#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int a[N], n, ans[N], b[N];
int main() {
  scanf("%d", &n);
  if (n == 1) {
    printf("1\n");
    return 0;
  }
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) {
    b[0] = 0;
    int to = (int)sqrt((double)a[i]), k = 0;
    for (int j = 1; j <= to; j++)
      if (a[i] % j == 0) {
        if (j > 1) b[++b[0]] = j, k = max(k, ans[j]);
        if (j * j != a[i]) b[++b[0]] = a[i] / j, k = max(k, ans[a[i] / j]);
      }
    for (int j = 1; j <= b[0]; j++) ans[b[j]] = max(ans[b[j]], k + 1);
  }
  for (int i = 1; i <= (int)1e5; i++) ans[0] = max(ans[0], ans[i]);
  printf("%d\n", ans[0]);
  return 0;
}

### Prompt

Your task is to create 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 N = 1e5 + 10;
int a[N], n, ans[N], b[N];
int main() {
  scanf("%d", &n);
  if (n == 1) {
    printf("1\n");
    return 0;
  }
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) {
    b[0] = 0;
    int to = (int)sqrt((double)a[i]), k = 0;
    for (int j = 1; j <= to; j++)
      if (a[i] % j == 0) {
        if (j > 1) b[++b[0]] = j, k = max(k, ans[j]);
        if (j * j != a[i]) b[++b[0]] = a[i] / j, k = max(k, ans[a[i] / j]);
      }
    for (int j = 1; j <= b[0]; j++) ans[b[j]] = max(ans[b[j]], k + 1);
  }
  for (int i = 1; i <= (int)1e5; i++) ans[0] = max(ans[0], ans[i]);
  printf("%d\n", ans[0]);
  return 0;
}

```