#include <bits/stdc++.h>
using namespace std;
long long dp[100005] = {0};
vector<long long> getPrimes(long long n) {
  vector<long long> vect;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      vect.push_back(i);
    }
    while (n % i == 0) n /= i;
  }
  if (n >= 2) vect.push_back(n);
  return vect;
}
signed main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long n;
  cin >> n;
  long long arr[n];
  for (long long i = 0; i < n; i++) cin >> arr[i];
  for (long long i = 0; i < n; i++) {
    long long vv = arr[i];
    vector<long long> v = getPrimes(vv);
    long long ans = -1;
    for (long long j = 0; j < v.size(); j++) {
      long long p = v[j];
      ans = max(ans, dp[p] + 1);
    }
    for (long long j = 0; j < v.size(); j++) {
      long long p = v[j];
      dp[p] = ans;
    }
  }
  long long res = 1;
  for (long long i = 0; i < 100002; i++) {
    res = max(res, dp[i]);
  }
  cout << res;
  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;
long long dp[100005] = {0};
vector<long long> getPrimes(long long n) {
  vector<long long> vect;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      vect.push_back(i);
    }
    while (n % i == 0) n /= i;
  }
  if (n >= 2) vect.push_back(n);
  return vect;
}
signed main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long n;
  cin >> n;
  long long arr[n];
  for (long long i = 0; i < n; i++) cin >> arr[i];
  for (long long i = 0; i < n; i++) {
    long long vv = arr[i];
    vector<long long> v = getPrimes(vv);
    long long ans = -1;
    for (long long j = 0; j < v.size(); j++) {
      long long p = v[j];
      ans = max(ans, dp[p] + 1);
    }
    for (long long j = 0; j < v.size(); j++) {
      long long p = v[j];
      dp[p] = ans;
    }
  }
  long long res = 1;
  for (long long i = 0; i < 100002; i++) {
    res = max(res, dp[i]);
  }
  cout << res;
  return 0;
}

```