#include<bits/stdc++.h>
using namespace std;
int dis[205][205],e[205][205];
signed main(){
	int n;char ch;
	cin>>n;
	memset(dis,0x3f,sizeof(dis));
	for(int i=1;i<=n;++i) dis[i][i]=0;
	for(int i=1;i<=n;++i)
		for(int j=1;j<=n;++j){
			cin>>ch;
			if(ch=='1') e[i][j]=dis[i][j]=1;
		}
	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]);
	int ans=-1;
	for(int u=1;u<=n;++u){
		int fl=0;
		for(int i=1;i<=n;++i)
			for(int j=1;j<=n;++j)
				if(e[i][j] && abs(dis[u][i]-dis[u][j])!=1) fl=1;
		if(fl) break;
		for(int i=1;i<=n;++i)
			ans=max(ans,dis[u][i]+1);
	}
	cout<<ans;
	return 0;
}
### Prompt

Generate 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;
int dis[205][205],e[205][205];
signed main(){
	int n;char ch;
	cin>>n;
	memset(dis,0x3f,sizeof(dis));
	for(int i=1;i<=n;++i) dis[i][i]=0;
	for(int i=1;i<=n;++i)
		for(int j=1;j<=n;++j){
			cin>>ch;
			if(ch=='1') e[i][j]=dis[i][j]=1;
		}
	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]);
	int ans=-1;
	for(int u=1;u<=n;++u){
		int fl=0;
		for(int i=1;i<=n;++i)
			for(int j=1;j<=n;++j)
				if(e[i][j] && abs(dis[u][i]-dis[u][j])!=1) fl=1;
		if(fl) break;
		for(int i=1;i<=n;++i)
			ans=max(ans,dis[u][i]+1);
	}
	cout<<ans;
	return 0;
}
```