#include <bits/stdc++.h>
int p[100007], a[100007];
int main() {
  int n, i, ans, j;
  while (~scanf("%d", &n)) {
    for (i = 0; i < n; i++) {
      scanf("%d", &a[i]);
      p[i + 1] = 0;
    }
    for (i = 0, ans = 0; i < n; i++) {
      int res = 0, tmp = a[i];
      for (j = 2; j * j <= tmp; j++)
        if (tmp % j == 0) {
          if (p[j] > res) res = p[j];
          while (tmp % j == 0) tmp /= j;
        }
      if (tmp > 1 && p[tmp] > res) res = p[tmp];
      res += 1;
      if (res > ans) ans = res;
      for (j = 2; j * j <= a[i]; j++)
        if (a[i] % j == 0) {
          p[j] = res;
          while (a[i] % j == 0) a[i] /= j;
        }
      if (a[i] > 1) p[a[i]] = res;
    }
    printf("%d\n", ans);
  }
}

### 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>
int p[100007], a[100007];
int main() {
  int n, i, ans, j;
  while (~scanf("%d", &n)) {
    for (i = 0; i < n; i++) {
      scanf("%d", &a[i]);
      p[i + 1] = 0;
    }
    for (i = 0, ans = 0; i < n; i++) {
      int res = 0, tmp = a[i];
      for (j = 2; j * j <= tmp; j++)
        if (tmp % j == 0) {
          if (p[j] > res) res = p[j];
          while (tmp % j == 0) tmp /= j;
        }
      if (tmp > 1 && p[tmp] > res) res = p[tmp];
      res += 1;
      if (res > ans) ans = res;
      for (j = 2; j * j <= a[i]; j++)
        if (a[i] % j == 0) {
          p[j] = res;
          while (a[i] % j == 0) a[i] /= j;
        }
      if (a[i] > 1) p[a[i]] = res;
    }
    printf("%d\n", ans);
  }
}

```