#include <bits/stdc++.h>
using namespace std;
long N, A[100000], f[100000];
long gdc(long a, long b) {
  if (a == 0)
    return b;
  else
    gdc(b % a, a);
}
int main() {
  cin >> N;
  long t;
  for (int i = 1; i <= N; i++) {
    cin >> A[i];
    t = 0;
    for (int j = max(1, i - 100); j <= i - 1; j++)
      if (gdc(A[i], A[j]) > 1) t = max(t, f[j]);
    f[i] = t + 1;
  }
  t = 0;
  for (long i = 1; i <= N; i++) {
    t = max(t, f[i]);
  }
  cout << t;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
long N, A[100000], f[100000];
long gdc(long a, long b) {
  if (a == 0)
    return b;
  else
    gdc(b % a, a);
}
int main() {
  cin >> N;
  long t;
  for (int i = 1; i <= N; i++) {
    cin >> A[i];
    t = 0;
    for (int j = max(1, i - 100); j <= i - 1; j++)
      if (gdc(A[i], A[j]) > 1) t = max(t, f[j]);
    f[i] = t + 1;
  }
  t = 0;
  for (long i = 1; i <= N; i++) {
    t = max(t, f[i]);
  }
  cout << t;
}

```