#include <bits/stdc++.h>
int n, x, f[110000 + 1], ans = 0, maxw, p[110000 + 1], tot = 0, s[110000 + 1],
                         w[110000 + 1];
bool v[110000 + 1], h[110000 + 1];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &x);
    h[x] = true, maxw = std::max(x, maxw);
  }
  for (int i = 2; i <= maxw; ++i) {
    if (!v[i]) p[tot++] = i, s[i] = i;
    for (int j = 0; j < tot && i * p[j] <= maxw; ++j) {
      v[i * p[j]] = true;
      s[i * p[j]] = p[j];
    }
  }
  for (int i = 1; i <= maxw; ++i) {
    if (!h[i]) continue;
    f[i] = 1;
    for (int x = i; x != 1; x /= s[x]) f[i] = std::max(f[i], w[s[x]] + 1);
    for (int x = i; x != 1; x /= s[x]) w[s[x]] = std::max(w[s[x]], f[i]);
  }
  for (int i = 1; i <= maxw; ++i) ans = std::max(ans, f[i]);
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please provide a CPP coded solution 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>
int n, x, f[110000 + 1], ans = 0, maxw, p[110000 + 1], tot = 0, s[110000 + 1],
                         w[110000 + 1];
bool v[110000 + 1], h[110000 + 1];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &x);
    h[x] = true, maxw = std::max(x, maxw);
  }
  for (int i = 2; i <= maxw; ++i) {
    if (!v[i]) p[tot++] = i, s[i] = i;
    for (int j = 0; j < tot && i * p[j] <= maxw; ++j) {
      v[i * p[j]] = true;
      s[i * p[j]] = p[j];
    }
  }
  for (int i = 1; i <= maxw; ++i) {
    if (!h[i]) continue;
    f[i] = 1;
    for (int x = i; x != 1; x /= s[x]) f[i] = std::max(f[i], w[s[x]] + 1);
    for (int x = i; x != 1; x /= s[x]) w[s[x]] = std::max(w[s[x]], f[i]);
  }
  for (int i = 1; i <= maxw; ++i) ans = std::max(ans, f[i]);
  printf("%d\n", ans);
  return 0;
}

```