#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=202;

char adjM[N][N];

int res_BFS(int v,int n);

int main()
{
	int n; scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%s",adjM[i]+1);
	int res = -1;
	for(int i=1;i<=n;i++) res = max(res,res_BFS(i,n));
	printf("%d\n",res);
}
///////
int res_BFS(int v,int n)
{
	queue<int> kol;
	int dist[N];
	for(int i=1;i<=n;i++) dist[i]=2*N;
	kol.push(v);
	dist[v]=0;

	while(!kol.empty())
	{
		int x = kol.front();
		kol.pop();

		for(int i=1;i<=n;i++)
		{
			if(adjM[x][i]=='1' && dist[i]>dist[x]+1)
			{
				dist[i]=dist[x]+1;
				kol.push(i);
			}
		}
	}

	int res = 0;
	for(int i=1;i<=n;i++)
	{
		res = max(res,dist[i]);
		for(int j=1;j<=n;j++)
		{
			if((dist[i]==dist[j] || abs(dist[i]-dist[j])>1) && adjM[i][j] == '1') return -1;
		}
	}

	return res+1;
}

### 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;
typedef long long ll;
const int N=202;

char adjM[N][N];

int res_BFS(int v,int n);

int main()
{
	int n; scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%s",adjM[i]+1);
	int res = -1;
	for(int i=1;i<=n;i++) res = max(res,res_BFS(i,n));
	printf("%d\n",res);
}
///////
int res_BFS(int v,int n)
{
	queue<int> kol;
	int dist[N];
	for(int i=1;i<=n;i++) dist[i]=2*N;
	kol.push(v);
	dist[v]=0;

	while(!kol.empty())
	{
		int x = kol.front();
		kol.pop();

		for(int i=1;i<=n;i++)
		{
			if(adjM[x][i]=='1' && dist[i]>dist[x]+1)
			{
				dist[i]=dist[x]+1;
				kol.push(i);
			}
		}
	}

	int res = 0;
	for(int i=1;i<=n;i++)
	{
		res = max(res,dist[i]);
		for(int j=1;j<=n;j++)
		{
			if((dist[i]==dist[j] || abs(dist[i]-dist[j])>1) && adjM[i][j] == '1') return -1;
		}
	}

	return res+1;
}

```