#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100000;
int n;
int D[MAXN + 3], a[MAXN + 3], p[MAXN + 3], stos[MAXN + 3];
void erato(int N) {
  int g = (int)sqrt(N);
  for (int i = 2; i <= g; i++)
    if (p[i] == 0) {
      p[i] = i;
      for (int k = i * i; k <= N; k += i)
        if (p[k] == 0) p[k] = i;
    }
  for (int i = 2; i <= N; i++)
    if (p[i] == 0) p[i] = i;
}
void dynamic() {
  for (int i = 1; i <= n; i++) {
    D[i] = 1;
    int k = a[i];
    while (k > 1) {
      int prime = p[k];
      if (stos[prime] && stos[prime] != i) D[i] = max(D[i], D[stos[prime]] + 1);
      stos[prime] = i;
      k /= prime;
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  erato(a[n]);
  dynamic();
  int wyn = 0;
  for (int i = 1; i <= n; i++)
    if (D[i] > wyn) wyn = D[i];
  printf("%d\n", wyn);
  return 0;
}

### Prompt

Your challenge is to write 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;
const int MAXN = 100000;
int n;
int D[MAXN + 3], a[MAXN + 3], p[MAXN + 3], stos[MAXN + 3];
void erato(int N) {
  int g = (int)sqrt(N);
  for (int i = 2; i <= g; i++)
    if (p[i] == 0) {
      p[i] = i;
      for (int k = i * i; k <= N; k += i)
        if (p[k] == 0) p[k] = i;
    }
  for (int i = 2; i <= N; i++)
    if (p[i] == 0) p[i] = i;
}
void dynamic() {
  for (int i = 1; i <= n; i++) {
    D[i] = 1;
    int k = a[i];
    while (k > 1) {
      int prime = p[k];
      if (stos[prime] && stos[prime] != i) D[i] = max(D[i], D[stos[prime]] + 1);
      stos[prime] = i;
      k /= prime;
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  erato(a[n]);
  dynamic();
  int wyn = 0;
  for (int i = 1; i <= n; i++)
    if (D[i] > wyn) wyn = D[i];
  printf("%d\n", wyn);
  return 0;
}

```