/*Lucky_Glass*/
#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int N=200;

int n,Vans;
vector<int> lnk[N+3];
char str[N+3];
int Etag[N+3];

void BFS(int st){
	for(int i=1;i<=n;i++) Etag[i]=0;
	queue<int> que;
	que.push(st);Etag[st]=1;
	while(!que.empty()){
		int u=que.front();que.pop();
		for(auto v : lnk[u]){
			if(Etag[v]){
				if(Etag[v]!=Etag[u]+1 && Etag[v]!=Etag[u]-1){
					printf("-1\n");
					exit(0);
				}
				else continue;
			}
			Etag[v]=Etag[u]+1;
			que.push(v);
			Vans=max(Vans,Etag[v]);
		}
	}
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%s",str+1);
		for(int j=1;j<=n;j++)
			if(str[j]=='1')
				lnk[i].push_back(j);
	}
	for(int i=1;i<=n;i++)
		BFS(i);
	printf("%d\n",Vans);
	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
/*Lucky_Glass*/
#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int N=200;

int n,Vans;
vector<int> lnk[N+3];
char str[N+3];
int Etag[N+3];

void BFS(int st){
	for(int i=1;i<=n;i++) Etag[i]=0;
	queue<int> que;
	que.push(st);Etag[st]=1;
	while(!que.empty()){
		int u=que.front();que.pop();
		for(auto v : lnk[u]){
			if(Etag[v]){
				if(Etag[v]!=Etag[u]+1 && Etag[v]!=Etag[u]-1){
					printf("-1\n");
					exit(0);
				}
				else continue;
			}
			Etag[v]=Etag[u]+1;
			que.push(v);
			Vans=max(Vans,Etag[v]);
		}
	}
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%s",str+1);
		for(int j=1;j<=n;j++)
			if(str[j]=='1')
				lnk[i].push_back(j);
	}
	for(int i=1;i<=n;i++)
		BFS(i);
	printf("%d\n",Vans);
	return 0;
}
```