#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
char s[205];
bool G[205][205];
int d[205],n;
queue<int> Q;
inline bool bfs(int u){
	int v;
	while(!Q.empty())Q.pop();
	d[u]=1;Q.push(u);
	while(!Q.empty()){
		u=Q.front();Q.pop();
		for(v=1;v<=n;++v)if(G[u][v])if(!d[v]){
			d[v]=d[u]+1;
			Q.push(v);
		}else if(d[v]!=d[u]+1&&d[v]!=d[u]-1)return 0;
	}
	return 1;
}
int main(){
	int i,j,ans=-1;
	scanf("%d",&n);
	for(i=1;i<=n;++i){
		scanf("%s",s+1);
		for(j=1;j<=n;++j)G[i][j]=s[j]=='1';
	}
	for(i=1;i<=n;++i){
		memset(d,0,sizeof(d));
		if(bfs(i))for(j=1;j<=n;++j)if(d[j]>ans)ans=d[j];
	}
	printf("%d\n",ans);
	return 0;
}
### 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<cstdio>
#include<cstring>
#include<queue>
using namespace std;
char s[205];
bool G[205][205];
int d[205],n;
queue<int> Q;
inline bool bfs(int u){
	int v;
	while(!Q.empty())Q.pop();
	d[u]=1;Q.push(u);
	while(!Q.empty()){
		u=Q.front();Q.pop();
		for(v=1;v<=n;++v)if(G[u][v])if(!d[v]){
			d[v]=d[u]+1;
			Q.push(v);
		}else if(d[v]!=d[u]+1&&d[v]!=d[u]-1)return 0;
	}
	return 1;
}
int main(){
	int i,j,ans=-1;
	scanf("%d",&n);
	for(i=1;i<=n;++i){
		scanf("%s",s+1);
		for(j=1;j<=n;++j)G[i][j]=s[j]=='1';
	}
	for(i=1;i<=n;++i){
		memset(d,0,sizeof(d));
		if(bfs(i))for(j=1;j<=n;++j)if(d[j]>ans)ans=d[j];
	}
	printf("%d\n",ans);
	return 0;
}
```