#include <bits/stdc++.h>
#define INF 1e9

using namespace std;

int n;
int G[210][210], dep[210];

void dfs(int p, int dpth)
{
	if (dep[p])
	{
		if (dep[p] % 2 != dpth % 2)
		{
			cout << -1 << endl;
			exit(0);
		}
		return;
	}
	dep[p] = dpth;
	for (int i = 1; i <= n; i++)
		if (G[p][i] == 1)
			dfs(i, dep[p] + 1);
}

int main()
{
	ios::sync_with_stdio(false);
	cin >> n;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++)
		{
			char ch; cin >> ch;
			G[i][j] = ch - '0';
			if (!G[i][j]) G[i][j] = INF;
		}
	dfs(1, 1);
	for (int k = 1; k <= n; k++)
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++)
				G[i][j] = min(G[i][j], G[i][k] + G[k][j]);
	int ans = 0;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j < i; j++)
			ans = max(ans, G[i][j]);
	cout << ans + 1 << endl;
	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>
#define INF 1e9

using namespace std;

int n;
int G[210][210], dep[210];

void dfs(int p, int dpth)
{
	if (dep[p])
	{
		if (dep[p] % 2 != dpth % 2)
		{
			cout << -1 << endl;
			exit(0);
		}
		return;
	}
	dep[p] = dpth;
	for (int i = 1; i <= n; i++)
		if (G[p][i] == 1)
			dfs(i, dep[p] + 1);
}

int main()
{
	ios::sync_with_stdio(false);
	cin >> n;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++)
		{
			char ch; cin >> ch;
			G[i][j] = ch - '0';
			if (!G[i][j]) G[i][j] = INF;
		}
	dfs(1, 1);
	for (int k = 1; k <= n; k++)
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++)
				G[i][j] = min(G[i][j], G[i][k] + G[k][j]);
	int ans = 0;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j < i; j++)
			ans = max(ans, G[i][j]);
	cout << ans + 1 << endl;
	return 0;
}

```