#include <bits/stdc++.h>
using namespace std;
const int INF = 1 << 29;
const double EPS = 1e-9;
const int dx[] = {1, 0, -1, 0}, dy[] = {0, -1, 0, 1};
int n;
int dp[100010];
int main() {
  cin >> n;
  memset(dp, 0, sizeof(dp));
  for (int i = 0; i < n; i++) {
    int a, aa;
    cin >> a;
    aa = a;
    int mx = 0;
    for (int i = 2; i * i <= a; i++) {
      if (a % i) continue;
      mx = max(mx, dp[i]);
      while (a % i == 0) a /= i;
    }
    mx = max(mx, dp[a]);
    for (int i = 2; i * i <= aa; i++) {
      if (aa % i) continue;
      dp[i] = mx + 1;
      while (aa % i == 0) aa /= i;
    }
    if (aa != 1) dp[aa] = mx + 1;
  }
  int res = 1;
  for (int i = 2; i <= 100000; i++) {
    res = max(res, dp[i]);
  }
  cout << res << endl;
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
const int INF = 1 << 29;
const double EPS = 1e-9;
const int dx[] = {1, 0, -1, 0}, dy[] = {0, -1, 0, 1};
int n;
int dp[100010];
int main() {
  cin >> n;
  memset(dp, 0, sizeof(dp));
  for (int i = 0; i < n; i++) {
    int a, aa;
    cin >> a;
    aa = a;
    int mx = 0;
    for (int i = 2; i * i <= a; i++) {
      if (a % i) continue;
      mx = max(mx, dp[i]);
      while (a % i == 0) a /= i;
    }
    mx = max(mx, dp[a]);
    for (int i = 2; i * i <= aa; i++) {
      if (aa % i) continue;
      dp[i] = mx + 1;
      while (aa % i == 0) aa /= i;
    }
    if (aa != 1) dp[aa] = mx + 1;
  }
  int res = 1;
  for (int i = 2; i <= 100000; i++) {
    res = max(res, dp[i]);
  }
  cout << res << endl;
  return 0;
}

```