#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int dp[N], a[N], n;
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  if (n == 1) {
    cout << 1;
    return 0;
  }
  for (int i = 0; i < n; ++i) {
    if (a[i] == 1) continue;
    int d = 2, mx = 0, x = a[i];
    while (d * d <= x)
      if (x % d == 0) {
        mx = max(mx, dp[d] + 1);
        x /= d;
      } else
        ++d;
    mx = max(mx, dp[x] + 1);
    d = 2;
    x = a[i];
    while (d * d <= x)
      if (x % d == 0) {
        dp[d] = max(dp[d], mx);
        x /= d;
      } else
        ++d;
    dp[x] = max(dp[x], mx);
  }
  int mx = 0;
  for (int i = 1; i < N; ++i) mx = max(mx, dp[i]);
  cout << mx;
  return 0;
}

### 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;
const int N = 1e5 + 5;
int dp[N], a[N], n;
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  if (n == 1) {
    cout << 1;
    return 0;
  }
  for (int i = 0; i < n; ++i) {
    if (a[i] == 1) continue;
    int d = 2, mx = 0, x = a[i];
    while (d * d <= x)
      if (x % d == 0) {
        mx = max(mx, dp[d] + 1);
        x /= d;
      } else
        ++d;
    mx = max(mx, dp[x] + 1);
    d = 2;
    x = a[i];
    while (d * d <= x)
      if (x % d == 0) {
        dp[d] = max(dp[d], mx);
        x /= d;
      } else
        ++d;
    dp[x] = max(dp[x], mx);
  }
  int mx = 0;
  for (int i = 1; i < N; ++i) mx = max(mx, dp[i]);
  cout << mx;
  return 0;
}

```