#include <bits/stdc++.h>
using namespace std;
vector<int> prim;
const int N = 1e5 + 9;
int n, dp[10000], a[N];
bool v[100000];
int pos[100000];
void pre() {
  for (int i = 2; i < 100000; i++)
    if (!v[i]) {
      prim.push_back(i);
      pos[i] = prim.size() - 1;
      int k = i << 1;
      while (k < 100000) v[k] = 1, k += i;
    }
}
int main() {
  pre();
  int ans = 1;
  cin >> n;
  for (int i = 1, now; i <= n; i++) {
    cin >> now;
    int org = now;
    int tmp = 0;
    for (int j = 0; j < 100; j++)
      if (now % prim[j] == 0) {
        while (now % prim[j] == 0) now /= prim[j];
        tmp = max(tmp, dp[j] + 1);
      }
    if (now > 1) tmp = max(tmp, dp[pos[now]] + 1);
    for (int j = 0; j < 100; j++)
      if (org % prim[j] == 0) {
        while (org % prim[j] == 0) org /= prim[j];
        dp[j] = tmp;
      }
    if (org > 1) dp[pos[now]] = tmp;
    ans = max(ans, tmp);
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Your task is to create 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;
vector<int> prim;
const int N = 1e5 + 9;
int n, dp[10000], a[N];
bool v[100000];
int pos[100000];
void pre() {
  for (int i = 2; i < 100000; i++)
    if (!v[i]) {
      prim.push_back(i);
      pos[i] = prim.size() - 1;
      int k = i << 1;
      while (k < 100000) v[k] = 1, k += i;
    }
}
int main() {
  pre();
  int ans = 1;
  cin >> n;
  for (int i = 1, now; i <= n; i++) {
    cin >> now;
    int org = now;
    int tmp = 0;
    for (int j = 0; j < 100; j++)
      if (now % prim[j] == 0) {
        while (now % prim[j] == 0) now /= prim[j];
        tmp = max(tmp, dp[j] + 1);
      }
    if (now > 1) tmp = max(tmp, dp[pos[now]] + 1);
    for (int j = 0; j < 100; j++)
      if (org % prim[j] == 0) {
        while (org % prim[j] == 0) org /= prim[j];
        dp[j] = tmp;
      }
    if (org > 1) dp[pos[now]] = tmp;
    ans = max(ans, tmp);
  }
  cout << ans << endl;
  return 0;
}

```