#include <bits/stdc++.h>
using namespace std;
const int M = 100010;
vector<int> v[M];
int rec[M], maxlen[M], n, ma = 1;
bool vis[M];
void getans() {
  for (int i = 0; i < n; i++) {
    int x = 0, tmp = rec[i];
    for (int j = 0; j < v[tmp].size(); j++) x = max(maxlen[v[tmp][j]], x);
    x++;
    for (int j = 0; j < v[tmp].size(); j++)
      maxlen[v[tmp][j]] = max(maxlen[v[tmp][j]], x);
    if (x > ma) ma = x;
  }
}
void init() {
  memset(vis, 0, sizeof(vis));
  memset(maxlen, 0, sizeof(maxlen));
  for (int i = 2; i < M; i++) {
    if (!vis[i]) {
      for (int j = i; j < M; j += i) {
        v[j].push_back(i);
        vis[j] = 1;
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  init();
  cin >> n;
  for (int i = 0; i < n; i++) cin >> rec[i];
  getans();
  cout << ma << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

Squirrel Liss is interested in sequences. She also has preferences of integers. She thinks n integers a1, a2, ..., an are good.

Now she is interested in good sequences. A sequence x1, x2, ..., xk is called good if it satisfies the following three conditions:

  * The sequence is strictly increasing, i.e. xi < xi + 1 for each i (1 ≤ i ≤ k - 1). 
  * No two adjacent elements are coprime, i.e. gcd(xi, xi + 1) > 1 for each i (1 ≤ i ≤ k - 1) (where gcd(p, q) denotes the greatest common divisor of the integers p and q). 
  * All elements of the sequence are good integers. 



Find the length of the longest good sequence.

Input

The input consists of two lines. The first line contains a single integer n (1 ≤ n ≤ 105) — the number of good integers. The second line contains a single-space separated list of good integers a1, a2, ..., an in strictly increasing order (1 ≤ ai ≤ 105; ai < ai + 1).

Output

Print a single integer — the length of the longest good sequence.

Examples

Input

5
2 3 4 6 9


Output

4


Input

9
1 2 3 5 6 7 8 9 10


Output

4

Note

In the first example, the following sequences are examples of good sequences: [2; 4; 6; 9], [2; 4; 6], [3; 9], [6]. The length of the longest good sequence is 4.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int M = 100010;
vector<int> v[M];
int rec[M], maxlen[M], n, ma = 1;
bool vis[M];
void getans() {
  for (int i = 0; i < n; i++) {
    int x = 0, tmp = rec[i];
    for (int j = 0; j < v[tmp].size(); j++) x = max(maxlen[v[tmp][j]], x);
    x++;
    for (int j = 0; j < v[tmp].size(); j++)
      maxlen[v[tmp][j]] = max(maxlen[v[tmp][j]], x);
    if (x > ma) ma = x;
  }
}
void init() {
  memset(vis, 0, sizeof(vis));
  memset(maxlen, 0, sizeof(maxlen));
  for (int i = 2; i < M; i++) {
    if (!vis[i]) {
      for (int j = i; j < M; j += i) {
        v[j].push_back(i);
        vis[j] = 1;
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  init();
  cin >> n;
  for (int i = 0; i < n; i++) cin >> rec[i];
  getans();
  cout << ma << endl;
  return 0;
}

```