#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n, a[200009], ct[200009], b[200009], m, ma = 0, k;
  string s;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  for (int i = 0; i < 100001; i++) {
    ct[i] = -1;
    b[i] = 0;
  }
  for (int i = 0; i < n; i++) {
    ma = 0;
    for (int j = 1; j <= sqrt(a[i]); j++) {
      if (a[i] % j == 0) {
        long long temp = a[i] / j;
        if (j != 1 && ct[j] != -1 && b[ct[j]] > ma) ma = b[ct[j]];
        if (ct[temp] != -1 && b[ct[temp]] > ma) ma = b[ct[temp]];
        ct[j] = i;
        ct[temp] = i;
        b[i] = ma + 1;
      }
    }
  }
  ma = 0;
  for (int i = 0; i < n; i++) ma = max(ma, b[i]);
  cout << ma << 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 main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n, a[200009], ct[200009], b[200009], m, ma = 0, k;
  string s;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  for (int i = 0; i < 100001; i++) {
    ct[i] = -1;
    b[i] = 0;
  }
  for (int i = 0; i < n; i++) {
    ma = 0;
    for (int j = 1; j <= sqrt(a[i]); j++) {
      if (a[i] % j == 0) {
        long long temp = a[i] / j;
        if (j != 1 && ct[j] != -1 && b[ct[j]] > ma) ma = b[ct[j]];
        if (ct[temp] != -1 && b[ct[temp]] > ma) ma = b[ct[temp]];
        ct[j] = i;
        ct[temp] = i;
        b[i] = ma + 1;
      }
    }
  }
  ma = 0;
  for (int i = 0; i < n; i++) ma = max(ma, b[i]);
  cout << ma << endl;
  return 0;
}

```