#include <bits/stdc++.h>
using namespace std;
int dp[100000 + 1];
int f[100000 + 1];
int last[100000 + 1];
int coin[100000 + 1];
vector<int> temp;
int main() {
  int i, n, j, p, q, mx;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", &coin[i]);
  for (i = 0; i < n; i++) {
    mx = 0;
    temp.clear();
    for (j = 2; j * j <= coin[i]; j++) {
      if (coin[i] % j == 0) {
        p = coin[i] / j;
        f[p]++;
        dp[coin[i]] = max(dp[coin[i]], f[p]);
        dp[coin[i]] = max(dp[coin[i]], last[p] + 1);
        temp.push_back(p);
        q = j;
        if (p != q) {
          f[q]++;
          temp.push_back(q);
          dp[coin[i]] = max(dp[coin[i]], f[q]);
          dp[coin[i]] = max(dp[coin[i]], last[q] + 1);
        }
      }
    }
    for (j = 0; j < temp.size(); j++) {
      last[temp[j]] = max(last[temp[j]], dp[coin[i]]);
    }
    if (dp[coin[i]] == 0) {
      dp[coin[i]] = 1;
      f[coin[i]]++;
      last[coin[i]] = 1;
    }
  }
  mx = 0;
  for (i = 1; i <= 100000; i++) {
    mx = max(mx, dp[i]);
  }
  cout << mx << "\n";
}

### Prompt

Please formulate 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;
int dp[100000 + 1];
int f[100000 + 1];
int last[100000 + 1];
int coin[100000 + 1];
vector<int> temp;
int main() {
  int i, n, j, p, q, mx;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", &coin[i]);
  for (i = 0; i < n; i++) {
    mx = 0;
    temp.clear();
    for (j = 2; j * j <= coin[i]; j++) {
      if (coin[i] % j == 0) {
        p = coin[i] / j;
        f[p]++;
        dp[coin[i]] = max(dp[coin[i]], f[p]);
        dp[coin[i]] = max(dp[coin[i]], last[p] + 1);
        temp.push_back(p);
        q = j;
        if (p != q) {
          f[q]++;
          temp.push_back(q);
          dp[coin[i]] = max(dp[coin[i]], f[q]);
          dp[coin[i]] = max(dp[coin[i]], last[q] + 1);
        }
      }
    }
    for (j = 0; j < temp.size(); j++) {
      last[temp[j]] = max(last[temp[j]], dp[coin[i]]);
    }
    if (dp[coin[i]] == 0) {
      dp[coin[i]] = 1;
      f[coin[i]]++;
      last[coin[i]] = 1;
    }
  }
  mx = 0;
  for (i = 1; i <= 100000; i++) {
    mx = max(mx, dp[i]);
  }
  cout << mx << "\n";
}

```