#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 5;
int dp[MAXN], n;
bool use[MAXN];
vector<int> p;
void init() {
  memset(use, false, sizeof(use));
  for (int i = 2; i < MAXN; i++)
    if (!use[i]) {
      p.push_back(i);
      for (int j = i + i; j < MAXN; j += i) use[j] = true;
    }
}
vector<int> cal(int n) {
  vector<int> d;
  for (int i = 0; p[i] * p[i] <= n; i++)
    if (n % p[i] == 0) {
      d.push_back(p[i]);
      while (n % p[i] == 0) n /= p[i];
    }
  if (n > 1) d.push_back(n);
  return d;
}
int main() {
  init();
  while (~scanf("%d", &n)) {
    memset(dp, 0, sizeof(dp));
    int ans = 0, x;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &x);
      vector<int> d = cal(x);
      int tmp = 0;
      for (unsigned j = 0; j < d.size(); j++) tmp = max(tmp, dp[d[j]]);
      for (unsigned j = 0; j < d.size(); j++) dp[d[j]] = tmp + 1;
      ans = max(ans, tmp + 1);
    }
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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;
const int MAXN = 1e5 + 5;
int dp[MAXN], n;
bool use[MAXN];
vector<int> p;
void init() {
  memset(use, false, sizeof(use));
  for (int i = 2; i < MAXN; i++)
    if (!use[i]) {
      p.push_back(i);
      for (int j = i + i; j < MAXN; j += i) use[j] = true;
    }
}
vector<int> cal(int n) {
  vector<int> d;
  for (int i = 0; p[i] * p[i] <= n; i++)
    if (n % p[i] == 0) {
      d.push_back(p[i]);
      while (n % p[i] == 0) n /= p[i];
    }
  if (n > 1) d.push_back(n);
  return d;
}
int main() {
  init();
  while (~scanf("%d", &n)) {
    memset(dp, 0, sizeof(dp));
    int ans = 0, x;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &x);
      vector<int> d = cal(x);
      int tmp = 0;
      for (unsigned j = 0; j < d.size(); j++) tmp = max(tmp, dp[d[j]]);
      for (unsigned j = 0; j < d.size(); j++) dp[d[j]] = tmp + 1;
      ans = max(ans, tmp + 1);
    }
    cout << ans << endl;
  }
  return 0;
}

```