#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
int main() {
  long long n;
  cin >> n;
  vector<long long> dp(100005, 0);
  long long x;
  long long ans = 1;
  while (n--) {
    cin >> x;
    vector<long long> d;
    for (long long i = 1; i * i <= x; i++) {
      if (x % i == 0 and i != 1) d.push_back(i);
      if (x % i == 0 and x / i != i) d.push_back(x / i);
    }
    vector<long long> gg(d.size(), 0);
    for (long long j = 0; j < d.size(); j++) {
      long long mx = 0;
      for (long long k = 0; k < d.size(); k++)
        if (j != k) mx = max(dp[d[k]], mx);
      gg[j] = max(mx, dp[d[j]]);
    }
    for (long long j = 0; j < d.size(); j++) dp[d[j]] = gg[j] + 1;
  }
  for (long long i = 2; i < 100005; i++) ans = max(ans, dp[i]);
  cout << ans;
  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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
int main() {
  long long n;
  cin >> n;
  vector<long long> dp(100005, 0);
  long long x;
  long long ans = 1;
  while (n--) {
    cin >> x;
    vector<long long> d;
    for (long long i = 1; i * i <= x; i++) {
      if (x % i == 0 and i != 1) d.push_back(i);
      if (x % i == 0 and x / i != i) d.push_back(x / i);
    }
    vector<long long> gg(d.size(), 0);
    for (long long j = 0; j < d.size(); j++) {
      long long mx = 0;
      for (long long k = 0; k < d.size(); k++)
        if (j != k) mx = max(dp[d[k]], mx);
      gg[j] = max(mx, dp[d[j]]);
    }
    for (long long j = 0; j < d.size(); j++) dp[d[j]] = gg[j] + 1;
  }
  for (long long i = 2; i < 100005; i++) ans = max(ans, dp[i]);
  cout << ans;
  return 0;
}

```