#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
vector<int> divs[N];
int dp[N], d[N], arr[N];
int main() {
  int n;
  scanf("%d", &n);
  for (int a = 1; a <= n; a++) scanf("%d", &arr[a]);
  for (int a = 2; a <= arr[n]; a++)
    for (int b = 2 * a; b <= arr[n]; b += a) divs[b].push_back(a);
  int ans = 1;
  for (int a = 1; a <= n; a++) {
    dp[a] = 1;
    for (int b = 0; b < divs[arr[a]].size(); b++)
      dp[a] = max(d[divs[arr[a]][b]] + 1, dp[a]);
    for (int b = 0; b < divs[arr[a]].size(); b++)
      d[divs[arr[a]][b]] = max(d[divs[arr[a]][b]], dp[a]);
    d[arr[a]] = max(d[arr[a]], dp[a]);
    ans = max(ans, dp[a]);
  }
  printf("%d\n", ans);
  return 0;
}

### 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>
using namespace std;
const int N = 1e5 + 5;
vector<int> divs[N];
int dp[N], d[N], arr[N];
int main() {
  int n;
  scanf("%d", &n);
  for (int a = 1; a <= n; a++) scanf("%d", &arr[a]);
  for (int a = 2; a <= arr[n]; a++)
    for (int b = 2 * a; b <= arr[n]; b += a) divs[b].push_back(a);
  int ans = 1;
  for (int a = 1; a <= n; a++) {
    dp[a] = 1;
    for (int b = 0; b < divs[arr[a]].size(); b++)
      dp[a] = max(d[divs[arr[a]][b]] + 1, dp[a]);
    for (int b = 0; b < divs[arr[a]].size(); b++)
      d[divs[arr[a]][b]] = max(d[divs[arr[a]][b]], dp[a]);
    d[arr[a]] = max(d[arr[a]], dp[a]);
    ans = max(ans, dp[a]);
  }
  printf("%d\n", ans);
  return 0;
}

```