	for(int k=0;k<N;k++)for(int i=0;i<N;i++)for(int j=0;j<N;j++)dist[i][j]=min(dist[i][j],dist[i][k]+dist[k][j]);
	
	int ans=0;
	for(int i=0;i<N;i++)for(int j=0;j<N;j++)if(dist[i][j]<9999999)ans=max(ans,dist[i][j]);
	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>
using namespace std;

int N;
vector<string> S(200);
vector<int> bw(200,0);
int dist[200][200];

int BFS(){
	queue<int> que;
	que.push(0);
	bw[0]=1;
	while(!que.empty()){
		int f=que.front();
		que.pop();
		for(int j=0;j<N;j++){
			if(S[f][j]=='1'){
				if(bw[f]==bw[j])return 0;
				if(bw[j]==0){
					bw[j]=bw[f]*-1;
					que.push(j);
				}
			}
		}
	}
	return 1;
}

int main(){
	cin >> N;
	for(int i=0;i<N;i++)cin >> S[i];
	
	if(BFS()==0){
		cout << "-1" << endl;
		return 0;
	}
	
	for(int i=0;i<N;i++)for(int j=0;j<N;j++){
		if(i==j)dist[i][j]=0;
		else if(S[i][j]=='1')dist[i][j]=1;
		else dist[i][j]=9999999;
	}
	
	for(int k=0;k<N;k++)for(int i=0;i<N;i++)for(int j=0;j<N;j++)dist[i][j]=min(dist[i][j],dist[i][k]+dist[k][j]);
	
	int ans=0;
	for(int i=0;i<N;i++)for(int j=0;j<N;j++)if(dist[i][j]<9999999)ans=max(ans,dist[i][j]);
	cout << ans+1 << endl;
	
	return 0;
}
```