#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 100;
bool mark[N];
int arr[N], maxSeq[N];
vector<int> primeDiv[N], prime;
int n, ans;
void sieve() {
  mark[1] = true;
  for (int i = 2; i <= N - 100; i++) {
    if (!mark[i]) {
      prime.push_back(i);
      primeDiv[i].push_back(i);
      for (int j = 2 * i; j <= N - 100; j += i) {
        mark[j] = true;
        primeDiv[j].push_back(i);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  prime.push_back(1);
  sieve();
  cin >> n;
  for (int i = 0; i < n; i++) cin >> arr[i];
  for (int i = 0; i < n; i++) {
    if (arr[i] == 1) {
      maxSeq[1] = 1;
      continue;
    }
    int a = -1;
    for (auto x : primeDiv[arr[i]])
      if (maxSeq[x] > a) a = maxSeq[x];
    for (auto x : primeDiv[arr[i]]) maxSeq[x] = a + 1;
  }
  for (auto x : prime) ans = max(ans, maxSeq[x]);
  return cout << ans, 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 N = 1e5 + 100;
bool mark[N];
int arr[N], maxSeq[N];
vector<int> primeDiv[N], prime;
int n, ans;
void sieve() {
  mark[1] = true;
  for (int i = 2; i <= N - 100; i++) {
    if (!mark[i]) {
      prime.push_back(i);
      primeDiv[i].push_back(i);
      for (int j = 2 * i; j <= N - 100; j += i) {
        mark[j] = true;
        primeDiv[j].push_back(i);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  prime.push_back(1);
  sieve();
  cin >> n;
  for (int i = 0; i < n; i++) cin >> arr[i];
  for (int i = 0; i < n; i++) {
    if (arr[i] == 1) {
      maxSeq[1] = 1;
      continue;
    }
    int a = -1;
    for (auto x : primeDiv[arr[i]])
      if (maxSeq[x] > a) a = maxSeq[x];
    for (auto x : primeDiv[arr[i]]) maxSeq[x] = a + 1;
  }
  for (auto x : prime) ans = max(ans, maxSeq[x]);
  return cout << ans, 0;
}

```