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

const int N = 200 + 10;
bool adj[N][N];
int h[N], n;

int bfs(int root) {
	fill(h, h + n, -1);
  	queue<int> q;
  	q.push(root);
  	h[root] = 1;
  	while (q.size()) {
      	int v = q.front();
      	q.pop();
      	for (int i = 0; i < n; i++) {
          	if (h[i] == -1 && adj[v][i]) {
				q.push(i);
              	h[i] = h[v] + 1;
            }
        }
    }
  	int res = 0;
  	for (int i = 0; i < n; i++)
      	res = max(res, h[i]);
  	for (int i = 0; i < n; i++)
      	for (int j = 0; j < n; j++)
          	if (adj[i][j] && abs(h[i] - h[j]) != 1)
              	res = -1;
	return res;
}

int main() {
  	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  	cin >> n;
  	string s;
  	for (int i = 0; i < n; i++) {
      	cin >> s;
    	for (int j = 0; j < n; j++)
          	adj[i][j] = s[j] - '0';
    }
  	int ans = -1;
  	for (int i = 0; i < n; i++)
      	ans = max(ans, bfs(i));
	cout << ans << "\n";
  	return 0;
}
### Prompt

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

const int N = 200 + 10;
bool adj[N][N];
int h[N], n;

int bfs(int root) {
	fill(h, h + n, -1);
  	queue<int> q;
  	q.push(root);
  	h[root] = 1;
  	while (q.size()) {
      	int v = q.front();
      	q.pop();
      	for (int i = 0; i < n; i++) {
          	if (h[i] == -1 && adj[v][i]) {
				q.push(i);
              	h[i] = h[v] + 1;
            }
        }
    }
  	int res = 0;
  	for (int i = 0; i < n; i++)
      	res = max(res, h[i]);
  	for (int i = 0; i < n; i++)
      	for (int j = 0; j < n; j++)
          	if (adj[i][j] && abs(h[i] - h[j]) != 1)
              	res = -1;
	return res;
}

int main() {
  	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  	cin >> n;
  	string s;
  	for (int i = 0; i < n; i++) {
      	cin >> s;
    	for (int j = 0; j < n; j++)
          	adj[i][j] = s[j] - '0';
    }
  	int ans = -1;
  	for (int i = 0; i < n; i++)
      	ans = max(ans, bfs(i));
	cout << ans << "\n";
  	return 0;
}
```