#include <bits/stdc++.h>
using namespace std;
long long int fast_exp(long long int base, long long int exp1) {
  long long int res = 1;
  while (exp1 > 0) {
    if (exp1 & 1) res = (res * base) % 1000000007;
    base = (base * base) % 1000000007;
    exp1 /= 2;
  }
  return res % 1000000007;
}
long long int pr[100001] = {0};
void isprime() {
  pr[0] = 1;
  pr[1] = 1;
  for (long long int a = 2; a <= 100001; a++) {
    if (!pr[a]) {
      for (long long int b = a; b < 100001; b += a) {
        pr[b] = a;
      }
    }
  }
}
long long int comp = 1, tim = 0;
struct vertex {
  vector<long long int> adj;
  long long int vis = 0;
  long long int parent = -1;
  long long int dist = 0;
  long long int component = 0;
  long long int in = 0;
  long long int out = 0;
  long long int low = 0;
  long long int child = 0;
  long long int indeg = 0;
  long long int outdeg = 0;
  bool AP = false;
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  isprime();
  long long int n;
  cin >> n;
  long long int x[n];
  long long int dp[100001] = {0};
  for (long long int a = 0; a < n; a++) {
    long long int b;
    cin >> b;
    long long int k = b;
    while (k > 1) {
      long long int p = pr[k];
      dp[b] = max(dp[b], dp[p] + 1);
      while (k % p == 0) k /= p;
    }
    k = b;
    while (k > 1) {
      long long int p = pr[k];
      dp[p] = max(dp[p], dp[b]);
      while (k % p == 0) k /= p;
    }
    if (b == 1) dp[1] = 1;
  }
  cout << *max_element(dp, dp + 100001) << endl;
  return 0;
}

### Prompt

Your task is to create 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;
long long int fast_exp(long long int base, long long int exp1) {
  long long int res = 1;
  while (exp1 > 0) {
    if (exp1 & 1) res = (res * base) % 1000000007;
    base = (base * base) % 1000000007;
    exp1 /= 2;
  }
  return res % 1000000007;
}
long long int pr[100001] = {0};
void isprime() {
  pr[0] = 1;
  pr[1] = 1;
  for (long long int a = 2; a <= 100001; a++) {
    if (!pr[a]) {
      for (long long int b = a; b < 100001; b += a) {
        pr[b] = a;
      }
    }
  }
}
long long int comp = 1, tim = 0;
struct vertex {
  vector<long long int> adj;
  long long int vis = 0;
  long long int parent = -1;
  long long int dist = 0;
  long long int component = 0;
  long long int in = 0;
  long long int out = 0;
  long long int low = 0;
  long long int child = 0;
  long long int indeg = 0;
  long long int outdeg = 0;
  bool AP = false;
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  isprime();
  long long int n;
  cin >> n;
  long long int x[n];
  long long int dp[100001] = {0};
  for (long long int a = 0; a < n; a++) {
    long long int b;
    cin >> b;
    long long int k = b;
    while (k > 1) {
      long long int p = pr[k];
      dp[b] = max(dp[b], dp[p] + 1);
      while (k % p == 0) k /= p;
    }
    k = b;
    while (k > 1) {
      long long int p = pr[k];
      dp[p] = max(dp[p], dp[b]);
      while (k % p == 0) k /= p;
    }
    if (b == 1) dp[1] = 1;
  }
  cout << *max_element(dp, dp + 100001) << endl;
  return 0;
}

```