#include <bits/stdc++.h>
using namespace std;
const int N = int(1e5) + 1;
int ans = 1, n, d[N];
bool used[N];
vector<int> arr[N];
void solve(int num) {
  ans = max(ans, d[num]);
  for (int i = 0; i < arr[num].size(); ++i) {
    int v = arr[num][i];
    int j;
    for (j = num + v; j < N && !used[j]; j += v)
      ;
    if (j < N) {
      d[j] = max(d[j], d[num] + 1);
      arr[j].push_back(v);
    }
  }
}
int main(int argc, char *argv[]) {
  scanf("%d", &n);
  memset(d, 0, sizeof(d));
  memset(used, false, sizeof(used));
  while (n--) {
    int q;
    scanf("%d", &q);
    d[q] = 1;
    used[q] = true;
  }
  for (int i = 2; i < N; ++i) {
    arr[i].push_back(i);
  }
  for (int j = 2; j < N; ++j) {
    solve(j);
  }
  printf("%d", ans);
  return 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 N = int(1e5) + 1;
int ans = 1, n, d[N];
bool used[N];
vector<int> arr[N];
void solve(int num) {
  ans = max(ans, d[num]);
  for (int i = 0; i < arr[num].size(); ++i) {
    int v = arr[num][i];
    int j;
    for (j = num + v; j < N && !used[j]; j += v)
      ;
    if (j < N) {
      d[j] = max(d[j], d[num] + 1);
      arr[j].push_back(v);
    }
  }
}
int main(int argc, char *argv[]) {
  scanf("%d", &n);
  memset(d, 0, sizeof(d));
  memset(used, false, sizeof(used));
  while (n--) {
    int q;
    scanf("%d", &q);
    d[q] = 1;
    used[q] = true;
  }
  for (int i = 2; i < N; ++i) {
    arr[i].push_back(i);
  }
  for (int j = 2; j < N; ++j) {
    solve(j);
  }
  printf("%d", ans);
  return 0;
}

```