#include<bits/stdc++.h>
using namespace std;
const int N=201,I=1e9;
string s[N];
bool g[N][N];
int n,f[N][N],v[N],A,q[N],t,X;
int main()
{
	cin>>n;
	for(int i=1;i<=n;i++)
		cin>>s[i];
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			f[i][j]=I;
	for(int i=1;i<=n;i++)
		for(int j=0;j<s[i].size();j++)
			if(s[i][j]=='1')
				f[i][j+1]=g[i][j+1]=1;
	v[1]=1,q[++t]=1;
	while(t)
	{
		X=q[t--];
		for(int i=1;i<=n;i++)
			if(g[X][i])
			{
				if(v[X]==v[i])
				{
					cout<<-1;
					return 0;
				}
				else if(!v[i])
					q[++t]=i;
				v[i]=-v[X];
			}
	}
	for(int k=1;k<=n;k++)
		for(int i=1;i<=n;i++)
			for(int j=1;j<=n;j++)
				f[i][j]=min(f[i][j],f[i][k]+f[k][j]);
	for(int i=1;i<=n;i++)
		for(int j=1;j<i;j++)
			if(f[i][j]!=I)
				A=max(A,f[i][j]);
	cout<<A+1;
	return 0;
}
### Prompt

Your challenge is to write 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=201,I=1e9;
string s[N];
bool g[N][N];
int n,f[N][N],v[N],A,q[N],t,X;
int main()
{
	cin>>n;
	for(int i=1;i<=n;i++)
		cin>>s[i];
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			f[i][j]=I;
	for(int i=1;i<=n;i++)
		for(int j=0;j<s[i].size();j++)
			if(s[i][j]=='1')
				f[i][j+1]=g[i][j+1]=1;
	v[1]=1,q[++t]=1;
	while(t)
	{
		X=q[t--];
		for(int i=1;i<=n;i++)
			if(g[X][i])
			{
				if(v[X]==v[i])
				{
					cout<<-1;
					return 0;
				}
				else if(!v[i])
					q[++t]=i;
				v[i]=-v[X];
			}
	}
	for(int k=1;k<=n;k++)
		for(int i=1;i<=n;i++)
			for(int j=1;j<=n;j++)
				f[i][j]=min(f[i][j],f[i][k]+f[k][j]);
	for(int i=1;i<=n;i++)
		for(int j=1;j<i;j++)
			if(f[i][j]!=I)
				A=max(A,f[i][j]);
	cout<<A+1;
	return 0;
}
```