#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100009;
int n, k, i, j, a[MAXN], now = 0, ans = 0, f[MAXN], ls[MAXN], m;
int su[MAXN], tot = 0;
bool sh[MAXN];
int main() {
  memset(sh, false, sizeof(sh));
  for (i = 2; i < 10000; i++)
    if (!sh[i]) {
      su[++tot] = i;
      if (i * i > 100000000) continue;
      for (j = i * i; j <= 10000; j += i) sh[j] = true;
    }
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  memset(f, 0, sizeof(f));
  int t;
  for (i = 1; i <= n; i++) {
    k = -1;
    m = 0;
    t = a[i];
    for (j = 1; su[j] * su[j] <= t; j++)
      if (a[i] % su[j] == 0) {
        ls[++m] = su[j];
        if (f[su[j]] > k) k = f[su[j]];
        while (a[i] % su[j] == 0) a[i] /= su[j];
      }
    if (a[i] != 1) {
      ls[++m] = a[i];
      if (f[a[i]] > k) k = f[a[i]];
    }
    for (j = 1; j <= m; j++) f[ls[j]] = max(f[ls[j]], k + 1);
  }
  int ans = 1;
  for (i = 1; i <= 100000; i++) ans = max(ans, f[i]);
  printf("%d\n", ans);
  return 0;
}

### Prompt

Your challenge is to write 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 = 100009;
int n, k, i, j, a[MAXN], now = 0, ans = 0, f[MAXN], ls[MAXN], m;
int su[MAXN], tot = 0;
bool sh[MAXN];
int main() {
  memset(sh, false, sizeof(sh));
  for (i = 2; i < 10000; i++)
    if (!sh[i]) {
      su[++tot] = i;
      if (i * i > 100000000) continue;
      for (j = i * i; j <= 10000; j += i) sh[j] = true;
    }
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  memset(f, 0, sizeof(f));
  int t;
  for (i = 1; i <= n; i++) {
    k = -1;
    m = 0;
    t = a[i];
    for (j = 1; su[j] * su[j] <= t; j++)
      if (a[i] % su[j] == 0) {
        ls[++m] = su[j];
        if (f[su[j]] > k) k = f[su[j]];
        while (a[i] % su[j] == 0) a[i] /= su[j];
      }
    if (a[i] != 1) {
      ls[++m] = a[i];
      if (f[a[i]] > k) k = f[a[i]];
    }
    for (j = 1; j <= m; j++) f[ls[j]] = max(f[ls[j]], k + 1);
  }
  int ans = 1;
  for (i = 1; i <= 100000; i++) ans = max(ans, f[i]);
  printf("%d\n", ans);
  return 0;
}

```