#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200000;
int n;
int a[MAXN];
int f[MAXN];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) {
    int tmp = 1;
    for (int j = 1; j * j <= a[i]; j++)
      if (a[i] % j == 0) {
        if (j != 1) tmp = max(tmp, f[j] + 1);
        if (a[i] / j != 1) tmp = max(tmp, f[a[i] / j] + 1);
      }
    for (int j = 1; j * j <= a[i]; j++)
      if (a[i] % j == 0) {
        f[j] = max(f[j], tmp);
        f[a[i] / j] = max(f[a[i] / j], tmp);
      }
  }
  int ans = 0;
  for (int i = 1; i <= 100000; i++) ans = max(ans, f[i]);
  printf("%d\n", ans);
  return 0;
}

### 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;
const int MAXN = 200000;
int n;
int a[MAXN];
int f[MAXN];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) {
    int tmp = 1;
    for (int j = 1; j * j <= a[i]; j++)
      if (a[i] % j == 0) {
        if (j != 1) tmp = max(tmp, f[j] + 1);
        if (a[i] / j != 1) tmp = max(tmp, f[a[i] / j] + 1);
      }
    for (int j = 1; j * j <= a[i]; j++)
      if (a[i] % j == 0) {
        f[j] = max(f[j], tmp);
        f[a[i] / j] = max(f[a[i] / j], tmp);
      }
  }
  int ans = 0;
  for (int i = 1; i <= 100000; i++) ans = max(ans, f[i]);
  printf("%d\n", ans);
  return 0;
}

```