#include <bits/stdc++.h>
using namespace std;
int n, num[100009], dp[100009];
int main() {
  while (scanf("%d", &n) == 1) {
    int maxi = -1, result = -1;
    ;
    for (int i = int(0); i < int(n); i++) {
      scanf("%d", &num[i]);
      int maxi = 0;
      for (int j = 2; j * j <= num[i]; j++) {
        if (num[i] % j == 0)
          maxi = max(maxi, dp[j]), maxi = max(maxi, dp[num[i] / j]);
      }
      maxi = max(maxi, dp[num[i]]);
      result = max(result, maxi + 1);
      for (int j = 2; j * j <= num[i]; j++) {
        if (num[i] % j == 0) dp[j] = maxi + 1, dp[num[i] / j] = maxi + 1;
      }
      dp[num[i]] = maxi + 1;
    }
    cout << result << endl;
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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 n, num[100009], dp[100009];
int main() {
  while (scanf("%d", &n) == 1) {
    int maxi = -1, result = -1;
    ;
    for (int i = int(0); i < int(n); i++) {
      scanf("%d", &num[i]);
      int maxi = 0;
      for (int j = 2; j * j <= num[i]; j++) {
        if (num[i] % j == 0)
          maxi = max(maxi, dp[j]), maxi = max(maxi, dp[num[i] / j]);
      }
      maxi = max(maxi, dp[num[i]]);
      result = max(result, maxi + 1);
      for (int j = 2; j * j <= num[i]; j++) {
        if (num[i] % j == 0) dp[j] = maxi + 1, dp[num[i] / j] = maxi + 1;
      }
      dp[num[i]] = maxi + 1;
    }
    cout << result << endl;
  }
  return 0;
}

```