#include<bits/stdc++.h>
#define ll long long
#define uint unsigned
#define ull unsigned ll
#define pii pair<int,int>
#define pll pair<ll,ll>
#define PB push_back
#define fi first
#define se second
#define For(i,j,k) for (int i=(int)(j);i<=(int)(k);i++)
#define Rep(i,j,k) for (int i=(int)(j);i>=(int)(k);i--)
#define CLR(a,v) memset(a,v,sizeof(a))
#define CPY(a,b) memcpy(a,b,sizeof(a))
#define y1 djsfah
using namespace std;
const int N=205;
char s[N][N];
int dis[N],q[N],n,ans;
void check(int S){
	For(i,1,n) dis[i]=-1;
	int h=0,t=1; q[1]=S; dis[S]=1;
	while (h!=t){
		int x=q[++h];
		For(i,1,n)
			if (s[x][i]=='1'&&dis[i]==-1){
				dis[i]=dis[x]+1;
				q[++t]=i;
			}
	}
	For(i,1,n)
		For(j,i+1,n)
			if (s[i][j]=='1')
				if (abs(dis[i]-dis[j])!=1)
					return;
	ans=max(ans,dis[q[n]]);
}
int main(){
	ans=-1;
	scanf("%d",&n);
	For(i,1,n) scanf("%s",s[i]+1);
	For(i,1,n) check(i);
	printf("%d\n",ans);
}
/*
6
2
1 3
4 5
*/
### Prompt

Please provide a cpp coded solution to the problem described below:

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
#define uint unsigned
#define ull unsigned ll
#define pii pair<int,int>
#define pll pair<ll,ll>
#define PB push_back
#define fi first
#define se second
#define For(i,j,k) for (int i=(int)(j);i<=(int)(k);i++)
#define Rep(i,j,k) for (int i=(int)(j);i>=(int)(k);i--)
#define CLR(a,v) memset(a,v,sizeof(a))
#define CPY(a,b) memcpy(a,b,sizeof(a))
#define y1 djsfah
using namespace std;
const int N=205;
char s[N][N];
int dis[N],q[N],n,ans;
void check(int S){
	For(i,1,n) dis[i]=-1;
	int h=0,t=1; q[1]=S; dis[S]=1;
	while (h!=t){
		int x=q[++h];
		For(i,1,n)
			if (s[x][i]=='1'&&dis[i]==-1){
				dis[i]=dis[x]+1;
				q[++t]=i;
			}
	}
	For(i,1,n)
		For(j,i+1,n)
			if (s[i][j]=='1')
				if (abs(dis[i]-dis[j])!=1)
					return;
	ans=max(ans,dis[q[n]]);
}
int main(){
	ans=-1;
	scanf("%d",&n);
	For(i,1,n) scanf("%s",s[i]+1);
	For(i,1,n) check(i);
	printf("%d\n",ans);
}
/*
6
2
1 3
4 5
*/
```