#include <bits/stdc++.h>
using namespace std;
const int nax = 100005, mod = 998244353, pot = 17;
int t[nax], n;
int best[nax];
int sito[nax];
int res = 1;
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  for (int i = 2; nax > i; ++i) {
    if (sito[i] == 0) {
      for (int j = i; nax > j; j += i) sito[j] = i;
    }
  }
  cin >> n;
  for (int i = 1; n >= i; ++i) {
    cin >> t[i];
    int ans = 0;
    int x = t[i];
    while (x != 1) {
      ans = max(ans, best[sito[x]]);
      x /= sito[x];
    }
    x = t[i];
    while (x != 1) {
      best[sito[x]] = max(best[sito[x]], ans + 1);
      x /= sito[x];
    }
    res = max(res, ans + 1);
  }
  cout << res;
  return 0;
}

### Prompt

Develop a solution in cpp 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;
const int nax = 100005, mod = 998244353, pot = 17;
int t[nax], n;
int best[nax];
int sito[nax];
int res = 1;
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  for (int i = 2; nax > i; ++i) {
    if (sito[i] == 0) {
      for (int j = i; nax > j; j += i) sito[j] = i;
    }
  }
  cin >> n;
  for (int i = 1; n >= i; ++i) {
    cin >> t[i];
    int ans = 0;
    int x = t[i];
    while (x != 1) {
      ans = max(ans, best[sito[x]]);
      x /= sito[x];
    }
    x = t[i];
    while (x != 1) {
      best[sito[x]] = max(best[sito[x]], ans + 1);
      x /= sito[x];
    }
    res = max(res, ans + 1);
  }
  cout << res;
  return 0;
}

```