#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define INF 1LL<<62
#define inf 1000000007

bool ch[202][202];
ll d[202][202];
int main() {
	ll n;
	cin>>n;
	for(ll i=0;i<n;i++){
		for(ll j=0;j<n;j++){
			d[i][j]=inf;
			char now;
			cin>>now;
			//cout <<now;
			ch[i][j]=now-'0';
			//cout << ch[i][j];
			if(ch[i][j]==1){
			d[i][j]=ch[i][j];
			}
			if(i==j){
				d[i][j]=0;
			}
		}
	}
	for(ll k=0;k<n;k++){
		for(ll i=0;i<n;i++){
			for(ll j=0;j<n;j++){
				d[i][j]=min(d[i][j],d[i][k]+d[k][j]);
			}
		}
	}
	//cout << d[0][0];
	ll ans=0;
	for(ll i=0;i<n;i++){
		for(ll j=0;j<n;j++){
			if(i==j){
				continue;
			}
			if(ch[i][j]){
				if(d[0][i]%2==d[0][j]%2){
					cout << -1;
					return 0;
				}
			}
			if(d[i][j]!=inf){
			ans=max(ans,d[i][j]);
			}
		}
	}
	cout <<ans+1;
	// your code goes here
	return 0;
}
### Prompt

Create a solution in cpp for 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>
using namespace std;
typedef long long ll;
#define INF 1LL<<62
#define inf 1000000007

bool ch[202][202];
ll d[202][202];
int main() {
	ll n;
	cin>>n;
	for(ll i=0;i<n;i++){
		for(ll j=0;j<n;j++){
			d[i][j]=inf;
			char now;
			cin>>now;
			//cout <<now;
			ch[i][j]=now-'0';
			//cout << ch[i][j];
			if(ch[i][j]==1){
			d[i][j]=ch[i][j];
			}
			if(i==j){
				d[i][j]=0;
			}
		}
	}
	for(ll k=0;k<n;k++){
		for(ll i=0;i<n;i++){
			for(ll j=0;j<n;j++){
				d[i][j]=min(d[i][j],d[i][k]+d[k][j]);
			}
		}
	}
	//cout << d[0][0];
	ll ans=0;
	for(ll i=0;i<n;i++){
		for(ll j=0;j<n;j++){
			if(i==j){
				continue;
			}
			if(ch[i][j]){
				if(d[0][i]%2==d[0][j]%2){
					cout << -1;
					return 0;
				}
			}
			if(d[i][j]!=inf){
			ans=max(ans,d[i][j]);
			}
		}
	}
	cout <<ans+1;
	// your code goes here
	return 0;
}
```