#include <bits/stdc++.h>
using namespace std;
const int maxn = 100005;
const int logn = 17;
const int inf = (int)1e9;
const long long mod = (int)1e9 + 7;
const long long l_inf = (long long)4e12;
const long double pi = acosl(-1.0);
const long double eps = 1e-12;
int n, ans = 0;
int a[maxn];
int dp[maxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.precision(15);
  cout << fixed;
  srand(566);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  for (int i = 0; i < n; i++) {
    int best = 0;
    vector<int> divs;
    for (int d = 2; d * d <= a[i]; d++) {
      if (a[i] % d == 0) {
        best = max(best, dp[d]);
        divs.push_back(d);
        while (a[i] % d == 0) a[i] /= d;
      }
    }
    if (a[i] != 1) {
      best = max(best, dp[a[i]]);
      divs.push_back(a[i]);
    }
    for (int d : divs) dp[d] = max(dp[d], best + 1);
    ans = max(ans, best + 1);
  }
  cout << ans;
  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;
const int maxn = 100005;
const int logn = 17;
const int inf = (int)1e9;
const long long mod = (int)1e9 + 7;
const long long l_inf = (long long)4e12;
const long double pi = acosl(-1.0);
const long double eps = 1e-12;
int n, ans = 0;
int a[maxn];
int dp[maxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.precision(15);
  cout << fixed;
  srand(566);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  for (int i = 0; i < n; i++) {
    int best = 0;
    vector<int> divs;
    for (int d = 2; d * d <= a[i]; d++) {
      if (a[i] % d == 0) {
        best = max(best, dp[d]);
        divs.push_back(d);
        while (a[i] % d == 0) a[i] /= d;
      }
    }
    if (a[i] != 1) {
      best = max(best, dp[a[i]]);
      divs.push_back(a[i]);
    }
    for (int d : divs) dp[d] = max(dp[d], best + 1);
    ans = max(ans, best + 1);
  }
  cout << ans;
  return 0;
}

```