#include <bits/stdc++.h>
using namespace std;
int a[100010], dp[100010], prime[100010], l[100010];
vector<int> pd[100010];
void pf() {
  int n = 1e5 + 10;
  for (int i = 1; i <= n; i++) prime[i] = 1;
  for (int i = 2; i <= n / 2 + 1; i++) {
    if (prime[i] == 1) {
      for (int j = i; j <= n; j += i) {
        if ((j % i) == 0) {
          prime[j] = 0;
          pd[j].push_back(i);
        }
      }
    }
  }
}
void solve(int n) {
  pf();
  int mx, res = 0;
  for (int i = 1; i <= n; i++) {
    mx = 0;
    for (int p : pd[a[i]]) mx = max(mx, l[p]);
    dp[i] = mx + 1;
    for (int p : pd[a[i]]) l[p] = max(l[p], dp[i]);
  }
  for (int i = 1; i <= n; i++) res = max(res, dp[i]);
  cout << res << endl;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  solve(n);
}

### Prompt

Create a solution in cpp for the following problem:

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 a[100010], dp[100010], prime[100010], l[100010];
vector<int> pd[100010];
void pf() {
  int n = 1e5 + 10;
  for (int i = 1; i <= n; i++) prime[i] = 1;
  for (int i = 2; i <= n / 2 + 1; i++) {
    if (prime[i] == 1) {
      for (int j = i; j <= n; j += i) {
        if ((j % i) == 0) {
          prime[j] = 0;
          pd[j].push_back(i);
        }
      }
    }
  }
}
void solve(int n) {
  pf();
  int mx, res = 0;
  for (int i = 1; i <= n; i++) {
    mx = 0;
    for (int p : pd[a[i]]) mx = max(mx, l[p]);
    dp[i] = mx + 1;
    for (int p : pd[a[i]]) l[p] = max(l[p], dp[i]);
  }
  for (int i = 1; i <= n; i++) res = max(res, dp[i]);
  cout << res << endl;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  solve(n);
}

```