#include <bits/stdc++.h>
using namespace std;
const int MX = 1 * 1e5 + 10;
const double PI = acos(-1.0), EPS = 1e-9;
int N, k;
int arr[MX];
int dp[MX];
int best[MX];
vector<int> fact;
int p[MX];
vector<int> ff(int n) {
  fact.clear();
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) fact.push_back(i);
    while (n % i == 0) {
      n /= i;
    }
  }
  if (n > 1) fact.push_back(n);
  sort(fact.begin(), fact.end());
  fact.resize(unique(fact.begin(), fact.end()) - fact.begin());
  return fact;
}
int main() {
  cin >> N;
  for (int i = 0; i < N; i++) cin >> arr[i];
  p[0] = p[1] = 1;
  for (int i = 2; i < MX; i++) {
    if (p[i]) continue;
    for (int j = i + i; j < MX; j += i) p[j]++;
  }
  for (int i = 0; i < N; i++) {
    if (!p[arr[i]]) {
      best[arr[i]] = 1;
      continue;
    }
    fact = ff(arr[i]);
    int mx = 0;
    for (int j = 0; j < fact.size(); j++) {
      best[fact[j]]++;
      mx = max(mx, best[fact[j]]);
    }
    for (int j = 0; j < fact.size(); j++) best[fact[j]] = mx;
  }
  int ans = 1;
  for (int i = 0; i < MX; i++) ans = max(ans, best[i]);
  cout << ans << "\n";
  return 0;
}

### Prompt

Please create a solution in cpp 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 int MX = 1 * 1e5 + 10;
const double PI = acos(-1.0), EPS = 1e-9;
int N, k;
int arr[MX];
int dp[MX];
int best[MX];
vector<int> fact;
int p[MX];
vector<int> ff(int n) {
  fact.clear();
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) fact.push_back(i);
    while (n % i == 0) {
      n /= i;
    }
  }
  if (n > 1) fact.push_back(n);
  sort(fact.begin(), fact.end());
  fact.resize(unique(fact.begin(), fact.end()) - fact.begin());
  return fact;
}
int main() {
  cin >> N;
  for (int i = 0; i < N; i++) cin >> arr[i];
  p[0] = p[1] = 1;
  for (int i = 2; i < MX; i++) {
    if (p[i]) continue;
    for (int j = i + i; j < MX; j += i) p[j]++;
  }
  for (int i = 0; i < N; i++) {
    if (!p[arr[i]]) {
      best[arr[i]] = 1;
      continue;
    }
    fact = ff(arr[i]);
    int mx = 0;
    for (int j = 0; j < fact.size(); j++) {
      best[fact[j]]++;
      mx = max(mx, best[fact[j]]);
    }
    for (int j = 0; j < fact.size(); j++) best[fact[j]] = mx;
  }
  int ans = 1;
  for (int i = 0; i < MX; i++) ans = max(ans, best[i]);
  cout << ans << "\n";
  return 0;
}

```