#include <bits/stdc++.h>
using namespace std;
int ans[100005];
vector<int> factorisation(int n) {
  vector<int> v;
  for (int i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      v.push_back(i);
      while (n % i == 0) n /= i;
    }
  }
  if (n > 1) {
    v.push_back(n);
  }
  return v;
}
int main() {
  int n;
  scanf("%d", &n);
  memset(ans, 0, sizeof(ans));
  int x;
  for (int i = 0; i <= (int)n - 1; ++i) {
    scanf("%d", &x);
    if (x == 1) {
      ans[1] = 1;
    }
    vector<int> v = factorisation(x);
    int mx = 1;
    for (int j = 0; j <= (int)(int)v.size() - 1; ++j) {
      mx = max(mx, ans[v[j]] + 1);
    }
    for (int j = 0; j <= (int)(int)v.size() - 1; ++j) {
      ans[v[j]] = mx;
    }
  }
  printf("%d", *max_element(ans, ans + x + 1));
}

### Prompt

Please formulate 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 ans[100005];
vector<int> factorisation(int n) {
  vector<int> v;
  for (int i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      v.push_back(i);
      while (n % i == 0) n /= i;
    }
  }
  if (n > 1) {
    v.push_back(n);
  }
  return v;
}
int main() {
  int n;
  scanf("%d", &n);
  memset(ans, 0, sizeof(ans));
  int x;
  for (int i = 0; i <= (int)n - 1; ++i) {
    scanf("%d", &x);
    if (x == 1) {
      ans[1] = 1;
    }
    vector<int> v = factorisation(x);
    int mx = 1;
    for (int j = 0; j <= (int)(int)v.size() - 1; ++j) {
      mx = max(mx, ans[v[j]] + 1);
    }
    for (int j = 0; j <= (int)(int)v.size() - 1; ++j) {
      ans[v[j]] = mx;
    }
  }
  printf("%d", *max_element(ans, ans + x + 1));
}

```