#include<bits/stdc++.h>
#define llong long long
using namespace std;

const int N = 200;
char g[N+3][N+3];
int dis[N+3][N+3];
int n;

int absl(int x) {return x<0?-x:x;}

int main()
{
	scanf("%d",&n);
	for(int i=1; i<=n; i++) scanf("%s",g[i]+1);
	memset(dis,42,sizeof(dis));
	for(int i=1; i<=n; i++) for(int j=1; j<=n; j++) {if(g[i][j]=='1') dis[i][j] = 1;}
	for(int i=1; i<=n; i++) dis[i][i] = 0;
	for(int k=1; k<=n; k++)
	{
		for(int i=1; i<=n; i++)
		{
			for(int j=1; j<=n; j++)
			{
				dis[i][j] = min(dis[i][j],dis[i][k]+dis[k][j]);
			}
		}
	}
//	for(int i=1; i<=n; i++) {for(int j=1; j<=n; j++) printf("%d ",dis[i][j]); puts("");}
	int ans = -1;
	for(int k=1; k<=n; k++)
	{
		bool ok = true;
		for(int i=1; i<=n; i++)
		{
			for(int j=1; j<=n; j++)
			{
				if(g[i][j]=='1'&&dis[k][i]==dis[k][j]) {ok = false; break;}
			}
		}
		if(ok)
		{
			for(int i=1; i<=n; i++)
			{
				ans = max(ans,dis[k][i]+1);
			}
		}
	}
	printf("%d\n",ans);
	return 0;
}
### Prompt

In cpp, your task is to solve 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 llong long long
using namespace std;

const int N = 200;
char g[N+3][N+3];
int dis[N+3][N+3];
int n;

int absl(int x) {return x<0?-x:x;}

int main()
{
	scanf("%d",&n);
	for(int i=1; i<=n; i++) scanf("%s",g[i]+1);
	memset(dis,42,sizeof(dis));
	for(int i=1; i<=n; i++) for(int j=1; j<=n; j++) {if(g[i][j]=='1') dis[i][j] = 1;}
	for(int i=1; i<=n; i++) dis[i][i] = 0;
	for(int k=1; k<=n; k++)
	{
		for(int i=1; i<=n; i++)
		{
			for(int j=1; j<=n; j++)
			{
				dis[i][j] = min(dis[i][j],dis[i][k]+dis[k][j]);
			}
		}
	}
//	for(int i=1; i<=n; i++) {for(int j=1; j<=n; j++) printf("%d ",dis[i][j]); puts("");}
	int ans = -1;
	for(int k=1; k<=n; k++)
	{
		bool ok = true;
		for(int i=1; i<=n; i++)
		{
			for(int j=1; j<=n; j++)
			{
				if(g[i][j]=='1'&&dis[k][i]==dis[k][j]) {ok = false; break;}
			}
		}
		if(ok)
		{
			for(int i=1; i<=n; i++)
			{
				ans = max(ans,dis[k][i]+1);
			}
		}
	}
	printf("%d\n",ans);
	return 0;
}
```