#include<bits/stdc++.h>
using namespace std;
const int N=205;
char s[N];
int a[N][N],flag[N],n;
void dfs(int x){
	for (int i=1;i<=n;i++)
		if (a[x][i]){
			if (!flag[i]){
				flag[i]=3-flag[x];
				dfs(i);
			}
			else if (flag[i]+flag[x]!=3){
				puts("-1");
				exit(0);
			}
		}
}
int main(){
	scanf("%d",&n);
	for (int i=1;i<=n;i++){
		scanf("%s",s+1);
		for (int j=1;j<=n;j++)a[i][j]=s[j]-'0';
	}
	flag[1]=1;dfs(1);
	for (int i=1;i<=n;i++)
		for (int j=1;j<=n;j++)
			if (i!=j&&!a[i][j])a[i][j]=1e9;	
	for (int k=1;k<=n;k++)
		for (int i=1;i<=n;i++)
			for (int j=1;j<=n;j++)a[i][j]=min(a[i][k]+a[k][j],a[i][j]);
	int ans=0;
	for (int i=1;i<=n;i++)
		for (int j=1;j<=n;j++)
			ans=max(ans,a[i][j]);
	printf("%d\n",ans+1);				
}
### 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=205;
char s[N];
int a[N][N],flag[N],n;
void dfs(int x){
	for (int i=1;i<=n;i++)
		if (a[x][i]){
			if (!flag[i]){
				flag[i]=3-flag[x];
				dfs(i);
			}
			else if (flag[i]+flag[x]!=3){
				puts("-1");
				exit(0);
			}
		}
}
int main(){
	scanf("%d",&n);
	for (int i=1;i<=n;i++){
		scanf("%s",s+1);
		for (int j=1;j<=n;j++)a[i][j]=s[j]-'0';
	}
	flag[1]=1;dfs(1);
	for (int i=1;i<=n;i++)
		for (int j=1;j<=n;j++)
			if (i!=j&&!a[i][j])a[i][j]=1e9;	
	for (int k=1;k<=n;k++)
		for (int i=1;i<=n;i++)
			for (int j=1;j<=n;j++)a[i][j]=min(a[i][k]+a[k][j],a[i][j]);
	int ans=0;
	for (int i=1;i<=n;i++)
		for (int j=1;j<=n;j++)
			ans=max(ans,a[i][j]);
	printf("%d\n",ans+1);				
}
```