#include <bits/stdc++.h>
using namespace std;
int T, N, B[18], F[100001], ans_wer = 1;
void crs() {
  memset(F, 0, sizeof(F));
  ans_wer = 1;
}
inline void pd(int k) {
  int l = 0, i = 2;
  while (i * i <= k) {
    if (k % i == 0) {
      B[++l] = i;
      do {
        k /= i;
      } while (k % i == 0);
    }
    i++;
  }
  if (k > 1) B[++l] = k;
  int tmp = 0;
  for (int i = 1; i <= l; i++) {
    F[B[i]]++;
    tmp = max(tmp, F[B[i]]);
  }
  for (int i = 1; i <= l; i++) F[B[i]] = tmp;
  ans_wer = max(ans_wer, tmp);
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) {
    int a;
    scanf("%d", &a);
    pd(a);
  }
  printf("%d\n", ans_wer);
  crs();
  return 0;
}

### Prompt

Please create a solution in CPP 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 T, N, B[18], F[100001], ans_wer = 1;
void crs() {
  memset(F, 0, sizeof(F));
  ans_wer = 1;
}
inline void pd(int k) {
  int l = 0, i = 2;
  while (i * i <= k) {
    if (k % i == 0) {
      B[++l] = i;
      do {
        k /= i;
      } while (k % i == 0);
    }
    i++;
  }
  if (k > 1) B[++l] = k;
  int tmp = 0;
  for (int i = 1; i <= l; i++) {
    F[B[i]]++;
    tmp = max(tmp, F[B[i]]);
  }
  for (int i = 1; i <= l; i++) F[B[i]] = tmp;
  ans_wer = max(ans_wer, tmp);
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) {
    int a;
    scanf("%d", &a);
    pd(a);
  }
  printf("%d\n", ans_wer);
  crs();
  return 0;
}

```