#include <bits/stdc++.h>
using namespace std;
const int N=205;
char mat[N][N];
int dist[N],n;
int BFS(int u)
{
	for(int i=1;i<=n;i++) dist[i]=-1;
	queue<int> q;
	q.push(u);
	dist[u]=1;
	while(q.size())
	{
		int x=q.front();
		q.pop();
		for(int y=1;y<=n;y++) if(mat[x][y]=='1')
		{
			if(dist[y]==-1)
			{
				dist[y]=dist[x]+1;
				q.push(y);
			}
			else if(dist[x]==dist[y])
			{
				printf("-1\n");
				exit(0);
			}
		}
	}
	int mx=0;
	for(int i=1;i<=n;i++) mx=max(mx,dist[i]);
	return mx;
}
int main()
{
	scanf("%i",&n);
	for(int i=1;i<=n;i++) scanf("%s",mat[i]+1);
	int ans=0;
	for(int i=1;i<=n;i++) ans=max(ans,BFS(i));
	printf("%i\n",ans);
	return 0;
}
### Prompt

Please formulate a cpp solution to the following problem:

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=205;
char mat[N][N];
int dist[N],n;
int BFS(int u)
{
	for(int i=1;i<=n;i++) dist[i]=-1;
	queue<int> q;
	q.push(u);
	dist[u]=1;
	while(q.size())
	{
		int x=q.front();
		q.pop();
		for(int y=1;y<=n;y++) if(mat[x][y]=='1')
		{
			if(dist[y]==-1)
			{
				dist[y]=dist[x]+1;
				q.push(y);
			}
			else if(dist[x]==dist[y])
			{
				printf("-1\n");
				exit(0);
			}
		}
	}
	int mx=0;
	for(int i=1;i<=n;i++) mx=max(mx,dist[i]);
	return mx;
}
int main()
{
	scanf("%i",&n);
	for(int i=1;i<=n;i++) scanf("%s",mat[i]+1);
	int ans=0;
	for(int i=1;i<=n;i++) ans=max(ans,BFS(i));
	printf("%i\n",ans);
	return 0;
}
```