#include <bits/stdc++.h>
using namespace std;
vector<pair<unsigned long long int, int> > factorization(
    unsigned long long int n) {
  vector<pair<unsigned long long int, int> > res;
  unsigned long long int radice = sqrt(n);
  unsigned long long int i = 2;
  while (i <= radice and n > 1) {
    if (n % i == 0) {
      pair<unsigned long long int, int> nuovo = make_pair(i, 0);
      while (n % i == 0) {
        nuovo.second++;
        n /= i;
      }
      res.push_back(nuovo);
      radice = sqrt(n);
    }
    i++;
  }
  if (n > 1) {
    pair<unsigned long long int, int> nuovo = make_pair(n, 1);
    res.push_back(nuovo);
  }
  return res;
}
int n;
int seq[100010];
int migliore = 0;
bool primi[100010];
int peso[100010];
void primizza() {
  primi[2] = false;
  for (int i = 2; i < 100010; i++) {
    if (!primi[i]) {
      int k = 2;
      while (k * i < 100010) {
        primi[k * i] = true;
        k++;
      }
    }
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> seq[i];
  sort(seq, seq + n);
  for (int i = 0; i < n; i++) {
    int x = seq[i];
    vector<pair<unsigned long long int, int> > fatt = factorization(x);
    int bb = 0;
    for (int j = 0; j < int(fatt.size()); j++)
      bb = max(bb, peso[fatt[j].first]);
    migliore = max(migliore, bb + 1);
    for (int j = 0; j < int(fatt.size()); j++) peso[fatt[j].first] = bb + 1;
  }
  cout << migliore << endl;
}

### Prompt

Your task is to create 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;
vector<pair<unsigned long long int, int> > factorization(
    unsigned long long int n) {
  vector<pair<unsigned long long int, int> > res;
  unsigned long long int radice = sqrt(n);
  unsigned long long int i = 2;
  while (i <= radice and n > 1) {
    if (n % i == 0) {
      pair<unsigned long long int, int> nuovo = make_pair(i, 0);
      while (n % i == 0) {
        nuovo.second++;
        n /= i;
      }
      res.push_back(nuovo);
      radice = sqrt(n);
    }
    i++;
  }
  if (n > 1) {
    pair<unsigned long long int, int> nuovo = make_pair(n, 1);
    res.push_back(nuovo);
  }
  return res;
}
int n;
int seq[100010];
int migliore = 0;
bool primi[100010];
int peso[100010];
void primizza() {
  primi[2] = false;
  for (int i = 2; i < 100010; i++) {
    if (!primi[i]) {
      int k = 2;
      while (k * i < 100010) {
        primi[k * i] = true;
        k++;
      }
    }
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> seq[i];
  sort(seq, seq + n);
  for (int i = 0; i < n; i++) {
    int x = seq[i];
    vector<pair<unsigned long long int, int> > fatt = factorization(x);
    int bb = 0;
    for (int j = 0; j < int(fatt.size()); j++)
      bb = max(bb, peso[fatt[j].first]);
    migliore = max(migliore, bb + 1);
    for (int j = 0; j < int(fatt.size()); j++) peso[fatt[j].first] = bb + 1;
  }
  cout << migliore << endl;
}

```