#include <bits/stdc++.h>
using namespace std;
const int LIM = 100000;
int divisor_of[LIM + 1];
void init() {
  divisor_of[1] = 1;
  for (int i = 2; i <= LIM; ++i) {
    if (divisor_of[i] == 0) {
      divisor_of[i] = i;
      for (long long j = (long long)i * i; j <= LIM; j += i) {
        if (divisor_of[j] == 0) divisor_of[j] = i;
      }
    }
  }
}
bool solve() {
  int N;
  if (!(cin >> N)) return false;
  if (!N) return false;
  vector<int> dp(100001, 0);
  dp[1] = 1;
  for (int i = 0; i < N; ++i) {
    int v;
    cin >> v;
    int best = 0;
    int tmp = v;
    while (tmp > 1) {
      best = max(best, dp[divisor_of[tmp]] + 1);
      tmp /= divisor_of[tmp];
    }
    tmp = v;
    while (tmp > 1) {
      dp[divisor_of[tmp]] = max(dp[divisor_of[tmp]], best);
      tmp /= divisor_of[tmp];
    }
  }
  cout << *max_element(dp.begin(), dp.end()) << endl;
  return true;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(0);
  init();
  while (solve())
    ;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 LIM = 100000;
int divisor_of[LIM + 1];
void init() {
  divisor_of[1] = 1;
  for (int i = 2; i <= LIM; ++i) {
    if (divisor_of[i] == 0) {
      divisor_of[i] = i;
      for (long long j = (long long)i * i; j <= LIM; j += i) {
        if (divisor_of[j] == 0) divisor_of[j] = i;
      }
    }
  }
}
bool solve() {
  int N;
  if (!(cin >> N)) return false;
  if (!N) return false;
  vector<int> dp(100001, 0);
  dp[1] = 1;
  for (int i = 0; i < N; ++i) {
    int v;
    cin >> v;
    int best = 0;
    int tmp = v;
    while (tmp > 1) {
      best = max(best, dp[divisor_of[tmp]] + 1);
      tmp /= divisor_of[tmp];
    }
    tmp = v;
    while (tmp > 1) {
      dp[divisor_of[tmp]] = max(dp[divisor_of[tmp]], best);
      tmp /= divisor_of[tmp];
    }
  }
  cout << *max_element(dp.begin(), dp.end()) << endl;
  return true;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(0);
  init();
  while (solve())
    ;
  return 0;
}

```