#include <bits/stdc++.h>
using namespace std;
vector<int> divisor[100010];
bool good[100010];
int dp[100010];
int main(void) {
  int ans = 0, N, i, j;
  cin >> N;
  for ((i) = 0; (i) < (int)(N); (i)++) {
    int x;
    scanf("%d", &x);
    good[x] = true;
  }
  for (i = 2; i <= 100000; i++)
    for (j = i; j <= 100000; j += i) divisor[j].push_back(i);
  for (i = 1; i <= 100000; i++)
    if (good[i]) {
      int tmp = 1;
      for ((j) = 0; (j) < (int)(divisor[i].size()); (j)++)
        tmp = max(tmp, dp[divisor[i][j]] + 1);
      ans = max(ans, tmp);
      for ((j) = 0; (j) < (int)(divisor[i].size()); (j)++)
        dp[divisor[i][j]] = max(dp[divisor[i][j]], tmp);
    }
  cout << ans << endl;
  return 0;
}

### Prompt

Your task is to create 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;
vector<int> divisor[100010];
bool good[100010];
int dp[100010];
int main(void) {
  int ans = 0, N, i, j;
  cin >> N;
  for ((i) = 0; (i) < (int)(N); (i)++) {
    int x;
    scanf("%d", &x);
    good[x] = true;
  }
  for (i = 2; i <= 100000; i++)
    for (j = i; j <= 100000; j += i) divisor[j].push_back(i);
  for (i = 1; i <= 100000; i++)
    if (good[i]) {
      int tmp = 1;
      for ((j) = 0; (j) < (int)(divisor[i].size()); (j)++)
        tmp = max(tmp, dp[divisor[i][j]] + 1);
      ans = max(ans, tmp);
      for ((j) = 0; (j) < (int)(divisor[i].size()); (j)++)
        dp[divisor[i][j]] = max(dp[divisor[i][j]], tmp);
    }
  cout << ans << endl;
  return 0;
}

```