#include <bits/stdc++.h>
using namespace std;
long long int last_occurence[100005], pr[100005], arr[100005], dp[100005];
vector<long long int> divi[100005];
void precompute() {
  long long int i, j;
  for (i = 2; i <= 100000; i++) {
    for (j = 2 * i; j <= 100000; j += i) {
      divi[j].push_back(i);
      pr[j] = 1;
    }
  }
  for (i = 2; i <= 100000; i++) {
    if (pr[i] == 0) divi[i].push_back(i);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long int n, i, j, ans = 1, k;
  cin >> n;
  precompute();
  for (i = 1; i <= n; i++) {
    cin >> arr[i];
  }
  for (i = 1; i <= n; i++) {
    j = arr[i];
    for (k = 0; k < divi[j].size(); k++) {
      dp[i] = max(dp[i], 1 + dp[last_occurence[divi[j][k]]]);
      last_occurence[divi[j][k]] = i;
      ans = max(ans, dp[i]);
    }
  }
  cout << ans;
  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;
long long int last_occurence[100005], pr[100005], arr[100005], dp[100005];
vector<long long int> divi[100005];
void precompute() {
  long long int i, j;
  for (i = 2; i <= 100000; i++) {
    for (j = 2 * i; j <= 100000; j += i) {
      divi[j].push_back(i);
      pr[j] = 1;
    }
  }
  for (i = 2; i <= 100000; i++) {
    if (pr[i] == 0) divi[i].push_back(i);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long int n, i, j, ans = 1, k;
  cin >> n;
  precompute();
  for (i = 1; i <= n; i++) {
    cin >> arr[i];
  }
  for (i = 1; i <= n; i++) {
    j = arr[i];
    for (k = 0; k < divi[j].size(); k++) {
      dp[i] = max(dp[i], 1 + dp[last_occurence[divi[j][k]]]);
      last_occurence[divi[j][k]] = i;
      ans = max(ans, dp[i]);
    }
  }
  cout << ans;
  return 0;
}

```