#include <bits/stdc++.h>
using namespace std;
void solve() {
  int prime[10001] = {0}, list[10001], k = 0;
  for (long long i = 2; i < 10001; i++) {
    if (!prime[i]) {
      list[k++] = i;
      for (long long j = i * i; j < 10001; j += i) prime[j] = 1;
    }
  }
  int n;
  cin >> n;
  vector<int> v(10001, 0);
  for (int i = (0); i < (n); i++) {
    int n1;
    scanf("%d", &n1);
    int maximum = -1;
    vector<int> v1;
    int j = 0;
    while (j < k && n1 != 1) {
      if (n1 % list[j] == 0) {
        v1.push_back(list[j]);
        maximum = max(maximum, v[list[j]] + 1);
        while (n1 % list[j] == 0) n1 /= list[j];
      }
      j++;
    }
    for (int k = (0); k < (v1.size()); k++) v[v1[k]] = maximum;
  }
  int maximum = -1;
  for (int i = (0); i < (10001); i++) maximum = max(maximum, v[i]);
  if (n == 1)
    cout << 1 << "\n";
  else
    cout << maximum << "\n";
}
int main() {
  solve();
  return 0;
}

### Prompt

Your challenge is to write 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;
void solve() {
  int prime[10001] = {0}, list[10001], k = 0;
  for (long long i = 2; i < 10001; i++) {
    if (!prime[i]) {
      list[k++] = i;
      for (long long j = i * i; j < 10001; j += i) prime[j] = 1;
    }
  }
  int n;
  cin >> n;
  vector<int> v(10001, 0);
  for (int i = (0); i < (n); i++) {
    int n1;
    scanf("%d", &n1);
    int maximum = -1;
    vector<int> v1;
    int j = 0;
    while (j < k && n1 != 1) {
      if (n1 % list[j] == 0) {
        v1.push_back(list[j]);
        maximum = max(maximum, v[list[j]] + 1);
        while (n1 % list[j] == 0) n1 /= list[j];
      }
      j++;
    }
    for (int k = (0); k < (v1.size()); k++) v[v1[k]] = maximum;
  }
  int maximum = -1;
  for (int i = (0); i < (10001); i++) maximum = max(maximum, v[i]);
  if (n == 1)
    cout << 1 << "\n";
  else
    cout << maximum << "\n";
}
int main() {
  solve();
  return 0;
}

```