#include <bits/stdc++.h>
using namespace std;
const int N = 112345;
int n, ans, dp[N];
int main() {
  scanf("%d", &n);
  for (int i = 0, x; i < n; ++i) {
    scanf("%d", &x);
    int curr = 0;
    vector<int> v;
    for (int p = 2; p * p <= x; ++p) {
      if (x % p != 0) continue;
      v.push_back(p);
      curr = max(curr, dp[p]);
      while (x % p == 0) x /= p;
    }
    if (x > 1) {
      v.push_back(x);
      curr = max(curr, dp[x]);
    }
    for (int p : v) dp[p] = curr + 1;
    ans = max(ans, curr + 1);
  }
  printf("%d\n", ans);
  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 N = 112345;
int n, ans, dp[N];
int main() {
  scanf("%d", &n);
  for (int i = 0, x; i < n; ++i) {
    scanf("%d", &x);
    int curr = 0;
    vector<int> v;
    for (int p = 2; p * p <= x; ++p) {
      if (x % p != 0) continue;
      v.push_back(p);
      curr = max(curr, dp[p]);
      while (x % p == 0) x /= p;
    }
    if (x > 1) {
      v.push_back(x);
      curr = max(curr, dp[x]);
    }
    for (int p : v) dp[p] = curr + 1;
    ans = max(ans, curr + 1);
  }
  printf("%d\n", ans);
  return 0;
}

```