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

const int N = 200 + 7;
int a[N][N];
int col[N];
bool mark[N], rid;
int n;

void dfs(int v)
{
	mark[v] = true;
	for (int i=0; i<n; i++)
		if (a[v][i] == 1)
		{
			if (!mark[i])
			{
				col[i] = col[v] ^ 1;
				dfs(i);
			}
			else
			{
				if (col[i] == col[v])
					rid = true;
			}
		}
}

int main()
{
	ios::sync_with_stdio(false);

	cin >> n;
	for (int i=0; i<n; i++)
	{
		string s;
		cin >> s;
		for (int j=0; j<n; j++)
		{
			a[i][j] = s[j] == '0' ? (i != j ? N : 0) : 1;
		}
	}


			for(int k=0; k<n; k++)
	for (int i=0; i<n; i++)
		for(int j=0; j<n; j++)
				a[i][j] = min(a[i][j], a[i][k] + a[k][j]);


	int d = 0;
	for (int i=0; i<n; i++)
		for (int j=0; j<n; j++)
			d = max(d, a[i][j]);

	for (int i=0; i<n; i++)
		for (int j=0; j<n; j++)
			for (int k=0; k<n; k++)
				if ((a[i][j] + a[j][k] + a[k][i]) % 2 != 0)
				{
					cout << "-1\n";
					return 0;
				}
	cout << d + 1 << "\n";

	return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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 N = 200 + 7;
int a[N][N];
int col[N];
bool mark[N], rid;
int n;

void dfs(int v)
{
	mark[v] = true;
	for (int i=0; i<n; i++)
		if (a[v][i] == 1)
		{
			if (!mark[i])
			{
				col[i] = col[v] ^ 1;
				dfs(i);
			}
			else
			{
				if (col[i] == col[v])
					rid = true;
			}
		}
}

int main()
{
	ios::sync_with_stdio(false);

	cin >> n;
	for (int i=0; i<n; i++)
	{
		string s;
		cin >> s;
		for (int j=0; j<n; j++)
		{
			a[i][j] = s[j] == '0' ? (i != j ? N : 0) : 1;
		}
	}


			for(int k=0; k<n; k++)
	for (int i=0; i<n; i++)
		for(int j=0; j<n; j++)
				a[i][j] = min(a[i][j], a[i][k] + a[k][j]);


	int d = 0;
	for (int i=0; i<n; i++)
		for (int j=0; j<n; j++)
			d = max(d, a[i][j]);

	for (int i=0; i<n; i++)
		for (int j=0; j<n; j++)
			for (int k=0; k<n; k++)
				if ((a[i][j] + a[j][k] + a[k][i]) % 2 != 0)
				{
					cout << "-1\n";
					return 0;
				}
	cout << d + 1 << "\n";

	return 0;
}
```