#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int n, finans = 0;
int a[N], ans[N], sieve[N], countprime[N];
vector<int> primes[N];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    ans[i] = 1;
  }
  for (int i = 2; i <= 1e5; i++) sieve[i] = 1;
  for (int i = 2; i <= 1e5; i++) {
    if (sieve[i]) {
      for (int j = 1; i * j <= 1e5; j++) {
        primes[i * j].push_back(i);
        sieve[i * j] = 0;
      }
    }
  }
  for (int i = n; i >= 1; i--) {
    for (auto it : primes[a[i]]) ans[i] = max(ans[i], countprime[it] + 1);
    for (auto it : primes[a[i]]) countprime[it] = max(countprime[it], ans[i]);
    finans = max(ans[i], finans);
  }
  cout << finans;
}

### Prompt

Please create a solution in Cpp 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;
const int N = 1e5 + 5;
int n, finans = 0;
int a[N], ans[N], sieve[N], countprime[N];
vector<int> primes[N];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    ans[i] = 1;
  }
  for (int i = 2; i <= 1e5; i++) sieve[i] = 1;
  for (int i = 2; i <= 1e5; i++) {
    if (sieve[i]) {
      for (int j = 1; i * j <= 1e5; j++) {
        primes[i * j].push_back(i);
        sieve[i * j] = 0;
      }
    }
  }
  for (int i = n; i >= 1; i--) {
    for (auto it : primes[a[i]]) ans[i] = max(ans[i], countprime[it] + 1);
    for (auto it : primes[a[i]]) countprime[it] = max(countprime[it], ans[i]);
    finans = max(ans[i], finans);
  }
  cout << finans;
}

```