#include<bits/stdc++.h>
#define fo(i,a,b)for(int i=a,_e=b;i<=_e;++i)
#define fd(i,a,b)for(int i=b,_e=a;i>=_e;--i)
#define ll long long
using namespace std;
const int N=205,mo=998244353;
int n,b[N][N];
int ans,c[N];
int h[N],d[N];
int ksm(int x,int y){
	int t=1;
	for(;y;y>>=1,x=(ll)x*x%mo)if(y&1)t=(ll)t*x%mo;
	return t;
}
bool pd(int x){
	fo(i,1,n)if(b[x][i]){
		if(c[i]){
			if(c[i]+c[x]!=3)return 0;
		}else{
			c[i]=3-c[x];
			if(!pd(i))return 0;
		}
	}
	return 1;
}
int main(){
//	freopen("a.out","w",stdout);
	cin>>n;
	fo(i,1,n){
		scanf("\n");
		fo(j,1,n)b[i][j]=getchar()-'0';
	}
	if(c[1]=1,!pd(1)){
		cout<<-1;return 0;
	}
	fo(i,1,n){
		int l=0,r=1;
		fo(j,1,n)h[j]=0;
		d[1]=i;h[i]=1;
		for(int x;x=d[++l],l<=r;)	
			fo(j,1,n)if(b[x][j]&&!h[j])
				d[++r]=j,h[j]=h[x]+1;
		ans=max(ans,h[d[n]]);
	}
	cout<<ans;
}
### 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<bits/stdc++.h>
#define fo(i,a,b)for(int i=a,_e=b;i<=_e;++i)
#define fd(i,a,b)for(int i=b,_e=a;i>=_e;--i)
#define ll long long
using namespace std;
const int N=205,mo=998244353;
int n,b[N][N];
int ans,c[N];
int h[N],d[N];
int ksm(int x,int y){
	int t=1;
	for(;y;y>>=1,x=(ll)x*x%mo)if(y&1)t=(ll)t*x%mo;
	return t;
}
bool pd(int x){
	fo(i,1,n)if(b[x][i]){
		if(c[i]){
			if(c[i]+c[x]!=3)return 0;
		}else{
			c[i]=3-c[x];
			if(!pd(i))return 0;
		}
	}
	return 1;
}
int main(){
//	freopen("a.out","w",stdout);
	cin>>n;
	fo(i,1,n){
		scanf("\n");
		fo(j,1,n)b[i][j]=getchar()-'0';
	}
	if(c[1]=1,!pd(1)){
		cout<<-1;return 0;
	}
	fo(i,1,n){
		int l=0,r=1;
		fo(j,1,n)h[j]=0;
		d[1]=i;h[i]=1;
		for(int x;x=d[++l],l<=r;)	
			fo(j,1,n)if(b[x][j]&&!h[j])
				d[++r]=j,h[j]=h[x]+1;
		ans=max(ans,h[d[n]]);
	}
	cout<<ans;
}
```