#include <bits/stdc++.h>
using namespace std;
int max_len[100000 + 1];
int min_p[100000 + 1];
vector<int> prime;
bitset<100000 + 1> is_prime;
void make_prime() {
  is_prime.set();
  is_prime[0] = false;
  is_prime[1] = false;
  int sq = sqrt(100000);
  for (int p = 2; p <= 100000; p++) {
    if (!is_prime[p]) continue;
    prime.push_back(p);
    if (p > sq) continue;
    for (int j = p * p; j <= 100000; j += p) {
      if (!is_prime[j]) continue;
      is_prime[j] = false;
      min_p[j] = p;
    }
  }
}
int n;
vector<int> a;
void input() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    a.push_back(x);
  }
}
vector<int> factor(int n) {
  if (n == 1) {
    vector<int> empty;
    return empty;
  }
  if (is_prime[n]) {
    vector<int> v;
    v.push_back(n);
    return v;
  }
  int p = min_p[n];
  vector<int> v = factor(n / p);
  if (v[v.size() - 1] != p) v.push_back(p);
  return v;
}
int main() {
  make_prime();
  input();
  int ans = 0;
  for (int i = 0; i < n; i++) {
    vector<int> v = factor(a[i]);
    int s = v.size();
    int len = 0;
    for (int j = 0; j < s; j++) {
      int p = v[j];
      if (max_len[p] > len) len = max_len[p];
    }
    len++;
    for (int j = 0; j < s; j++) {
      int p = v[j];
      max_len[p] = len;
    }
    if (ans < len) ans = len;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to 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;
int max_len[100000 + 1];
int min_p[100000 + 1];
vector<int> prime;
bitset<100000 + 1> is_prime;
void make_prime() {
  is_prime.set();
  is_prime[0] = false;
  is_prime[1] = false;
  int sq = sqrt(100000);
  for (int p = 2; p <= 100000; p++) {
    if (!is_prime[p]) continue;
    prime.push_back(p);
    if (p > sq) continue;
    for (int j = p * p; j <= 100000; j += p) {
      if (!is_prime[j]) continue;
      is_prime[j] = false;
      min_p[j] = p;
    }
  }
}
int n;
vector<int> a;
void input() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    a.push_back(x);
  }
}
vector<int> factor(int n) {
  if (n == 1) {
    vector<int> empty;
    return empty;
  }
  if (is_prime[n]) {
    vector<int> v;
    v.push_back(n);
    return v;
  }
  int p = min_p[n];
  vector<int> v = factor(n / p);
  if (v[v.size() - 1] != p) v.push_back(p);
  return v;
}
int main() {
  make_prime();
  input();
  int ans = 0;
  for (int i = 0; i < n; i++) {
    vector<int> v = factor(a[i]);
    int s = v.size();
    int len = 0;
    for (int j = 0; j < s; j++) {
      int p = v[j];
      if (max_len[p] > len) len = max_len[p];
    }
    len++;
    for (int j = 0; j < s; j++) {
      int p = v[j];
      max_len[p] = len;
    }
    if (ans < len) ans = len;
  }
  cout << ans << endl;
  return 0;
}

```