#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
using namespace std;
const int N=310;
int n,f[N][N],vi[N];
char s[N][N];
bool bl;
void dfs(int x)
{
	for (int i=1;i<=n;i++)
	{
		if (s[x][i]=='0') continue;
		if (vi[i]==vi[x])
		{
			bl=0;
			continue;
		}
		if (vi[i]) continue;
		vi[i]=3-vi[x];
		dfs(i);
	}
}
int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++) scanf("%s",s[i]+1);
	vi[1]=1;bl=1;
	dfs(1);
	if (!bl)
	{
		printf("-1\n");
		return 0;
	}
	for (int i=1;i<=n;i++)
	{
		for (int j=1;j<=n;j++)
		{
			if (s[i][j]=='0') f[i][j]=inf;
			else f[i][j]=1;
		}
	}
	for (int i=1;i<=n;i++) f[i][i]=0;
	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]);
		}
	}
	int ans=0;
	for (int i=1;i<=n;i++)
	{
		for (int j=1;j<=n;j++) ans=max(ans,f[i][j]);
	}
	ans++;
	printf("%d\n",ans);
}
### 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>
#define inf 0x3f3f3f3f
using namespace std;
const int N=310;
int n,f[N][N],vi[N];
char s[N][N];
bool bl;
void dfs(int x)
{
	for (int i=1;i<=n;i++)
	{
		if (s[x][i]=='0') continue;
		if (vi[i]==vi[x])
		{
			bl=0;
			continue;
		}
		if (vi[i]) continue;
		vi[i]=3-vi[x];
		dfs(i);
	}
}
int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++) scanf("%s",s[i]+1);
	vi[1]=1;bl=1;
	dfs(1);
	if (!bl)
	{
		printf("-1\n");
		return 0;
	}
	for (int i=1;i<=n;i++)
	{
		for (int j=1;j<=n;j++)
		{
			if (s[i][j]=='0') f[i][j]=inf;
			else f[i][j]=1;
		}
	}
	for (int i=1;i<=n;i++) f[i][i]=0;
	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]);
		}
	}
	int ans=0;
	for (int i=1;i<=n;i++)
	{
		for (int j=1;j<=n;j++) ans=max(ans,f[i][j]);
	}
	ans++;
	printf("%d\n",ans);
}
```