#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  int t = 1;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> dp(100005);
    long long a[n];
    for (long long i = 0; i < n; i++) cin >> a[i];
    long long mx = 1;
    for (long long i = 0; i < n; i++) {
      set<long long> s;
      for (long long j = 2; j * j <= a[i]; j++) {
        while (a[i] % j == 0) {
          s.insert(j);
          a[i] = a[i] / j;
        }
      }
      if (a[i] > 1) s.insert(a[i]);
      long long ans = 1;
      for (auto j : s) {
        ans = max(ans, dp[j] + 1);
      }
      mx = max(mx, ans);
      for (auto j : s) {
        dp[j] = ans;
      }
    }
    cout << mx << 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 main() {
  ios::sync_with_stdio(0);
  int t = 1;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> dp(100005);
    long long a[n];
    for (long long i = 0; i < n; i++) cin >> a[i];
    long long mx = 1;
    for (long long i = 0; i < n; i++) {
      set<long long> s;
      for (long long j = 2; j * j <= a[i]; j++) {
        while (a[i] % j == 0) {
          s.insert(j);
          a[i] = a[i] / j;
        }
      }
      if (a[i] > 1) s.insert(a[i]);
      long long ans = 1;
      for (auto j : s) {
        ans = max(ans, dp[j] + 1);
      }
      mx = max(mx, ans);
      for (auto j : s) {
        dp[j] = ans;
      }
    }
    cout << mx << endl;
  }
  return 0;
}

```