#include<cstdio>
int N,Q[210],D[210];
bool vis[210];
char S[210][210];
int main(){
	scanf("%d",&N);
	for(int i=0;i<N;i++)scanf("%s",S[i]);
	int ans=0;
	for(int o=0;o<N;o++){
		for(int i=0;i<N;i++)vis[i]=0;
		int*h=Q,*t=Q;vis[*t++=o]=1;D[o]=0;
		for(;h<t;h++){
			for(int j=0;j<N;j++)if(j!=*h&&S[*h][j]=='1'){
				if(!vis[j])vis[*t++=j]=1,D[j]=D[*h]+1,D[j]>ans?ans=D[j]:1;
				else if(D[j]==D[*h])return puts("-1"),0;
			}
		}
	}
	printf("%d\n",ans+1);
}
### Prompt

In CPP, your task is to solve 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>
int N,Q[210],D[210];
bool vis[210];
char S[210][210];
int main(){
	scanf("%d",&N);
	for(int i=0;i<N;i++)scanf("%s",S[i]);
	int ans=0;
	for(int o=0;o<N;o++){
		for(int i=0;i<N;i++)vis[i]=0;
		int*h=Q,*t=Q;vis[*t++=o]=1;D[o]=0;
		for(;h<t;h++){
			for(int j=0;j<N;j++)if(j!=*h&&S[*h][j]=='1'){
				if(!vis[j])vis[*t++=j]=1,D[j]=D[*h]+1,D[j]>ans?ans=D[j]:1;
				else if(D[j]==D[*h])return puts("-1"),0;
			}
		}
	}
	printf("%d\n",ans+1);
}
```