#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=205;
int n,ans,F[maxn][maxn],vis[maxn];
char mp[maxn][maxn];
bool DFS(int x,int y){
	if(vis[x]) return vis[x]==y;
	vis[x]=y;
	for (int i=1;i<=n;i++)
	if(mp[x][i]=='1')
	if(!DFS(i,y^3)) return 0;
	return 1;
}
int main(){
	scanf("%d",&n);
	for (int i=1;i<=n;i++) scanf("%s",mp[i]+1);
	if(!DFS(1,1)) return puts("-1")&0;
	for (int i=1;i<=n;i++)
	for (int j=1;j<=n;j++) if(i^j)
	F[i][j]=mp[i][j]^'0'?1:1e9;
	for (int k=1;k<=n;k++)
	for (int i=1;i<=n;i++) if(i!=k)
	for (int j=1;j<=n;j++) if(j!=k&&j!=i)
	F[i][j]=min(F[i][j],F[i][k]+F[k][j]);
	for (int i=1;i<=n;i++)
	for (int j=1;j<=n;j++) ans=max(ans,F[i][j]);
	printf("%d\n",ans+1);
	return 0;
}
### Prompt

Create a solution in Cpp for 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<cstdio>
#include<algorithm>
using namespace std;
const int maxn=205;
int n,ans,F[maxn][maxn],vis[maxn];
char mp[maxn][maxn];
bool DFS(int x,int y){
	if(vis[x]) return vis[x]==y;
	vis[x]=y;
	for (int i=1;i<=n;i++)
	if(mp[x][i]=='1')
	if(!DFS(i,y^3)) return 0;
	return 1;
}
int main(){
	scanf("%d",&n);
	for (int i=1;i<=n;i++) scanf("%s",mp[i]+1);
	if(!DFS(1,1)) return puts("-1")&0;
	for (int i=1;i<=n;i++)
	for (int j=1;j<=n;j++) if(i^j)
	F[i][j]=mp[i][j]^'0'?1:1e9;
	for (int k=1;k<=n;k++)
	for (int i=1;i<=n;i++) if(i!=k)
	for (int j=1;j<=n;j++) if(j!=k&&j!=i)
	F[i][j]=min(F[i][j],F[i][k]+F[k][j]);
	for (int i=1;i<=n;i++)
	for (int j=1;j<=n;j++) ans=max(ans,F[i][j]);
	printf("%d\n",ans+1);
	return 0;
}
```