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

long long n,ans,tmp;
char c;
long long dist[205],cont;
bool vis[205][2],grafo[205][205];
vector<int> v;

void dfs(int nodo,int mode){
	vis[nodo][mode]=true;
	
	for(int i=0;i<n;i++){
		if(grafo[nodo][i] && vis[i][mode^1]==false)dfs(i,mode^1);
	}
}

void bfs(int nodo){
	tmp=max(tmp,dist[nodo]);
	for(int i=0;i<n;i++){
		if(grafo[nodo][i] && dist[i]==-1){
			dist[i]=dist[nodo]+1;
			v.push_back(i);
		}
	}
}

int main(){

	cin>>n;
	
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){
			cin>>c;
			if(c=='1')grafo[i][j]=true;
		}
	}
	
	dfs(0,0);
	
	for(int i=0;i<n;i++){
		if(vis[i][0] && vis[i][1]){
			cout<<"-1\n";
			return 0;
		}
	}
	
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){
			dist[j]=-1;
		}
		dist[i]=0;
		tmp=0;
		cont=0;
		v.clear();
		v.push_back(i);
		
		while(cont<v.size()){
			bfs(v[cont]);
			cont++;
		}
		
		ans=max(ans,tmp+1);
	}
	
	cout<<ans<<"\n";
	

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;

long long n,ans,tmp;
char c;
long long dist[205],cont;
bool vis[205][2],grafo[205][205];
vector<int> v;

void dfs(int nodo,int mode){
	vis[nodo][mode]=true;
	
	for(int i=0;i<n;i++){
		if(grafo[nodo][i] && vis[i][mode^1]==false)dfs(i,mode^1);
	}
}

void bfs(int nodo){
	tmp=max(tmp,dist[nodo]);
	for(int i=0;i<n;i++){
		if(grafo[nodo][i] && dist[i]==-1){
			dist[i]=dist[nodo]+1;
			v.push_back(i);
		}
	}
}

int main(){

	cin>>n;
	
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){
			cin>>c;
			if(c=='1')grafo[i][j]=true;
		}
	}
	
	dfs(0,0);
	
	for(int i=0;i<n;i++){
		if(vis[i][0] && vis[i][1]){
			cout<<"-1\n";
			return 0;
		}
	}
	
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){
			dist[j]=-1;
		}
		dist[i]=0;
		tmp=0;
		cont=0;
		v.clear();
		v.push_back(i);
		
		while(cont<v.size()){
			bfs(v[cont]);
			cont++;
		}
		
		ans=max(ans,tmp+1);
	}
	
	cout<<ans<<"\n";
	

return 0;
}

```