#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int pri[maxn], tot;
bool bk[maxn];
vector<int> v[maxn];
int num[maxn];
void pre() {
  for (int i = 2; i < maxn; i++) {
    if (!bk[i]) pri[tot++] = i;
    for (int j = 0; j < tot; j++) {
      if (i * pri[j] >= maxn) break;
      bk[i * pri[j]] = true;
      if (i % pri[j] == 0) break;
    }
  }
}
void get(int n) {
  int t = num[n];
  for (int i = 0; i < tot && pri[i] * pri[i] <= t; i++) {
    if (t % pri[i] == 0) {
      v[n].push_back(pri[i]);
      while (t % pri[i] == 0) t /= pri[i];
    }
  }
  if (t != 1) v[n].push_back(t);
}
int cnt[maxn];
int main() {
  int n;
  scanf("%d", &n);
  pre();
  for (int i = 1; i <= n; i++) {
    scanf("%d", &num[i]);
    get(i);
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    int mx = 0;
    for (int j = 0; j < v[i].size(); j++) {
      mx = max(mx, cnt[v[i][j]]);
    }
    for (int j = 0; j < v[i].size(); j++) {
      cnt[v[i][j]] = mx + 1;
    }
    ans = max(ans, mx + 1);
  }
  cout << ans << endl;
}

### Prompt

Develop a solution in Cpp 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 maxn = 1e5 + 10;
int pri[maxn], tot;
bool bk[maxn];
vector<int> v[maxn];
int num[maxn];
void pre() {
  for (int i = 2; i < maxn; i++) {
    if (!bk[i]) pri[tot++] = i;
    for (int j = 0; j < tot; j++) {
      if (i * pri[j] >= maxn) break;
      bk[i * pri[j]] = true;
      if (i % pri[j] == 0) break;
    }
  }
}
void get(int n) {
  int t = num[n];
  for (int i = 0; i < tot && pri[i] * pri[i] <= t; i++) {
    if (t % pri[i] == 0) {
      v[n].push_back(pri[i]);
      while (t % pri[i] == 0) t /= pri[i];
    }
  }
  if (t != 1) v[n].push_back(t);
}
int cnt[maxn];
int main() {
  int n;
  scanf("%d", &n);
  pre();
  for (int i = 1; i <= n; i++) {
    scanf("%d", &num[i]);
    get(i);
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    int mx = 0;
    for (int j = 0; j < v[i].size(); j++) {
      mx = max(mx, cnt[v[i][j]]);
    }
    for (int j = 0; j < v[i].size(); j++) {
      cnt[v[i][j]] = mx + 1;
    }
    ans = max(ans, mx + 1);
  }
  cout << ans << endl;
}

```