#include <bits/stdc++.h>
using namespace std;
long long a[100009], dp[100009];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long i = 2, j;
  while (i <= 100009) {
    if (a[i] == 0) {
      j = i;
      while (j <= 100009) {
        a[j] = i;
        j += i;
      }
    }
    i++;
  }
  long long n, d, mx, ans = 0;
  cin >> n;
  while (n--) {
    cin >> d;
    vector<long long> v;
    while (d > 1) {
      v.push_back(a[d]);
      mx = a[d];
      while (d % mx == 0) d /= mx;
    }
    mx = 0;
    for (auto x : v) mx = max(mx, dp[x]);
    mx++;
    ans = max(ans, mx);
    for (auto x : v) dp[x] = max(dp[x], mx);
  }
  cout << ans << "\n";
}

### Prompt

Please formulate 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;
long long a[100009], dp[100009];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long i = 2, j;
  while (i <= 100009) {
    if (a[i] == 0) {
      j = i;
      while (j <= 100009) {
        a[j] = i;
        j += i;
      }
    }
    i++;
  }
  long long n, d, mx, ans = 0;
  cin >> n;
  while (n--) {
    cin >> d;
    vector<long long> v;
    while (d > 1) {
      v.push_back(a[d]);
      mx = a[d];
      while (d % mx == 0) d /= mx;
    }
    mx = 0;
    for (auto x : v) mx = max(mx, dp[x]);
    mx++;
    ans = max(ans, mx);
    for (auto x : v) dp[x] = max(dp[x], mx);
  }
  cout << ans << "\n";
}

```