#include <bits/stdc++.h>
using namespace std;
void cmax(int& x, const int& y) { x = x > y ? x : y; }
void cmin(int& x, const int& y) { x = x < y ? x : y; }
const int N = 1e5 + 1;
int rev[N];
vector<int> g[N];
int dis[N];
int get(int x) {
  if (~dis[x]) return dis[x];
  dis[x] = 1;
  for (auto y : g[x]) cmax(dis[x], get(y) + 1);
  return dis[x];
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    rev[x] = i;
  }
  memset(dis, -1, sizeof(dis));
  for (int i = 2; i <= N - 1; i++) {
    int lst = 0;
    for (int j = i; j < N; j += i) {
      if (rev[j]) {
        if (lst) g[rev[j]].push_back(lst);
        lst = rev[j];
      }
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) cmax(ans, get(i));
  cout << ans << '\n';
  return 0;
}

### Prompt

Your task is to create 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;
void cmax(int& x, const int& y) { x = x > y ? x : y; }
void cmin(int& x, const int& y) { x = x < y ? x : y; }
const int N = 1e5 + 1;
int rev[N];
vector<int> g[N];
int dis[N];
int get(int x) {
  if (~dis[x]) return dis[x];
  dis[x] = 1;
  for (auto y : g[x]) cmax(dis[x], get(y) + 1);
  return dis[x];
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    rev[x] = i;
  }
  memset(dis, -1, sizeof(dis));
  for (int i = 2; i <= N - 1; i++) {
    int lst = 0;
    for (int j = i; j < N; j += i) {
      if (rev[j]) {
        if (lst) g[rev[j]].push_back(lst);
        lst = rev[j];
      }
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) cmax(ans, get(i));
  cout << ans << '\n';
  return 0;
}

```