#include <bits/stdc++.h>
using namespace std;
vector<int> factorize(int x) {
  vector<int> ret;
  for (int i = 1; i <= sqrt(x); i++) {
    if (x % i == 0) {
      if (x / i == i) {
        ret.push_back(i);
      } else {
        ret.push_back(i);
        ret.push_back(x / i);
      }
    }
  }
  return ret;
}
int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (auto& x : a) cin >> x;
  unordered_map<int, int> last;
  vector<int> dp(n);
  for (int i = 0; i < n; i++) {
    int x = a[i];
    vector<int> factors = factorize(x);
    dp[i] = 0;
    for (int factor : factors) {
      if (factor != 1) {
        if (last.find(factor) != last.end())
          dp[i] = max(dp[i], dp[last[factor]]);
        last[factor] = i;
      }
    }
    dp[i]++;
  }
  cout << *max_element(dp.begin(), dp.end()) << endl;
  return 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;
vector<int> factorize(int x) {
  vector<int> ret;
  for (int i = 1; i <= sqrt(x); i++) {
    if (x % i == 0) {
      if (x / i == i) {
        ret.push_back(i);
      } else {
        ret.push_back(i);
        ret.push_back(x / i);
      }
    }
  }
  return ret;
}
int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (auto& x : a) cin >> x;
  unordered_map<int, int> last;
  vector<int> dp(n);
  for (int i = 0; i < n; i++) {
    int x = a[i];
    vector<int> factors = factorize(x);
    dp[i] = 0;
    for (int factor : factors) {
      if (factor != 1) {
        if (last.find(factor) != last.end())
          dp[i] = max(dp[i], dp[last[factor]]);
        last[factor] = i;
      }
    }
    dp[i]++;
  }
  cout << *max_element(dp.begin(), dp.end()) << endl;
  return 0;
}

```