#include <bits/stdc++.h>

using namespace std;

const int DIM = 207;

vector <int> v[DIM];
int col[DIM];

int n;

int get(int start)
{
	memset(col, 0, sizeof(col));
	
	queue <int> q;
	
	col[start] = 1;
	q.push(start);
	
	int res = 1;
	
	while(!q.empty())
	{
		int nod = q.front();
		q.pop();
		
		for(auto i : v[nod])
			if(col[i] != 0 && abs(col[i] - col[nod]) != 1)
			{
				return -1;
			}
			else
				if(col[i] == 0)
				{
					col[i] = col[nod] + 1;
					res = max(res, col[i]);
					q.push(i);
				}
	}
	
	return res;
}

int main()
{
	cin >> n;
	
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= n; j++)
		{
			char c;
			cin >> c;
			
			if(c == '1')
				v[i].push_back(j);
		}
	
	int ans = -1;
	
	for(int i = 1; i <= n; i++)
		ans = max(ans, get(i));
	
	cout << ans;
}
### 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 <bits/stdc++.h>

using namespace std;

const int DIM = 207;

vector <int> v[DIM];
int col[DIM];

int n;

int get(int start)
{
	memset(col, 0, sizeof(col));
	
	queue <int> q;
	
	col[start] = 1;
	q.push(start);
	
	int res = 1;
	
	while(!q.empty())
	{
		int nod = q.front();
		q.pop();
		
		for(auto i : v[nod])
			if(col[i] != 0 && abs(col[i] - col[nod]) != 1)
			{
				return -1;
			}
			else
				if(col[i] == 0)
				{
					col[i] = col[nod] + 1;
					res = max(res, col[i]);
					q.push(i);
				}
	}
	
	return res;
}

int main()
{
	cin >> n;
	
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= n; j++)
		{
			char c;
			cin >> c;
			
			if(c == '1')
				v[i].push_back(j);
		}
	
	int ans = -1;
	
	for(int i = 1; i <= n; i++)
		ans = max(ans, get(i));
	
	cout << ans;
}
```