#include <bits/stdc++.h>
using namespace std;
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, (a % b));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> v(n);
  for (long long int i = 0; i < n; i++) cin >> v[i];
  vector<int> dp(100001, 0);
  for (long long int i = 0; i < n; i++) {
    vector<int> div;
    int mx = 0;
    for (long long int j = 2; j < sqrt(v[i]) + 1; j++) {
      if (v[i] % j == 0) {
        if (v[i] != j) {
          div.push_back(v[i] / j);
          div.push_back(j);
          mx = max(dp[j], mx);
          mx = max(dp[v[i] / j], mx);
        } else {
          div.push_back(j);
          mx = max(dp[j], mx);
        }
      }
    }
    if (int(div.size()) <= 1) dp[v[i]] = mx + 1;
    for (long long int j = 0; j < int(div.size()); j++) {
      dp[div[j]] = mx + 1;
    }
  }
  cout << *max_element(dp.begin(), dp.end()) << endl;
  return 0;
}

### Prompt

Generate 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;
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, (a % b));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> v(n);
  for (long long int i = 0; i < n; i++) cin >> v[i];
  vector<int> dp(100001, 0);
  for (long long int i = 0; i < n; i++) {
    vector<int> div;
    int mx = 0;
    for (long long int j = 2; j < sqrt(v[i]) + 1; j++) {
      if (v[i] % j == 0) {
        if (v[i] != j) {
          div.push_back(v[i] / j);
          div.push_back(j);
          mx = max(dp[j], mx);
          mx = max(dp[v[i] / j], mx);
        } else {
          div.push_back(j);
          mx = max(dp[j], mx);
        }
      }
    }
    if (int(div.size()) <= 1) dp[v[i]] = mx + 1;
    for (long long int j = 0; j < int(div.size()); j++) {
      dp[div[j]] = mx + 1;
    }
  }
  cout << *max_element(dp.begin(), dp.end()) << endl;
  return 0;
}

```