#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5;
int n, Max, res, a[N], p[N], t[N];
bool not_prime[N];
void Input() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  Max = *max_element(a + 1, a + n + 1);
}
void Sieve() {
  for (int i = 2; i <= Max; i++) {
    if (!not_prime[i]) {
      for (int j = i; j <= Max; j += i) {
        p[j] = i;
        not_prime[j + i] = true;
      }
    }
  }
}
void Solve() {
  for (int i = 1; i <= n; i++) {
    int c_res = 0;
    for (int j = a[i]; j > 1; j /= p[j]) c_res = max(c_res, t[p[j]]);
    for (int j = a[i]; j > 1; j /= p[j]) t[p[j]] = c_res + 1;
    res = max(res, c_res + 1);
  }
  cout << res << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  Input();
  Sieve();
  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;
const int N = 3e5 + 5;
int n, Max, res, a[N], p[N], t[N];
bool not_prime[N];
void Input() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  Max = *max_element(a + 1, a + n + 1);
}
void Sieve() {
  for (int i = 2; i <= Max; i++) {
    if (!not_prime[i]) {
      for (int j = i; j <= Max; j += i) {
        p[j] = i;
        not_prime[j + i] = true;
      }
    }
  }
}
void Solve() {
  for (int i = 1; i <= n; i++) {
    int c_res = 0;
    for (int j = a[i]; j > 1; j /= p[j]) c_res = max(c_res, t[p[j]]);
    for (int j = a[i]; j > 1; j /= p[j]) t[p[j]] = c_res + 1;
    res = max(res, c_res + 1);
  }
  cout << res << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  Input();
  Sieve();
  Solve();
  return 0;
}

```