#include<bits/stdc++.h>
using namespace std;

#define INF int(1e9)
#define rep(i,n) for(int i=0;i<(n);i++)
#define REP(i,n) for(int i=1;i<=(n);i++)
#define mp make_pair
#define pb push_back
#define fst first
#define snd second
typedef long long ll;
typedef pair<int,int> pii;

const int maxn=205;
int n,ql,qr,ans;
int dist[maxn],que[maxn];
char e[maxn][maxn];

int solve(int s){
	rep(i,n)dist[i]=INF;
	dist[s]=0;
	ql=qr=0;
	que[qr++]=s;
	while(ql<qr){
		int x=que[ql++];
		rep(i,n)if(e[x][i]=='1'){
			if(dist[i]==INF){
				dist[i]=dist[x]+1;
				que[qr++]=i;
			}
			else{
				int cur=abs(dist[i]-dist[x]);
				if(cur!=1)return -1;
			}
		}
	}
	int mx=0;
	rep(i,n)mx=max(mx,dist[i]+1);
	return mx;
}

int main(){
	scanf("%d",&n);
	rep(i,n)scanf("%s",e[i]);
	ans=-1;
	rep(i,n)ans=max(ans,solve(i));
	printf("%d\n",ans);
	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>
using namespace std;

#define INF int(1e9)
#define rep(i,n) for(int i=0;i<(n);i++)
#define REP(i,n) for(int i=1;i<=(n);i++)
#define mp make_pair
#define pb push_back
#define fst first
#define snd second
typedef long long ll;
typedef pair<int,int> pii;

const int maxn=205;
int n,ql,qr,ans;
int dist[maxn],que[maxn];
char e[maxn][maxn];

int solve(int s){
	rep(i,n)dist[i]=INF;
	dist[s]=0;
	ql=qr=0;
	que[qr++]=s;
	while(ql<qr){
		int x=que[ql++];
		rep(i,n)if(e[x][i]=='1'){
			if(dist[i]==INF){
				dist[i]=dist[x]+1;
				que[qr++]=i;
			}
			else{
				int cur=abs(dist[i]-dist[x]);
				if(cur!=1)return -1;
			}
		}
	}
	int mx=0;
	rep(i,n)mx=max(mx,dist[i]+1);
	return mx;
}

int main(){
	scanf("%d",&n);
	rep(i,n)scanf("%s",e[i]);
	ans=-1;
	rep(i,n)ans=max(ans,solve(i));
	printf("%d\n",ans);
	return 0;
}
```