
#include <bits/stdc++.h>
typedef long long ll;
#define FOR(i,a,b) for(ll i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)
using namespace std;
const ll MAX_N=200,INF=1e18;
ll N;
ll G[MAX_N][MAX_N];
ll col[MAX_N];
bool dfs(ll v,ll c){
	col[v]=c;
	REP(i,N){
		if(i==v)continue;
		if(G[i][v]==1){
			if(col[i]!=0&&col[i]==c){
				return false;
			}
			if(col[i]==0&&!dfs(i,-c))return false;
		}
	}
	return true;
}
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin>>N;
	REP(i,N){
		string S;
		cin>>S;
		REP(j,N){
			G[i][j]=(S[j]=='1'?1:INF);
		}
	}
	REP(i,N)G[i][i]=0;
	if(dfs(0,1)){
		REP(k,N)REP(i,N)REP(j,N){
			if(G[i][k]+G[k][j]<G[i][j]){
				G[i][j]=G[i][k]+G[k][j];
			}
		}
		ll ans=0;
		REP(i,N)REP(j,N){
			ans=max(ans,G[i][j]+1);
		}
		cout<<ans<<endl;
	}else{
		cout<<-1<<endl;
	}
}
### Prompt

Please formulate 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>
typedef long long ll;
#define FOR(i,a,b) for(ll i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)
using namespace std;
const ll MAX_N=200,INF=1e18;
ll N;
ll G[MAX_N][MAX_N];
ll col[MAX_N];
bool dfs(ll v,ll c){
	col[v]=c;
	REP(i,N){
		if(i==v)continue;
		if(G[i][v]==1){
			if(col[i]!=0&&col[i]==c){
				return false;
			}
			if(col[i]==0&&!dfs(i,-c))return false;
		}
	}
	return true;
}
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin>>N;
	REP(i,N){
		string S;
		cin>>S;
		REP(j,N){
			G[i][j]=(S[j]=='1'?1:INF);
		}
	}
	REP(i,N)G[i][i]=0;
	if(dfs(0,1)){
		REP(k,N)REP(i,N)REP(j,N){
			if(G[i][k]+G[k][j]<G[i][j]){
				G[i][j]=G[i][k]+G[k][j];
			}
		}
		ll ans=0;
		REP(i,N)REP(j,N){
			ans=max(ans,G[i][j]+1);
		}
		cout<<ans<<endl;
	}else{
		cout<<-1<<endl;
	}
}
```