#include <bits/stdc++.h>
using namespace std;
int prime[100050];
void sieve() {
  prime[0] = prime[1] = 0;
  int i, j;
  for (i = 2; i < 100050; i++) prime[i] = 1;
  for (i = 2; i < sqrt(100050); i++)
    if (prime[i])
      for (j = i; j * i < 100050; j++) prime[i * j] = 0;
}
int main() {
  sieve();
  vector<int> div;
  int n, t, pos;
  int a[100050];
  int dp[100050];
  int d[100050];
  int memo[100050];
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 2; i <= a[n]; i++) {
    if (prime[i]) div.push_back(i);
  }
  int sz = div.size();
  if (n == 1) {
    cout << 1 << endl;
    return 0;
  }
  int Max = -1;
  for (int i = 1; i <= n; i++) {
    t = a[i];
    pos = 0;
    for (int j = 0; j < sz && div[j] <= t; j++)
      if (t % div[j] == 0) {
        if (dp[a[i]] < d[j] + 1) {
          dp[a[i]] = d[j] + 1;
        }
        memo[pos] = j;
        pos++;
        while (t % div[j] == 0) t /= div[j];
      }
    for (int k = 0; k < pos; k++)
      if (dp[a[i]] > d[memo[k]]) {
        d[memo[k]] = dp[a[i]];
      }
    if (dp[a[i]] > Max) Max = dp[a[i]];
  }
  cout << Max << 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 prime[100050];
void sieve() {
  prime[0] = prime[1] = 0;
  int i, j;
  for (i = 2; i < 100050; i++) prime[i] = 1;
  for (i = 2; i < sqrt(100050); i++)
    if (prime[i])
      for (j = i; j * i < 100050; j++) prime[i * j] = 0;
}
int main() {
  sieve();
  vector<int> div;
  int n, t, pos;
  int a[100050];
  int dp[100050];
  int d[100050];
  int memo[100050];
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 2; i <= a[n]; i++) {
    if (prime[i]) div.push_back(i);
  }
  int sz = div.size();
  if (n == 1) {
    cout << 1 << endl;
    return 0;
  }
  int Max = -1;
  for (int i = 1; i <= n; i++) {
    t = a[i];
    pos = 0;
    for (int j = 0; j < sz && div[j] <= t; j++)
      if (t % div[j] == 0) {
        if (dp[a[i]] < d[j] + 1) {
          dp[a[i]] = d[j] + 1;
        }
        memo[pos] = j;
        pos++;
        while (t % div[j] == 0) t /= div[j];
      }
    for (int k = 0; k < pos; k++)
      if (dp[a[i]] > d[memo[k]]) {
        d[memo[k]] = dp[a[i]];
      }
    if (dp[a[i]] > Max) Max = dp[a[i]];
  }
  cout << Max << endl;
  return 0;
}

```