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

using ll = long long;

int main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	int n;
	cin >> n;
	vector <int> g[n+1];
	for (int i=1; i<=n; i++) {
		for (int j=1; j<=n; j++) {
			char x;
			cin >> x;
			if (x == '1') {
				g[i].push_back(j);
			}
		}
	}
	int ans = -1;
	for (int i=1; i<=n; i++) {
		queue <int> bfs;
		vector <int> d(n+1);
		d[i] = 1;
		bfs.push(i);
		bool can = 1;
		while (!bfs.empty()) {
			int v = bfs.front();
			bfs.pop();
			for (auto j:g[v]) {
				if (!d[j]) {
					d[j] = d[v]+1;
					bfs.push(j);
				}
				if (d[j] == d[v]) {
					can = 0;
				}
			}
			if (bfs.empty() && can) {
				ans = max(ans, d[v]);
			}
		}
	}
	cout << ans;
	return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

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;

using ll = long long;

int main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	int n;
	cin >> n;
	vector <int> g[n+1];
	for (int i=1; i<=n; i++) {
		for (int j=1; j<=n; j++) {
			char x;
			cin >> x;
			if (x == '1') {
				g[i].push_back(j);
			}
		}
	}
	int ans = -1;
	for (int i=1; i<=n; i++) {
		queue <int> bfs;
		vector <int> d(n+1);
		d[i] = 1;
		bfs.push(i);
		bool can = 1;
		while (!bfs.empty()) {
			int v = bfs.front();
			bfs.pop();
			for (auto j:g[v]) {
				if (!d[j]) {
					d[j] = d[v]+1;
					bfs.push(j);
				}
				if (d[j] == d[v]) {
					can = 0;
				}
			}
			if (bfs.empty() && can) {
				ans = max(ans, d[v]);
			}
		}
	}
	cout << ans;
	return 0;
}

```