#include <bits/stdc++.h>
using namespace std;
int isp[100010], p[100010];
int x[100010];
vector<int> d[100010];
int last[10010];
int sieve(int n) {
  fill(isp, isp + n + 1, 1);
  isp[0] = isp[1] = 0;
  for (int i = 0; i < (n); i++) x[i] = i;
  int c = 0;
  for (int i = 2; i < (n + 1); i++)
    if (isp[i]) {
      isp[i] = 1;
      p[c] = i;
      d[i].push_back(c);
      for (int j = i * 2; j <= n; j += i) {
        isp[j] = 0;
        d[j].push_back(c);
      }
      c++;
    }
  return c;
}
int n, a[100010];
int dp[100010];
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < (n); i++) cin >> a[i];
  int P = sieve(100000);
  int ans = 0;
  for (int i = 0; i < (n); i++) {
    dp[i] = 1;
    for (int j = 0; j < (((int)d[a[i]].size())); j++) {
      dp[i] = max(dp[i], 1 + last[d[a[i]][j]]);
    }
    ans = max(dp[i], ans);
    for (int j = 0; j < (((int)d[a[i]].size())); j++) {
      last[d[a[i]][j]] = max(last[d[a[i]][j]], dp[i]);
    }
  }
  cout << ans << endl;
}

### Prompt

Generate 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 isp[100010], p[100010];
int x[100010];
vector<int> d[100010];
int last[10010];
int sieve(int n) {
  fill(isp, isp + n + 1, 1);
  isp[0] = isp[1] = 0;
  for (int i = 0; i < (n); i++) x[i] = i;
  int c = 0;
  for (int i = 2; i < (n + 1); i++)
    if (isp[i]) {
      isp[i] = 1;
      p[c] = i;
      d[i].push_back(c);
      for (int j = i * 2; j <= n; j += i) {
        isp[j] = 0;
        d[j].push_back(c);
      }
      c++;
    }
  return c;
}
int n, a[100010];
int dp[100010];
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < (n); i++) cin >> a[i];
  int P = sieve(100000);
  int ans = 0;
  for (int i = 0; i < (n); i++) {
    dp[i] = 1;
    for (int j = 0; j < (((int)d[a[i]].size())); j++) {
      dp[i] = max(dp[i], 1 + last[d[a[i]][j]]);
    }
    ans = max(dp[i], ans);
    for (int j = 0; j < (((int)d[a[i]].size())); j++) {
      last[d[a[i]][j]] = max(last[d[a[i]][j]], dp[i]);
    }
  }
  cout << ans << endl;
}

```