#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100005;
vector<int> ld;
void sieve() {
  ld.resize(MAXN);
  ld[0] = 1;
  for (int i = 1; i < MAXN; i++) ld[i] = i;
  for (int i = 2; i * i < MAXN; i++)
    for (int j = i * i; j < MAXN; j += i)
      if (ld[j] == j) ld[j] = i;
}
vector<int> getlist(int n) {
  vector<int> ans;
  while (n > 1) {
    if (ans.size() == 0 || ans.back() != ld[n]) ans.push_back(ld[n]);
    n /= ld[n];
  }
  return ans;
}
vector<vector<int> > adj(MAXN);
vector<int> vec;
int memo[100005];
int solve(int idx) {
  int &ret = memo[idx];
  if (ret != -1) return ret;
  ret = 1;
  vector<int> list = getlist(vec[idx]);
  for (int k = 0; k < (int)list.size(); k++) {
    int up = lower_bound(adj[list[k]].begin(), adj[list[k]].end(), idx) -
             adj[list[k]].begin() - 1;
    if (up >= 0) {
      up = adj[list[k]][up];
      ret = max(ret, solve(up) + 1);
    }
  }
  return ret;
}
int main() {
  sieve();
  int n;
  cin >> n;
  vec.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> vec[i];
    vector<int> list = getlist(vec[i]);
    for (int k = 0; k < (int)list.size(); k++) {
      adj[list[k]].push_back(i);
    }
  }
  memset(memo, -1, sizeof memo);
  int ans = 0;
  for (int i = 0; i < n; i++) ans = max(ans, solve(i));
  cout << ans << endl;
}

### 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;
const int MAXN = 100005;
vector<int> ld;
void sieve() {
  ld.resize(MAXN);
  ld[0] = 1;
  for (int i = 1; i < MAXN; i++) ld[i] = i;
  for (int i = 2; i * i < MAXN; i++)
    for (int j = i * i; j < MAXN; j += i)
      if (ld[j] == j) ld[j] = i;
}
vector<int> getlist(int n) {
  vector<int> ans;
  while (n > 1) {
    if (ans.size() == 0 || ans.back() != ld[n]) ans.push_back(ld[n]);
    n /= ld[n];
  }
  return ans;
}
vector<vector<int> > adj(MAXN);
vector<int> vec;
int memo[100005];
int solve(int idx) {
  int &ret = memo[idx];
  if (ret != -1) return ret;
  ret = 1;
  vector<int> list = getlist(vec[idx]);
  for (int k = 0; k < (int)list.size(); k++) {
    int up = lower_bound(adj[list[k]].begin(), adj[list[k]].end(), idx) -
             adj[list[k]].begin() - 1;
    if (up >= 0) {
      up = adj[list[k]][up];
      ret = max(ret, solve(up) + 1);
    }
  }
  return ret;
}
int main() {
  sieve();
  int n;
  cin >> n;
  vec.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> vec[i];
    vector<int> list = getlist(vec[i]);
    for (int k = 0; k < (int)list.size(); k++) {
      adj[list[k]].push_back(i);
    }
  }
  memset(memo, -1, sizeof memo);
  int ans = 0;
  for (int i = 0; i < n; i++) ans = max(ans, solve(i));
  cout << ans << endl;
}

```