#include <bits/stdc++.h>
using namespace std;
vector<long long> prime_factors[200005];
long long pr[500005], ans[500005];
void solve() {
  for (long long i = 2; i <= 100000; i++) {
    if (pr[i] == 0) {
      prime_factors[i].push_back(i);
      for (long long j = i * 2; j <= 100000; j += i)
        pr[j] = 1, prime_factors[j].push_back(i);
    }
  }
  long long n;
  cin >> n;
  long long a[n];
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < n; i++) {
    long long mx = -1;
    for (long long j = 0; j < prime_factors[a[i]].size(); j++)
      mx = max(mx, ans[prime_factors[a[i]][j]]);
    for (long long j = 0; j < prime_factors[a[i]].size(); j++)
      ans[prime_factors[a[i]][j]] = mx + 1;
  }
  long long res = 0;
  for (long long i = 2; i <= 100000; i++) res = max(res, ans[i]);
  res = (res < 1) ? 1 : res;
  cout << res << '\n';
  return;
}
signed main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  long long T = 1;
  while (T--) {
    solve();
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
vector<long long> prime_factors[200005];
long long pr[500005], ans[500005];
void solve() {
  for (long long i = 2; i <= 100000; i++) {
    if (pr[i] == 0) {
      prime_factors[i].push_back(i);
      for (long long j = i * 2; j <= 100000; j += i)
        pr[j] = 1, prime_factors[j].push_back(i);
    }
  }
  long long n;
  cin >> n;
  long long a[n];
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < n; i++) {
    long long mx = -1;
    for (long long j = 0; j < prime_factors[a[i]].size(); j++)
      mx = max(mx, ans[prime_factors[a[i]][j]]);
    for (long long j = 0; j < prime_factors[a[i]].size(); j++)
      ans[prime_factors[a[i]][j]] = mx + 1;
  }
  long long res = 0;
  for (long long i = 2; i <= 100000; i++) res = max(res, ans[i]);
  res = (res < 1) ? 1 : res;
  cout << res << '\n';
  return;
}
signed main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  long long T = 1;
  while (T--) {
    solve();
  }
  return 0;
}

```