#include <bits/stdc++.h>
using namespace std;
const int N = 1e5, M = 3e5, OO = 0x3f3f3f3f, MOD = 1e9 + 7;
int primes[N][7], arr[N], len[N], ans[N];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", arr + i);
    int cnt = 0, x = arr[i];
    for (int j = 2; j <= arr[i] / j; j += 1 + (j & 1)) {
      if (!(x % j)) {
        primes[i][cnt++] = j;
        while (!(x % j)) x /= j;
      }
    }
    if (x > 1) primes[i][cnt++] = x;
    len[i] = cnt;
  }
  int best, mx = 1;
  for (int i = 0; i < n; i++) {
    best = 1;
    for (int j = 0; j < len[i]; j++) best = max(best, ans[primes[i][j]]);
    mx = max(best, mx);
    for (int j = 0; j < len[i]; j++)
      ans[primes[i][j]] = max(best + 1, ans[primes[i][j]]);
  }
  printf("%d", mx);
  return 0;
}

### Prompt

Generate 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 N = 1e5, M = 3e5, OO = 0x3f3f3f3f, MOD = 1e9 + 7;
int primes[N][7], arr[N], len[N], ans[N];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", arr + i);
    int cnt = 0, x = arr[i];
    for (int j = 2; j <= arr[i] / j; j += 1 + (j & 1)) {
      if (!(x % j)) {
        primes[i][cnt++] = j;
        while (!(x % j)) x /= j;
      }
    }
    if (x > 1) primes[i][cnt++] = x;
    len[i] = cnt;
  }
  int best, mx = 1;
  for (int i = 0; i < n; i++) {
    best = 1;
    for (int j = 0; j < len[i]; j++) best = max(best, ans[primes[i][j]]);
    mx = max(best, mx);
    for (int j = 0; j < len[i]; j++)
      ans[primes[i][j]] = max(best + 1, ans[primes[i][j]]);
  }
  printf("%d", mx);
  return 0;
}

```