#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
int x, memo[100001];
vector<int> ans[100001];
vector<int> factors(int a) {
  vector<int> pf;
  for (int i = 2; i <= sqrt(a); i += (1 + (i & 1))) {
    if (a % i == 0) {
      pf.push_back(i);
      while (a % i == 0) a /= i;
    }
  }
  if (a > 1) pf.push_back(a);
  return pf;
}
int solve(int a) {
  int &cur = memo[a];
  if (~cur) return cur;
  cur = 1;
  vector<int> pf = factors(a);
  for (int i = 0; i < pf.size(); i++) {
    int r = upper_bound(ans[pf[i]].begin(), ans[pf[i]].end(), a) -
            ans[pf[i]].begin();
    if (r < ans[pf[i]].size()) cur = max(cur, 1 + solve(ans[pf[i]][r]));
  }
  return cur;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  memset(memo, -1, sizeof(memo));
  cin >> x;
  vector<int> vec(x);
  for (int i = 0; i < x; i++) {
    cin >> vec[i];
    vector<int> pf = factors(vec[i]);
    for (int c = 0; c < pf.size(); c++) ans[pf[c]].push_back(vec[i]);
  }
  int a = 0;
  for (int i = 0; i < x; i++) a = max(a, solve(vec[i]));
  cout << a;
}

### Prompt

Please create a solution in CPP 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 mod = 1000000007;
int x, memo[100001];
vector<int> ans[100001];
vector<int> factors(int a) {
  vector<int> pf;
  for (int i = 2; i <= sqrt(a); i += (1 + (i & 1))) {
    if (a % i == 0) {
      pf.push_back(i);
      while (a % i == 0) a /= i;
    }
  }
  if (a > 1) pf.push_back(a);
  return pf;
}
int solve(int a) {
  int &cur = memo[a];
  if (~cur) return cur;
  cur = 1;
  vector<int> pf = factors(a);
  for (int i = 0; i < pf.size(); i++) {
    int r = upper_bound(ans[pf[i]].begin(), ans[pf[i]].end(), a) -
            ans[pf[i]].begin();
    if (r < ans[pf[i]].size()) cur = max(cur, 1 + solve(ans[pf[i]][r]));
  }
  return cur;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  memset(memo, -1, sizeof(memo));
  cin >> x;
  vector<int> vec(x);
  for (int i = 0; i < x; i++) {
    cin >> vec[i];
    vector<int> pf = factors(vec[i]);
    for (int c = 0; c < pf.size(); c++) ans[pf[c]].push_back(vec[i]);
  }
  int a = 0;
  for (int i = 0; i < x; i++) a = max(a, solve(vec[i]));
  cout << a;
}

```