#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 99;
int n, ans, a[N], dp[N], mark[N];
vector<int> g[N];
int main() {
  cin >> n;
  for (int i = 1; i < n + 1; i++) {
    cin >> a[i];
    mark[a[i]] = i;
  }
  sort(a + 1, a + 1 + n);
  for (int i = 2; i < N; i++) {
    vector<int> v;
    for (int j = i; j < N; j += i)
      if (mark[j]) v.push_back(mark[j]);
    for (int j = 0; j < max(int(v.size() - 1), 0); j++)
      g[v[j]].push_back(v[j + 1]);
  }
  for (int i = n; i >= 1; i--) {
    for (int j = 0; j < g[i].size(); j++) dp[i] = max(dp[i], dp[g[i][j]] + 1);
    ;
    ans = max(ans, dp[i]);
    ;
  }
  cout << ans + 1;
}

### 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>
using namespace std;
const int N = 1e5 + 99;
int n, ans, a[N], dp[N], mark[N];
vector<int> g[N];
int main() {
  cin >> n;
  for (int i = 1; i < n + 1; i++) {
    cin >> a[i];
    mark[a[i]] = i;
  }
  sort(a + 1, a + 1 + n);
  for (int i = 2; i < N; i++) {
    vector<int> v;
    for (int j = i; j < N; j += i)
      if (mark[j]) v.push_back(mark[j]);
    for (int j = 0; j < max(int(v.size() - 1), 0); j++)
      g[v[j]].push_back(v[j + 1]);
  }
  for (int i = n; i >= 1; i--) {
    for (int j = 0; j < g[i].size(); j++) dp[i] = max(dp[i], dp[g[i][j]] + 1);
    ;
    ans = max(ans, dp[i]);
    ;
  }
  cout << ans + 1;
}

```