#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main() {
	int n;
	cin>>n;
	vector<string> g(n);
	for(auto &s:g)
		cin>>s;
	int res=-1;
	for(int u=0;u<n;++u){
		queue<int> q;
		q.push(u);
		vector<int> cost(n,-1);
		vector<int> clr(n,-1);
		clr[u]=0;
		cost[u]=0;
		while(!q.empty()){
			int x=q.front();
			q.pop();
			for(int v=0;v<n;++v)
				if(g[x][v]=='1' && clr[v]==-1){
					clr[v]=!clr[x];
					q.push(v);
					cost[v]=cost[x]+1;
				}
		}
		int cur=*max_element(cost.begin(),cost.end());
		if(u==0)
			for(int i=0;i<n;++i)
				for(int j=0;j<n;++j)
					if(g[i][j]=='1' && clr[i]==clr[j]){
						cout<<-1<<endl;
						return 0;
					}
		res=max(res,cur+1);
	}
	cout<<res<<endl;
	return 0;
}
### 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>
using namespace std;
typedef long long ll;
int main() {
	int n;
	cin>>n;
	vector<string> g(n);
	for(auto &s:g)
		cin>>s;
	int res=-1;
	for(int u=0;u<n;++u){
		queue<int> q;
		q.push(u);
		vector<int> cost(n,-1);
		vector<int> clr(n,-1);
		clr[u]=0;
		cost[u]=0;
		while(!q.empty()){
			int x=q.front();
			q.pop();
			for(int v=0;v<n;++v)
				if(g[x][v]=='1' && clr[v]==-1){
					clr[v]=!clr[x];
					q.push(v);
					cost[v]=cost[x]+1;
				}
		}
		int cur=*max_element(cost.begin(),cost.end());
		if(u==0)
			for(int i=0;i<n;++i)
				for(int j=0;j<n;++j)
					if(g[i][j]=='1' && clr[i]==clr[j]){
						cout<<-1<<endl;
						return 0;
					}
		res=max(res,cur+1);
	}
	cout<<res<<endl;
	return 0;
}
```