#include<iostream>
#include<cstring>
#include<cassert>
#include<cmath>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<time.h>
#include<algorithm>
using namespace std;
#define REP(i,x,y) for(int i=x;i<=y;i++)
#define rep(i,n) REP(i,1,n)
#define rep0(i,n) REP(i,0,n-1)
#define repG(i,x) for(int i=pos[x];~i;i=e[i].next)
#define ll long long
#define db double
const int N=207;
const int INF=1e9+7;
char s[N];
int d[N][N];
vector<int>E[N];
int n;

int main(){
	scanf("%d",&n);
	rep(i,n)rep(j,n)d[i][j]=(i==j)?0:INF;
	rep(i,n){
		scanf("%s",s+1);
		rep(j,n){
			if(s[j]=='1'){
				d[i][j]=1;
				E[i].push_back(j);
			}
		}
	}
	rep(k,n)rep(i,n)rep(j,n)d[i][j]=min(d[i][j],d[i][k]+d[k][j]);
	int mx=-1;
	rep(i,n){
		bool fl=1;
		rep(j,n)rep0(k,E[j].size())if(d[i][j]==d[i][E[j][k]])fl=0;
		if(fl){
			int mn=0;
			rep(j,n)mn=max(mn,d[i][j]);
			mx=max(mx,mn);
		}
	}
	if(mx>=0)printf("%d\n",mx+1);
	else puts("-1");
	return 0;
}

### Prompt

Develop a solution in CPP 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<iostream>
#include<cstring>
#include<cassert>
#include<cmath>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<time.h>
#include<algorithm>
using namespace std;
#define REP(i,x,y) for(int i=x;i<=y;i++)
#define rep(i,n) REP(i,1,n)
#define rep0(i,n) REP(i,0,n-1)
#define repG(i,x) for(int i=pos[x];~i;i=e[i].next)
#define ll long long
#define db double
const int N=207;
const int INF=1e9+7;
char s[N];
int d[N][N];
vector<int>E[N];
int n;

int main(){
	scanf("%d",&n);
	rep(i,n)rep(j,n)d[i][j]=(i==j)?0:INF;
	rep(i,n){
		scanf("%s",s+1);
		rep(j,n){
			if(s[j]=='1'){
				d[i][j]=1;
				E[i].push_back(j);
			}
		}
	}
	rep(k,n)rep(i,n)rep(j,n)d[i][j]=min(d[i][j],d[i][k]+d[k][j]);
	int mx=-1;
	rep(i,n){
		bool fl=1;
		rep(j,n)rep0(k,E[j].size())if(d[i][j]==d[i][E[j][k]])fl=0;
		if(fl){
			int mn=0;
			rep(j,n)mn=max(mn,d[i][j]);
			mx=max(mx,mn);
		}
	}
	if(mx>=0)printf("%d\n",mx+1);
	else puts("-1");
	return 0;
}

```