#include <bits/stdc++.h>
using namespace std;
constexpr auto EPS = 1e-9;
clock_t T;
void GOAT() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
}
long long gcd(int a, int b) { return !b ? abs(a) : gcd(b, a % b); }
int n, ar[100005], mem[100005];
vector<int> divisors(int n) {
  vector<int> ret;
  for (int i = 2; 1ll * i * i <= n; ++i) {
    if (n % i == 0) {
      ret.push_back(i);
      if (i != n / i) ret.push_back(n / i);
    }
  }
  ret.push_back(n);
  return ret;
}
int main() {
  GOAT();
  memset(mem, 0, sizeof mem);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> ar[i];
  int mx = 1;
  for (int i = 0; i < n; i++) {
    vector<int> v = divisors(ar[i]);
    int ans = 1;
    for (auto it : v) {
      ans = max(ans, mem[it] + 1);
    }
    mx = max(mx, ans);
    for (auto it : v) mem[it] = ans;
  }
  cout << mx << "\n";
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
constexpr auto EPS = 1e-9;
clock_t T;
void GOAT() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
}
long long gcd(int a, int b) { return !b ? abs(a) : gcd(b, a % b); }
int n, ar[100005], mem[100005];
vector<int> divisors(int n) {
  vector<int> ret;
  for (int i = 2; 1ll * i * i <= n; ++i) {
    if (n % i == 0) {
      ret.push_back(i);
      if (i != n / i) ret.push_back(n / i);
    }
  }
  ret.push_back(n);
  return ret;
}
int main() {
  GOAT();
  memset(mem, 0, sizeof mem);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> ar[i];
  int mx = 1;
  for (int i = 0; i < n; i++) {
    vector<int> v = divisors(ar[i]);
    int ans = 1;
    for (auto it : v) {
      ans = max(ans, mem[it] + 1);
    }
    mx = max(mx, ans);
    for (auto it : v) mem[it] = ans;
  }
  cout << mx << "\n";
}

```