#include <iostream>
#include <string>
#include <vector>
#include <algorithm>

using namespace std;

int solve(int N, const vector<string> &A, int s)
{
	//cout << "s=" << s << endl;

	vector<string> T = A;

	vector<bool> visited(N, false);

	vector<int> v;
	v.push_back(s);

	int ans = 0;
	while (!v.empty()) {
		++ans;
		vector<int> next;

		for (auto x : v) {
			visited[x] = true;
		}
		for (auto x : v) {
			for (int i = 0; i < N; ++i) {
				if (T[x][i] == '1') {
					if (visited[i]) {
						//cout << "ng" << endl;
						return -1;
					}
					next.push_back(i);
					//cout << i << " ";
					T[x][i] = '0';
					T[i][x] = '0';
				}
			}
		}
		v = move(next);
	}
	return ans;
}

int solve(int N, const vector<string> &A)
{
	int ans = -1;
	int s;
	for (s = 0; s < N; ++s) {
		ans = max(ans, solve(N, A, s));
	}
	return ans;
}

int main()
{
	int N;
	cin >> N;
	vector<string> S(N);
	for (auto &x : S) {
		cin >> x;
	}

	cout << solve(N, S) << endl;
}

### Prompt

Your challenge is to write a cpp solution 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 <iostream>
#include <string>
#include <vector>
#include <algorithm>

using namespace std;

int solve(int N, const vector<string> &A, int s)
{
	//cout << "s=" << s << endl;

	vector<string> T = A;

	vector<bool> visited(N, false);

	vector<int> v;
	v.push_back(s);

	int ans = 0;
	while (!v.empty()) {
		++ans;
		vector<int> next;

		for (auto x : v) {
			visited[x] = true;
		}
		for (auto x : v) {
			for (int i = 0; i < N; ++i) {
				if (T[x][i] == '1') {
					if (visited[i]) {
						//cout << "ng" << endl;
						return -1;
					}
					next.push_back(i);
					//cout << i << " ";
					T[x][i] = '0';
					T[i][x] = '0';
				}
			}
		}
		v = move(next);
	}
	return ans;
}

int solve(int N, const vector<string> &A)
{
	int ans = -1;
	int s;
	for (s = 0; s < N; ++s) {
		ans = max(ans, solve(N, A, s));
	}
	return ans;
}

int main()
{
	int N;
	cin >> N;
	vector<string> S(N);
	for (auto &x : S) {
		cin >> x;
	}

	cout << solve(N, S) << endl;
}

```