#include <bits/stdc++.h>
#define MAX 234
using namespace std;

int N;
char str[MAX];
int color[MAX];
int graph[MAX][MAX];
int depth[MAX];

int bfs(int cur)
{
	queue<int> q;
	q.push(cur);
	color[cur] = 0;
	depth[cur] = 0;
	int ret = 0;
	while (!q.empty())
	{
		cur = q.front();
		q.pop();
		for (int i = 0; i < N; i++)
		{
			if (graph[cur][i])
			{
				if (color[i] == color[cur])
				{
					cout << -1 << endl;
					exit(0);
				}
				if (color[i] < 0)
				{
					q.push(i);
					color[i] = !color[cur];
					depth[i] = depth[cur] + 1;
					ret = depth[i];
				}
			}
		}
	}
	return ret;
}

int main(void)
{
	cin >> N;
	for (int i = 0; i < N; i++)
	{
		cin >> str;
		for (int j = 0; j < N; j++) graph[i][j] = str[j] - '0';
	}
	int mmax = -1;
	for (int i = 0; i < N; i++)
	{
		for (int j = 0; j < N; j++)
		{
			depth[j] = -1;
			color[j] = -1;
		}
		mmax = max(mmax, bfs(i));
	}
	cout << mmax + 1 << 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>
#define MAX 234
using namespace std;

int N;
char str[MAX];
int color[MAX];
int graph[MAX][MAX];
int depth[MAX];

int bfs(int cur)
{
	queue<int> q;
	q.push(cur);
	color[cur] = 0;
	depth[cur] = 0;
	int ret = 0;
	while (!q.empty())
	{
		cur = q.front();
		q.pop();
		for (int i = 0; i < N; i++)
		{
			if (graph[cur][i])
			{
				if (color[i] == color[cur])
				{
					cout << -1 << endl;
					exit(0);
				}
				if (color[i] < 0)
				{
					q.push(i);
					color[i] = !color[cur];
					depth[i] = depth[cur] + 1;
					ret = depth[i];
				}
			}
		}
	}
	return ret;
}

int main(void)
{
	cin >> N;
	for (int i = 0; i < N; i++)
	{
		cin >> str;
		for (int j = 0; j < N; j++) graph[i][j] = str[j] - '0';
	}
	int mmax = -1;
	for (int i = 0; i < N; i++)
	{
		for (int j = 0; j < N; j++)
		{
			depth[j] = -1;
			color[j] = -1;
		}
		mmax = max(mmax, bfs(i));
	}
	cout << mmax + 1 << endl;
	return 0;
}

```