#include <bits/stdc++.h>
using namespace std;
vector<int> v;
void cal_div(int n) {
  v.resize(0);
  int root = sqrt(n);
  for (int i = 2; i <= root; i++) {
    if (n % i == 0) {
      v.push_back(i);
      if (n / i != i) {
        v.push_back(n / i);
      }
    }
  }
  v.push_back(n);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int t = 1;
  while (t--) {
    int n;
    cin >> n;
    long long int a[n + 3];
    for (int i = 0; i < n; i++) cin >> a[i];
    long long int d[100005];
    long long int dp[100005];
    memset(d, 0, sizeof d);
    memset(dp, 0, sizeof dp);
    long long int ans = INT_MIN;
    for (int i = 0; i < n; i++) {
      cal_div(a[i]);
      for (int j = 0; j < v.size(); j++) {
        dp[i] = max(dp[i], 1 + d[v[j]]);
      }
      for (int j = 0; j < v.size(); j++) {
        d[v[j]] = dp[i];
      }
      ans = max(ans, dp[i]);
    }
    cout << ans << "\n";
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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<int> v;
void cal_div(int n) {
  v.resize(0);
  int root = sqrt(n);
  for (int i = 2; i <= root; i++) {
    if (n % i == 0) {
      v.push_back(i);
      if (n / i != i) {
        v.push_back(n / i);
      }
    }
  }
  v.push_back(n);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int t = 1;
  while (t--) {
    int n;
    cin >> n;
    long long int a[n + 3];
    for (int i = 0; i < n; i++) cin >> a[i];
    long long int d[100005];
    long long int dp[100005];
    memset(d, 0, sizeof d);
    memset(dp, 0, sizeof dp);
    long long int ans = INT_MIN;
    for (int i = 0; i < n; i++) {
      cal_div(a[i]);
      for (int j = 0; j < v.size(); j++) {
        dp[i] = max(dp[i], 1 + d[v[j]]);
      }
      for (int j = 0; j < v.size(); j++) {
        d[v[j]] = dp[i];
      }
      ans = max(ans, dp[i]);
    }
    cout << ans << "\n";
  }
  return 0;
}

```