#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
long long int dp[100005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int a, b, i, ans = 0, j;
  cin >> a;
  for (i = 0; i < a; i++) {
    cin >> b;
    dp[b] = 1;
    for (j = 2; j * j <= b; j++)
      if (b % j == 0) dp[b] = max(dp[b], max(dp[j], dp[b / j]) + 1);
    for (j = 2; j * j <= b; j++)
      if (b % j == 0) dp[j] = dp[b / j] = dp[b];
    ans = max(dp[b], ans);
  }
  cout << ans << endl;
}

### Prompt

Create a solution in cpp for 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>
#pragma GCC optimize("O3")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
long long int dp[100005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int a, b, i, ans = 0, j;
  cin >> a;
  for (i = 0; i < a; i++) {
    cin >> b;
    dp[b] = 1;
    for (j = 2; j * j <= b; j++)
      if (b % j == 0) dp[b] = max(dp[b], max(dp[j], dp[b / j]) + 1);
    for (j = 2; j * j <= b; j++)
      if (b % j == 0) dp[j] = dp[b / j] = dp[b];
    ans = max(dp[b], ans);
  }
  cout << ans << endl;
}

```