#include <bits/stdc++.h>
bool prime(int x) {
  for (int i = 2; i * i <= x; ++i)
    if ((x % i) == 0) return false;
  return true;
}
int n, a[100010], q[100010], f[100010], g[100010], p[100010], pri[100010],
    h = 0, t = 0, c = 0, ans = 0, max = 0;
int main() {
  int k;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    if (a[i] > max) max = a[i];
  }
  memset(pri, 0, sizeof(pri));
  for (int i = 2; i <= max; ++i)
    if (prime(i)) {
      p[++c] = i;
      pri[i] = 1;
    }
  pri[1] = 1;
  memset(f, 0, sizeof(f));
  for (int i = 1; i <= n; ++i) {
    k = a[i];
    g[i] = 1;
    if (pri[k]) {
      g[i] = 1;
      f[k] = 1;
      if (ans < 1) ans = 1;
      continue;
    }
    for (int j = 1; p[j] * p[j] <= a[i]; ++j)
      if ((k % p[j]) == 0) {
        while ((k % p[j]) == 0) k /= p[j];
        if (f[p[j]] + 1 > g[i]) g[i] = f[p[j]] + 1;
      }
    if (k > 1) {
      if (f[k] + 1 > g[i]) g[i] = f[k] + 1;
      f[k] = g[i];
    }
    if (g[i] > ans) ans = g[i];
    k = a[i];
    for (int j = 1; p[j] * p[j] <= a[i]; ++j)
      if (k % p[j] == 0) f[p[j]] = g[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>
bool prime(int x) {
  for (int i = 2; i * i <= x; ++i)
    if ((x % i) == 0) return false;
  return true;
}
int n, a[100010], q[100010], f[100010], g[100010], p[100010], pri[100010],
    h = 0, t = 0, c = 0, ans = 0, max = 0;
int main() {
  int k;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    if (a[i] > max) max = a[i];
  }
  memset(pri, 0, sizeof(pri));
  for (int i = 2; i <= max; ++i)
    if (prime(i)) {
      p[++c] = i;
      pri[i] = 1;
    }
  pri[1] = 1;
  memset(f, 0, sizeof(f));
  for (int i = 1; i <= n; ++i) {
    k = a[i];
    g[i] = 1;
    if (pri[k]) {
      g[i] = 1;
      f[k] = 1;
      if (ans < 1) ans = 1;
      continue;
    }
    for (int j = 1; p[j] * p[j] <= a[i]; ++j)
      if ((k % p[j]) == 0) {
        while ((k % p[j]) == 0) k /= p[j];
        if (f[p[j]] + 1 > g[i]) g[i] = f[p[j]] + 1;
      }
    if (k > 1) {
      if (f[k] + 1 > g[i]) g[i] = f[k] + 1;
      f[k] = g[i];
    }
    if (g[i] > ans) ans = g[i];
    k = a[i];
    for (int j = 1; p[j] * p[j] <= a[i]; ++j)
      if (k % p[j] == 0) f[p[j]] = g[i];
  }
  printf("%d\n", ans);
  return 0;
}

```