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

int main()
{
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<string> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  int ans = 1;
  for (int i = 0; i < n; ++i) {
    queue<int> q;
    q.push(i);
    vector<int> vis(n);
    vis[i] = 1;
    int k = 1;
    while (!q.empty()) {
      int cur = q.front();
      q.pop();
      k = vis[cur];
      for (int j = 0; j < n; ++j) {
        if (a[cur][j] - '0') {
          if (!vis[j]) {
            vis[j] = vis[cur] + 1;
            q.push(j);
          } else if (j != cur && vis[j] % 2 == vis[cur] % 2) {
            cout << -1 << endl;
            return 0;
          }
        }
      }
    }
    ans = max(ans, k);
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;

int main()
{
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<string> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  int ans = 1;
  for (int i = 0; i < n; ++i) {
    queue<int> q;
    q.push(i);
    vector<int> vis(n);
    vis[i] = 1;
    int k = 1;
    while (!q.empty()) {
      int cur = q.front();
      q.pop();
      k = vis[cur];
      for (int j = 0; j < n; ++j) {
        if (a[cur][j] - '0') {
          if (!vis[j]) {
            vis[j] = vis[cur] + 1;
            q.push(j);
          } else if (j != cur && vis[j] % 2 == vis[cur] % 2) {
            cout << -1 << endl;
            return 0;
          }
        }
      }
    }
    ans = max(ans, k);
  }
  cout << ans << endl;
  return 0;
}

```