#include <bits/stdc++.h>
using namespace std;
long long int spf[100010];
void seive() {
  spf[1] = 1;
  for (long long int i = 2; i < 100010; i += 1) {
    spf[i] = i;
  }
  for (long long int i = 2; i < 100010; i += 2) {
    spf[i] = 2;
  }
  for (long long int i = 3; i * i < 100010; i += 1) {
    if (spf[i] == i) {
      for (long long int j = i * i; j < 100010; j += i) {
        if (spf[j] == j) spf[j] = i;
      }
    }
  }
}
set<long long int> get_fact(long long int x) {
  set<long long int> an;
  while (x != 1) {
    an.insert(spf[x]);
    x /= spf[x];
  }
  return an;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  seive();
  long long int n;
  cin >> n;
  long long int A[n];
  for (long long int i = 0; i < n; i += 1) {
    cin >> A[i];
  }
  if (n == 1) {
    cout << "1";
    return 0;
  }
  set<long long int> fac;
  map<long long int, long long int> cnt;
  for (long long int i = n - 1; i >= 0; i--) {
    if (A[i] == 1) continue;
    fac.clear();
    long long int ans = std::numeric_limits<long long int>::min();
    fac = get_fact(A[i]);
    for (set<long long int>::iterator itr = fac.begin(); itr != fac.end();
         itr++)
      ans = max(ans, cnt[(*itr)]);
    ans++;
    for (set<long long int>::iterator itr = fac.begin(); itr != fac.end();
         itr++)
      cnt[(*itr)] = ans;
  }
  long long int acc = std::numeric_limits<long long int>::min();
  for (map<long long int, long long int>::iterator itr = cnt.begin();
       itr != cnt.end(); itr++)
    acc = max(acc, itr->second);
  cout << acc;
  return 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;
long long int spf[100010];
void seive() {
  spf[1] = 1;
  for (long long int i = 2; i < 100010; i += 1) {
    spf[i] = i;
  }
  for (long long int i = 2; i < 100010; i += 2) {
    spf[i] = 2;
  }
  for (long long int i = 3; i * i < 100010; i += 1) {
    if (spf[i] == i) {
      for (long long int j = i * i; j < 100010; j += i) {
        if (spf[j] == j) spf[j] = i;
      }
    }
  }
}
set<long long int> get_fact(long long int x) {
  set<long long int> an;
  while (x != 1) {
    an.insert(spf[x]);
    x /= spf[x];
  }
  return an;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  seive();
  long long int n;
  cin >> n;
  long long int A[n];
  for (long long int i = 0; i < n; i += 1) {
    cin >> A[i];
  }
  if (n == 1) {
    cout << "1";
    return 0;
  }
  set<long long int> fac;
  map<long long int, long long int> cnt;
  for (long long int i = n - 1; i >= 0; i--) {
    if (A[i] == 1) continue;
    fac.clear();
    long long int ans = std::numeric_limits<long long int>::min();
    fac = get_fact(A[i]);
    for (set<long long int>::iterator itr = fac.begin(); itr != fac.end();
         itr++)
      ans = max(ans, cnt[(*itr)]);
    ans++;
    for (set<long long int>::iterator itr = fac.begin(); itr != fac.end();
         itr++)
      cnt[(*itr)] = ans;
  }
  long long int acc = std::numeric_limits<long long int>::min();
  for (map<long long int, long long int>::iterator itr = cnt.begin();
       itr != cnt.end(); itr++)
    acc = max(acc, itr->second);
  cout << acc;
  return 0;
}

```