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.
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  long long a[n];
  long long divisor[100001] = {-1};
  long long dp[100001] = {0};
  long long max_l = 1;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == 1) continue;
    long long longest = 1;
    long long root = sqrt(a[i]);
    divisor[a[i]] = a[i];
    for (long long j = 2; j < root + 1; j++) {
      if (a[i] % j == 0) {
        if (divisor[j] != (-1)) {
          longest = max(longest, dp[divisor[j]] + 1);
        }
        if (divisor[a[i] / j] != (-1)) {
          longest = max(longest, dp[divisor[a[i] / j]] + 1);
        }
        divisor[j] = a[i];
        divisor[a[i] / j] = a[i];
      }
    }
    dp[a[i]] = longest;
    max_l = max(max_l, longest);
  }
  cout << max_l;
}
