#include <bits/stdc++.h>
using namespace std;
int N;
int dp[100005];
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N;
  for (int i = 1; i <= N; i++) {
    int n;
    cin >> n;
    int mx = dp[n];
    for (int p = 2; p <= sqrt(n); p++) {
      if (n % p == 0) {
        mx = max(mx, dp[p]);
        mx = max(mx, dp[n / p]);
      }
    }
    mx++;
    for (int p = 2; p <= sqrt(n); p++) {
      if (n % p == 0) {
        dp[p] = dp[n / p] = mx;
      }
    }
    dp[n] = mx;
  }
  cout << *max_element(dp + 1, dp + 100001) << "\n";
}

### Prompt

Generate a cpp solution to 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 N;
int dp[100005];
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N;
  for (int i = 1; i <= N; i++) {
    int n;
    cin >> n;
    int mx = dp[n];
    for (int p = 2; p <= sqrt(n); p++) {
      if (n % p == 0) {
        mx = max(mx, dp[p]);
        mx = max(mx, dp[n / p]);
      }
    }
    mx++;
    for (int p = 2; p <= sqrt(n); p++) {
      if (n % p == 0) {
        dp[p] = dp[n / p] = mx;
      }
    }
    dp[n] = mx;
  }
  cout << *max_element(dp + 1, dp + 100001) << "\n";
}

```