#include <bits/stdc++.h>
using namespace std;
int A[100005], P[100005];
map<int, int> M;
int main() {
  int n, i, j, d;
  cin >> n;
  for (i = 0; i < (n); i++) {
    cin >> A[i];
  }
  int mx, ans = 0;
  for (i = 0; i < (n); i++) {
    mx = 0;
    for (j = 2; j * j <= A[i]; j++) {
      if (A[i] % j == 0) {
        d = A[i] / j;
        mx = ((M[d]) > (mx) ? (M[d]) : (mx));
        mx = ((M[j]) > (mx) ? (M[j]) : (mx));
      }
    }
    for (j = 2; j * j <= A[i]; j++) {
      if (A[i] % j == 0) {
        d = A[i] / j;
        M[d] = 1 + mx;
        M[j] = 1 + mx;
      }
    }
    M[A[i]] = mx + 1;
    ans = ((ans) > (mx + 1) ? (ans) : (mx + 1));
  }
  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;
int A[100005], P[100005];
map<int, int> M;
int main() {
  int n, i, j, d;
  cin >> n;
  for (i = 0; i < (n); i++) {
    cin >> A[i];
  }
  int mx, ans = 0;
  for (i = 0; i < (n); i++) {
    mx = 0;
    for (j = 2; j * j <= A[i]; j++) {
      if (A[i] % j == 0) {
        d = A[i] / j;
        mx = ((M[d]) > (mx) ? (M[d]) : (mx));
        mx = ((M[j]) > (mx) ? (M[j]) : (mx));
      }
    }
    for (j = 2; j * j <= A[i]; j++) {
      if (A[i] % j == 0) {
        d = A[i] / j;
        M[d] = 1 + mx;
        M[j] = 1 + mx;
      }
    }
    M[A[i]] = mx + 1;
    ans = ((ans) > (mx + 1) ? (ans) : (mx + 1));
  }
  cout << ans << endl;
  return 0;
}

```