#include<bits/stdc++.h>
#define ll long long
using namespace std;
int n,m,k;
bool b[205][205];
int q[205],h,t;int dis[205],vis[205];
char c[205];
int bfs(int x){
	h=0;q[t=1]=x;
	memset(dis,10,sizeof(dis));
	memset(vis,0,sizeof(vis));
	dis[x]=0;vis[x]=1;
	while(h<t){
		int u=q[++h];
		for(int i=1;i<=n;i++)if(i!=u&&b[u][i]){
			if(vis[i]&&(dis[i]+dis[u]+1)&1){
				cout<<-1;
				exit(0);
			}
			if(!vis[i])dis[i]=dis[u]+1,vis[i]=1,q[++t]=i;
		}
	}int all=0;
	for(int i=1;i<=n;i++)if(vis[i])all=max(all,dis[i]);
	return all;
}
int main(){
	cin>>n;
	for(int i=1;i<=n;i++){
		scanf("%s",c+1);
		for(int j=1;j<=n;j++)b[i][j]=(c[j]=='1');
	}
	int ret=0;
	for(int i=1;i<=n;i++)ret=max(ret,bfs(i));cout<<ret+1;
    return 0;
}







### Prompt

Construct a Cpp code solution to the problem outlined:

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>
#define ll long long
using namespace std;
int n,m,k;
bool b[205][205];
int q[205],h,t;int dis[205],vis[205];
char c[205];
int bfs(int x){
	h=0;q[t=1]=x;
	memset(dis,10,sizeof(dis));
	memset(vis,0,sizeof(vis));
	dis[x]=0;vis[x]=1;
	while(h<t){
		int u=q[++h];
		for(int i=1;i<=n;i++)if(i!=u&&b[u][i]){
			if(vis[i]&&(dis[i]+dis[u]+1)&1){
				cout<<-1;
				exit(0);
			}
			if(!vis[i])dis[i]=dis[u]+1,vis[i]=1,q[++t]=i;
		}
	}int all=0;
	for(int i=1;i<=n;i++)if(vis[i])all=max(all,dis[i]);
	return all;
}
int main(){
	cin>>n;
	for(int i=1;i<=n;i++){
		scanf("%s",c+1);
		for(int j=1;j<=n;j++)b[i][j]=(c[j]=='1');
	}
	int ret=0;
	for(int i=1;i<=n;i++)ret=max(ret,bfs(i));cout<<ret+1;
    return 0;
}







```