#include <bits/stdc++.h>
using namespace std;
int n, a[100001], dp[100001];
int main() {
  memset(dp, 0, sizeof dp);
  cin >> n;
  int ans = 1;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    vector<int> ps;
    for (int d = 2; d * d <= a[i]; d++) {
      if (a[i] % d == 0) {
        ps.push_back(d);
        while (a[i] % d == 0) {
          a[i] /= d;
        }
      }
    }
    if (a[i] != 1) ps.push_back(a[i]);
    int temp = 0;
    for (int pf : ps) {
      temp = max(temp, dp[pf] + 1);
    }
    ans = max(ans, temp);
    for (int pf : ps) {
      dp[pf] = temp;
    }
  }
  cout << ans << 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;
int n, a[100001], dp[100001];
int main() {
  memset(dp, 0, sizeof dp);
  cin >> n;
  int ans = 1;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    vector<int> ps;
    for (int d = 2; d * d <= a[i]; d++) {
      if (a[i] % d == 0) {
        ps.push_back(d);
        while (a[i] % d == 0) {
          a[i] /= d;
        }
      }
    }
    if (a[i] != 1) ps.push_back(a[i]);
    int temp = 0;
    for (int pf : ps) {
      temp = max(temp, dp[pf] + 1);
    }
    ans = max(ans, temp);
    for (int pf : ps) {
      dp[pf] = temp;
    }
  }
  cout << ans << endl;
  return 0;
}

```