#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;

int d[201];
char ss[201][202];
int main(){
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%s",ss[i]+1);
	int ans=0;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++)
			d[j]=0;
		queue<int>q;
		q.push(i);
		d[i]=1;
		while(!q.empty()){
			int u=q.front();
			q.pop();
			for(int j=1;j<=n;j++)
				if(ss[u][j]=='1'){
					if(!d[j]){
						d[j]=d[u]+1;
						q.push(j);
					}
					else if((d[u]&1)==(d[j]&1)){
						printf("-1\n");
						return 0;
					}
			}
		}
		for(int j=1;j<=n;j++)
			ans=max(ans,d[j]);
	}
	printf("%d\n",ans);
}

### Prompt

Develop a solution in cpp 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<cstdio>
#include<algorithm>
#include<queue>
using namespace std;

int d[201];
char ss[201][202];
int main(){
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%s",ss[i]+1);
	int ans=0;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++)
			d[j]=0;
		queue<int>q;
		q.push(i);
		d[i]=1;
		while(!q.empty()){
			int u=q.front();
			q.pop();
			for(int j=1;j<=n;j++)
				if(ss[u][j]=='1'){
					if(!d[j]){
						d[j]=d[u]+1;
						q.push(j);
					}
					else if((d[u]&1)==(d[j]&1)){
						printf("-1\n");
						return 0;
					}
			}
		}
		for(int j=1;j<=n;j++)
			ans=max(ans,d[j]);
	}
	printf("%d\n",ans);
}

```