#include <bits/stdc++.h>
using namespace std;
int n, m, i, j, k, l, o, p, a[100010], x[100010], b[100010], mx, mm, u;
vector<int> v;
int main() {
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", &a[i]);
  u = a[n - 1];
  for (i = 2; i <= 100000; i++) {
    if (b[i] == 0)
      for (j = i + i; j <= 100000; j += i)
        if (b[j] == 0) b[j] = i;
  }
  for (i = 0; i < n; i++) {
    mx = -1;
    p = -1;
    while (b[a[i]] != 0) {
      v.push_back(b[a[i]]);
      if (x[b[a[i]]] > mx) mx = x[b[a[i]]];
      a[i] = a[i] / b[a[i]];
    }
    v.push_back(a[i]);
    if (x[a[i]] > mx) mx = x[a[i]];
    while (!v.empty()) {
      x[v.back()] = mx + 1;
      v.pop_back();
    }
  }
  for (i = 0; i <= a[n - 1]; i++) mm = max(mm, x[i]);
  cout << mm;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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, m, i, j, k, l, o, p, a[100010], x[100010], b[100010], mx, mm, u;
vector<int> v;
int main() {
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", &a[i]);
  u = a[n - 1];
  for (i = 2; i <= 100000; i++) {
    if (b[i] == 0)
      for (j = i + i; j <= 100000; j += i)
        if (b[j] == 0) b[j] = i;
  }
  for (i = 0; i < n; i++) {
    mx = -1;
    p = -1;
    while (b[a[i]] != 0) {
      v.push_back(b[a[i]]);
      if (x[b[a[i]]] > mx) mx = x[b[a[i]]];
      a[i] = a[i] / b[a[i]];
    }
    v.push_back(a[i]);
    if (x[a[i]] > mx) mx = x[a[i]];
    while (!v.empty()) {
      x[v.back()] = mx + 1;
      v.pop_back();
    }
  }
  for (i = 0; i <= a[n - 1]; i++) mm = max(mm, x[i]);
  cout << mm;
}

```