#include <bits/stdc++.h>
using namespace std;
const long long int maxN = 1e5 + 10;
vector<long long int> primes, pf, maxAns;
bool isPrime[maxN];
void init() {
  for (long long int i = 0; i < maxN; i++) isPrime[i] = true;
  isPrime[1] = false;
  for (long long int i = 2; i < maxN; i++) {
    if (isPrime[i]) primes.push_back(i);
    for (long long int p : primes) {
      if (p * i >= maxN) break;
      isPrime[p * i] = false;
      if (!(i % p)) break;
    }
  }
}
void getPF(long long int x) {
  pf.clear();
  for (long long int p : primes) {
    if (isPrime[x]) {
      pf.push_back(x);
      break;
    }
    if (x == 1) break;
    if (!(x % p)) {
      pf.push_back(p);
      while (!(x % p)) x /= p;
    }
  }
}
long long int N, x;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  init();
  cin >> N;
  maxAns.clear();
  maxAns.resize(maxN);
  fill(maxAns.begin(), maxAns.end(), 0);
  long long int ans = 1;
  for (long long int i = 0; i < N; i++) {
    cin >> x;
    if (x == 1) continue;
    pf.clear();
    getPF(x);
    long long int current = 1;
    for (long long int p : pf) {
      current = max(current, maxAns[p] + 1);
      ans = max(ans, maxAns[p] + 1);
    }
    for (long long int p : pf) {
      maxAns[p] = current;
    }
    ans = max(ans, current);
  }
  cout << ans << endl;
}

### 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 long long int maxN = 1e5 + 10;
vector<long long int> primes, pf, maxAns;
bool isPrime[maxN];
void init() {
  for (long long int i = 0; i < maxN; i++) isPrime[i] = true;
  isPrime[1] = false;
  for (long long int i = 2; i < maxN; i++) {
    if (isPrime[i]) primes.push_back(i);
    for (long long int p : primes) {
      if (p * i >= maxN) break;
      isPrime[p * i] = false;
      if (!(i % p)) break;
    }
  }
}
void getPF(long long int x) {
  pf.clear();
  for (long long int p : primes) {
    if (isPrime[x]) {
      pf.push_back(x);
      break;
    }
    if (x == 1) break;
    if (!(x % p)) {
      pf.push_back(p);
      while (!(x % p)) x /= p;
    }
  }
}
long long int N, x;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  init();
  cin >> N;
  maxAns.clear();
  maxAns.resize(maxN);
  fill(maxAns.begin(), maxAns.end(), 0);
  long long int ans = 1;
  for (long long int i = 0; i < N; i++) {
    cin >> x;
    if (x == 1) continue;
    pf.clear();
    getPF(x);
    long long int current = 1;
    for (long long int p : pf) {
      current = max(current, maxAns[p] + 1);
      ans = max(ans, maxAns[p] + 1);
    }
    for (long long int p : pf) {
      maxAns[p] = current;
    }
    ans = max(ans, current);
  }
  cout << ans << endl;
}

```