#include <bits/stdc++.h>
using namespace std;
vector<int> p[100100];
vector<int> p2[100100];
vector<int>::iterator it;
bool v[100100];
bool v2[100100];
bool num[100100];
int dp[100100];
int dfs(int x) {
  if (dp[x] != -1) return dp[x];
  int r = 0;
  int n = p[x].size();
  int m = 0;
  for (int i = 0; i < n; i++) {
    it = upper_bound(p2[p[x][i]].begin(), p2[p[x][i]].end(), x);
    if (it != p2[p[x][i]].end()) r = max(r, dfs(*it));
  }
  return dp[x] = r + 1;
}
int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    num[a[i]] = 1;
  }
  for (int i = 2; i <= 100001; i++) {
    if (!v[i]) {
      for (int j = 1; (long long)j * i <= 100001; j++) {
        v[i * j] = 1;
        if (num[i * j]) {
          p2[i].push_back(i * j);
          p[i * j].push_back(i);
        }
      }
    }
  }
  memset(dp, -1, sizeof(dp));
  int maxx = 0;
  for (int i = 0; i < n; i++) {
    maxx = max(maxx, dfs(a[i]));
  }
  cout << maxx;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
vector<int> p[100100];
vector<int> p2[100100];
vector<int>::iterator it;
bool v[100100];
bool v2[100100];
bool num[100100];
int dp[100100];
int dfs(int x) {
  if (dp[x] != -1) return dp[x];
  int r = 0;
  int n = p[x].size();
  int m = 0;
  for (int i = 0; i < n; i++) {
    it = upper_bound(p2[p[x][i]].begin(), p2[p[x][i]].end(), x);
    if (it != p2[p[x][i]].end()) r = max(r, dfs(*it));
  }
  return dp[x] = r + 1;
}
int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    num[a[i]] = 1;
  }
  for (int i = 2; i <= 100001; i++) {
    if (!v[i]) {
      for (int j = 1; (long long)j * i <= 100001; j++) {
        v[i * j] = 1;
        if (num[i * j]) {
          p2[i].push_back(i * j);
          p[i * j].push_back(i);
        }
      }
    }
  }
  memset(dp, -1, sizeof(dp));
  int maxx = 0;
  for (int i = 0; i < n; i++) {
    maxx = max(maxx, dfs(a[i]));
  }
  cout << maxx;
  return 0;
}

```