#include<bits/stdc++.h>
using namespace std;
const int maxn=210;
int n,a[maxn][maxn];
vector<int> ver[maxn];
int flag[maxn];
inline void dfs(int x,int v)
{
	if(v==1) v=2;else v=1;
	if(flag[x]){if(flag[x]!=v){puts("-1");exit(0);}return;}
	flag[x]=v;
	for(int i=0;i<(int)ver[x].size();i++) dfs(ver[x][i],v);
}
int main()
{
	cin>>n;
	for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) a[i][j]=i==j?0:(n+1);
	for(int i=1;i<=n;i++) for(int j=1;j<=n;j++)
	{
		char c;cin>>c;if(i==j) continue;
		if(c=='1') a[i][j]=1,ver[i].push_back(j),ver[j].push_back(i);
	}
	dfs(1,0);
	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][j],a[i][k]+a[k][j]);
	int ans=0;
	for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) ans=max(ans,a[i][j]+1);
	cout<<ans<<endl;
}
### Prompt

Develop a solution in cpp to the problem described below:

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 maxn=210;
int n,a[maxn][maxn];
vector<int> ver[maxn];
int flag[maxn];
inline void dfs(int x,int v)
{
	if(v==1) v=2;else v=1;
	if(flag[x]){if(flag[x]!=v){puts("-1");exit(0);}return;}
	flag[x]=v;
	for(int i=0;i<(int)ver[x].size();i++) dfs(ver[x][i],v);
}
int main()
{
	cin>>n;
	for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) a[i][j]=i==j?0:(n+1);
	for(int i=1;i<=n;i++) for(int j=1;j<=n;j++)
	{
		char c;cin>>c;if(i==j) continue;
		if(c=='1') a[i][j]=1,ver[i].push_back(j),ver[j].push_back(i);
	}
	dfs(1,0);
	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][j],a[i][k]+a[k][j]);
	int ans=0;
	for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) ans=max(ans,a[i][j]+1);
	cout<<ans<<endl;
}
```