#include <bits/stdc++.h>
using namespace std;
const int OO = (int)1e9;
const double eps = 1e-10;
const int MX = 1e5 + 5;
int n, a[MX], mem[MX];
vector<vector<int> > PFs(MX), indices(MX);
int dp(int idx) {
  int& ret = mem[idx];
  if (ret != -1) return ret;
  int curr = a[idx];
  ret = 1;
  for (auto& PF : PFs[curr]) {
    int nxt = upper_bound(indices[PF].begin(), indices[PF].end(), idx) -
              indices[PF].begin();
    if (nxt != int(indices[PF].size()))
      ret = max(ret, 1 + dp(indices[PF][nxt]));
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cout.precision(10);
  for (int i = 2; i < int(MX); ++i) {
    int curr = i;
    for (int j = 2; j * j <= curr; ++j) {
      if (curr % j == 0) PFs[i].push_back(j);
      while (curr % j == 0) curr /= j;
    }
    if (curr > 1) PFs[i].push_back(curr);
  }
  cin >> n;
  for (int i = 0; i < int(n); ++i) {
    cin >> a[i];
    for (auto& PF : PFs[a[i]]) indices[PF].push_back(i);
  }
  int ans = 1;
  memset(mem, -1, sizeof(mem));
  for (int i = 2; i < int(MX); ++i)
    if (!indices[i].empty()) ans = max(ans, dp(indices[i][0]));
  cout << ans;
  return 0;
}

### Prompt

Create a solution in cpp for 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 OO = (int)1e9;
const double eps = 1e-10;
const int MX = 1e5 + 5;
int n, a[MX], mem[MX];
vector<vector<int> > PFs(MX), indices(MX);
int dp(int idx) {
  int& ret = mem[idx];
  if (ret != -1) return ret;
  int curr = a[idx];
  ret = 1;
  for (auto& PF : PFs[curr]) {
    int nxt = upper_bound(indices[PF].begin(), indices[PF].end(), idx) -
              indices[PF].begin();
    if (nxt != int(indices[PF].size()))
      ret = max(ret, 1 + dp(indices[PF][nxt]));
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cout.precision(10);
  for (int i = 2; i < int(MX); ++i) {
    int curr = i;
    for (int j = 2; j * j <= curr; ++j) {
      if (curr % j == 0) PFs[i].push_back(j);
      while (curr % j == 0) curr /= j;
    }
    if (curr > 1) PFs[i].push_back(curr);
  }
  cin >> n;
  for (int i = 0; i < int(n); ++i) {
    cin >> a[i];
    for (auto& PF : PFs[a[i]]) indices[PF].push_back(i);
  }
  int ans = 1;
  memset(mem, -1, sizeof(mem));
  for (int i = 2; i < int(MX); ++i)
    if (!indices[i].empty()) ans = max(ans, dp(indices[i][0]));
  cout << ans;
  return 0;
}

```