#include <bits/stdc++.h>
using namespace std;
int a[100010];
int dp[100010];
int n;
bool vis[1001];
int p[1001], pn = 0;
int addr[100010];
int f[11], fn;
void fen(int x) {
  fn = 0;
  for (int i = 0; i < pn; i++)
    if (x % p[i] == 0) {
      f[fn++] = p[i];
      while (x % p[i] == 0) x /= p[i];
    }
  if (x != 1) f[fn++] = x;
  return;
}
void update_addr(int pos) {
  fen(a[pos]);
  for (int i = 0; i < fn; i++)
    if (addr[f[i]] == -1 || dp[pos] > dp[addr[f[i]]]) addr[f[i]] = pos;
}
int get_val(int pos) {
  fen(a[pos]);
  int ret = 1;
  for (int i = 0; i < fn; i++)
    if (addr[f[i]] != -1) ret = max(ret, dp[addr[f[i]]] + 1);
  return ret;
}
int main() {
  for (int i = 2; i <= 1000; i++)
    if (!vis[i]) {
      for (int j = i + i; j <= 1000; j += i) vis[j] = 1;
    }
  for (int i = 2; i <= 1000; i++)
    if (!vis[i]) p[pn++] = i;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  memset(addr, -1, sizeof(addr));
  dp[0] = 1;
  int ans = 1;
  for (int i = 1; i < n; i++) {
    update_addr(i - 1);
    dp[i] = get_val(i);
    ans = max(ans, dp[i]);
  }
  cout << ans << endl;
  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;
int a[100010];
int dp[100010];
int n;
bool vis[1001];
int p[1001], pn = 0;
int addr[100010];
int f[11], fn;
void fen(int x) {
  fn = 0;
  for (int i = 0; i < pn; i++)
    if (x % p[i] == 0) {
      f[fn++] = p[i];
      while (x % p[i] == 0) x /= p[i];
    }
  if (x != 1) f[fn++] = x;
  return;
}
void update_addr(int pos) {
  fen(a[pos]);
  for (int i = 0; i < fn; i++)
    if (addr[f[i]] == -1 || dp[pos] > dp[addr[f[i]]]) addr[f[i]] = pos;
}
int get_val(int pos) {
  fen(a[pos]);
  int ret = 1;
  for (int i = 0; i < fn; i++)
    if (addr[f[i]] != -1) ret = max(ret, dp[addr[f[i]]] + 1);
  return ret;
}
int main() {
  for (int i = 2; i <= 1000; i++)
    if (!vis[i]) {
      for (int j = i + i; j <= 1000; j += i) vis[j] = 1;
    }
  for (int i = 2; i <= 1000; i++)
    if (!vis[i]) p[pn++] = i;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  memset(addr, -1, sizeof(addr));
  dp[0] = 1;
  int ans = 1;
  for (int i = 1; i < n; i++) {
    update_addr(i - 1);
    dp[i] = get_val(i);
    ans = max(ans, dp[i]);
  }
  cout << ans << endl;
  return 0;
}

```