#include <bits/stdc++.h>
using namespace std;
int n, v[(int)10e5 + 1], dp[(int)10e5 + 1];
vector<int> fatores;
vector<int> sieve(int n) {
  vector<int> ans;
  bool prime[n + 1];
  memset(&prime, true, sizeof(prime));
  prime[0] = false;
  prime[1] = false;
  int m = (int)sqrt(n);
  for (int i = 2; i <= m; i++) {
    if (prime[i]) {
      for (int k = i * i; k <= n; k += i) {
        prime[k] = false;
      }
    }
  }
  for (int i = 0; i <= n; i++) {
    if (prime[i]) ans.push_back(i);
  }
  return ans;
}
int main(void) {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &v[i]);
  }
  fatores = sieve((int)10e5 + 1);
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    vector<int> primes;
    int j = 0;
    while (v[i] > 1) {
      if (v[i] % fatores[j] == 0) {
        primes.push_back(fatores[j]);
      }
      while (v[i] % fatores[j] == 0) {
        v[i] /= fatores[j];
      }
      j++;
    }
    int best = 0;
    for (int i = 0; i < primes.size(); i++) {
      best = max(best, dp[primes[i]]);
    }
    for (int i = 0; i < primes.size(); i++) {
      dp[primes[i]] = best + 1;
    }
    ans = max(best + 1, ans);
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 n, v[(int)10e5 + 1], dp[(int)10e5 + 1];
vector<int> fatores;
vector<int> sieve(int n) {
  vector<int> ans;
  bool prime[n + 1];
  memset(&prime, true, sizeof(prime));
  prime[0] = false;
  prime[1] = false;
  int m = (int)sqrt(n);
  for (int i = 2; i <= m; i++) {
    if (prime[i]) {
      for (int k = i * i; k <= n; k += i) {
        prime[k] = false;
      }
    }
  }
  for (int i = 0; i <= n; i++) {
    if (prime[i]) ans.push_back(i);
  }
  return ans;
}
int main(void) {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &v[i]);
  }
  fatores = sieve((int)10e5 + 1);
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    vector<int> primes;
    int j = 0;
    while (v[i] > 1) {
      if (v[i] % fatores[j] == 0) {
        primes.push_back(fatores[j]);
      }
      while (v[i] % fatores[j] == 0) {
        v[i] /= fatores[j];
      }
      j++;
    }
    int best = 0;
    for (int i = 0; i < primes.size(); i++) {
      best = max(best, dp[primes[i]]);
    }
    for (int i = 0; i < primes.size(); i++) {
      dp[primes[i]] = best + 1;
    }
    ans = max(best + 1, ans);
  }
  printf("%d\n", ans);
  return 0;
}

```