#include <bits/stdc++.h>
using namespace std;
const long long maxn = 1e5 + 10;
long long a[maxn], dp[maxn], mx[maxn], dv[maxn];
vector<long long> p[maxn];
int32_t main() {
  ios_base::sync_with_stdio(false);
  for (long long i = 2; i < maxn; i++)
    if (dv[i] == 0)
      for (long long j = i; j < maxn; j += i) dv[j] = 1, p[j].push_back(i);
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i];
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    dp[i] = 1;
    for (auto x : p[a[i]]) dp[i] = max(dp[i], mx[x] + 1);
    for (auto x : p[a[i]]) mx[x] = max(mx[x], dp[i]);
    ans = max(ans, dp[i]);
  }
  cout << ans << endl;
}

### 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;
const long long maxn = 1e5 + 10;
long long a[maxn], dp[maxn], mx[maxn], dv[maxn];
vector<long long> p[maxn];
int32_t main() {
  ios_base::sync_with_stdio(false);
  for (long long i = 2; i < maxn; i++)
    if (dv[i] == 0)
      for (long long j = i; j < maxn; j += i) dv[j] = 1, p[j].push_back(i);
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i];
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    dp[i] = 1;
    for (auto x : p[a[i]]) dp[i] = max(dp[i], mx[x] + 1);
    for (auto x : p[a[i]]) mx[x] = max(mx[x], dp[i]);
    ans = max(ans, dp[i]);
  }
  cout << ans << endl;
}

```