#include <bits/stdc++.h>
using namespace std;
int dx[] = {0, 1, 0, -1, 1, 1, -1, -1};
int dy[] = {1, 0, -1, 0, 1, -1, 1, -1};
long long fix_mod(long long x, long long y) { return (y + x % y) % y; }
void fast() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
int idx[(int)1e5 + 10];
vector<vector<int>> edg;
int mem[(int)1e5 + 10];
int solve(int i) {
  int& ret = mem[i];
  if (~ret) return ret;
  ret = 0;
  for (auto v : edg[i]) {
    ret = max(ret, solve(v) + 1);
  }
  return ret;
}
int main() {
  fast();
  int n;
  cin >> n;
  vector<int> v(n);
  edg = vector<vector<int>>(n);
  memset(idx, -1, sizeof idx);
  memset(mem, -1, sizeof mem);
  for (int i = 0; i < n; i++) cin >> v[i], idx[v[i]] = i;
  for (int i = 2; i <= 1e5; i++) {
    int p = -1;
    for (int j = i; j <= 1e5; j += i) {
      if (idx[j] != -1) {
        if (p != -1) edg[p].push_back(idx[j]);
        p = idx[j];
      }
    }
  }
  int mx = 0;
  for (int i = 0; i < n; i++) mx = max(mx, solve(i) + 1);
  return cout << mx << "\n", 0;
  ;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 dx[] = {0, 1, 0, -1, 1, 1, -1, -1};
int dy[] = {1, 0, -1, 0, 1, -1, 1, -1};
long long fix_mod(long long x, long long y) { return (y + x % y) % y; }
void fast() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
int idx[(int)1e5 + 10];
vector<vector<int>> edg;
int mem[(int)1e5 + 10];
int solve(int i) {
  int& ret = mem[i];
  if (~ret) return ret;
  ret = 0;
  for (auto v : edg[i]) {
    ret = max(ret, solve(v) + 1);
  }
  return ret;
}
int main() {
  fast();
  int n;
  cin >> n;
  vector<int> v(n);
  edg = vector<vector<int>>(n);
  memset(idx, -1, sizeof idx);
  memset(mem, -1, sizeof mem);
  for (int i = 0; i < n; i++) cin >> v[i], idx[v[i]] = i;
  for (int i = 2; i <= 1e5; i++) {
    int p = -1;
    for (int j = i; j <= 1e5; j += i) {
      if (idx[j] != -1) {
        if (p != -1) edg[p].push_back(idx[j]);
        p = idx[j];
      }
    }
  }
  int mx = 0;
  for (int i = 0; i < n; i++) mx = max(mx, solve(i) + 1);
  return cout << mx << "\n", 0;
  ;
}

```