#include <bits/stdc++.h>
using namespace std;
const int M = 1e5 + 5;
vector<int> pr[M];
int dp[M], a[M];
bool mark[M];
void pre() {
  for (int i = 2; i < M; i++)
    if (!mark[i])
      for (int j = 2 * i; j < M; j += i) pr[j].push_back(i), mark[j] = 1;
}
int32_t main() {
  pre();
  int n, ans = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  if (n == 1 && a[0] == 1) return cout << 1 << endl, 0;
  for (int i = 0; i < n; i++) {
    for (auto p : pr[a[i]]) ((dp[a[i]]) = max((dp[a[i]]), (dp[p])));
    dp[a[i]]++;
    for (auto p : pr[a[i]]) ((dp[p]) = max((dp[p]), (dp[a[i]])));
    ((ans) = max((ans), (dp[a[i]])));
  }
  return cout << ans << endl, 0;
}

### Prompt

Create a solution in cpp for 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 M = 1e5 + 5;
vector<int> pr[M];
int dp[M], a[M];
bool mark[M];
void pre() {
  for (int i = 2; i < M; i++)
    if (!mark[i])
      for (int j = 2 * i; j < M; j += i) pr[j].push_back(i), mark[j] = 1;
}
int32_t main() {
  pre();
  int n, ans = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  if (n == 1 && a[0] == 1) return cout << 1 << endl, 0;
  for (int i = 0; i < n; i++) {
    for (auto p : pr[a[i]]) ((dp[a[i]]) = max((dp[a[i]]), (dp[p])));
    dp[a[i]]++;
    for (auto p : pr[a[i]]) ((dp[p]) = max((dp[p]), (dp[a[i]])));
    ((ans) = max((ans), (dp[a[i]])));
  }
  return cout << ans << endl, 0;
}

```