#include<bits/stdc++.h>
using namespace std;
const int N=222;
char s[N][N];
int n;
int dis[N];
void bfs(int f){
	memset(dis,0x3f,sizeof(dis));
	queue<int> q;
	q.push(f);
	dis[f]=1;
	while(!q.empty()){
		int x=q.front();
		q.pop();
		for(int i=0;i<n;i++){
			if(s[x][i]=='1'&&dis[x]+1<dis[i]){
				dis[i]=dis[x]+1;
				q.push(i);
				// cerr<<i<<" "<<dis[i]<<endl;
			}
		}
	}
}
int main(){
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){
			while(!isdigit(s[i][j]=getchar()))
				;
		}
	}
	bfs(0);
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){
			if(s[i][j]=='1'&&!((dis[i]^dis[j])&1)){
				// cerr<<i<<" "<<j<<endl;
				puts("-1");
				exit(0);
			}
		}
	}
	int ans=0;
	for(int i=0;i<n;i++){
		bfs(i);
		for(int j=0;j<n;j++){
			ans=max(ans,dis[j]);
		}
	}
	cout<<ans<<endl;
	return 0;
}
### 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;
const int N=222;
char s[N][N];
int n;
int dis[N];
void bfs(int f){
	memset(dis,0x3f,sizeof(dis));
	queue<int> q;
	q.push(f);
	dis[f]=1;
	while(!q.empty()){
		int x=q.front();
		q.pop();
		for(int i=0;i<n;i++){
			if(s[x][i]=='1'&&dis[x]+1<dis[i]){
				dis[i]=dis[x]+1;
				q.push(i);
				// cerr<<i<<" "<<dis[i]<<endl;
			}
		}
	}
}
int main(){
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){
			while(!isdigit(s[i][j]=getchar()))
				;
		}
	}
	bfs(0);
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){
			if(s[i][j]=='1'&&!((dis[i]^dis[j])&1)){
				// cerr<<i<<" "<<j<<endl;
				puts("-1");
				exit(0);
			}
		}
	}
	int ans=0;
	for(int i=0;i<n;i++){
		bfs(i);
		for(int j=0;j<n;j++){
			ans=max(ans,dis[j]);
		}
	}
	cout<<ans<<endl;
	return 0;
}
```