#include <bits/stdc++.h>
using namespace std;
int p[1000005];
int fnd(int x) { return p[x] = p[x] == x ? x : fnd(p[x]); }
void unn(int x, int y) { p[fnd(y)] = fnd(x); }
int dir[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
int gcd(int x, int y) { return x % y ? gcd(y, x % y) : y; }
int pq[5144];
vector<int> pr;
int net[100005];
int a[100005];
int main() {
  int n;
  scanf("%d", &n);
  if (n == 1) return printf("1\n"), 0;
  for (int i = 2; i <= 10000; i++) {
    if (net[i]) continue;
    pr.push_back(i);
    for (int j = i * i; j <= 100000; j += i) net[j] = 1;
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    int mx = 0;
    for (int j = 0; j < pr.size(); j++) {
      if (a[i] % pr[j] == 0) {
        mx = max(mx, pq[j]);
      }
    }
    for (int j = 0; j < pr.size(); j++) {
      if (a[i] % pr[j] == 0) {
        pq[j] = max(mx + 1, pq[j]);
      }
    }
  }
  int mx = 1;
  for (int i = 0; i < pr.size(); i++) mx = max(mx, pq[i]);
  printf("%d", mx);
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
int p[1000005];
int fnd(int x) { return p[x] = p[x] == x ? x : fnd(p[x]); }
void unn(int x, int y) { p[fnd(y)] = fnd(x); }
int dir[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
int gcd(int x, int y) { return x % y ? gcd(y, x % y) : y; }
int pq[5144];
vector<int> pr;
int net[100005];
int a[100005];
int main() {
  int n;
  scanf("%d", &n);
  if (n == 1) return printf("1\n"), 0;
  for (int i = 2; i <= 10000; i++) {
    if (net[i]) continue;
    pr.push_back(i);
    for (int j = i * i; j <= 100000; j += i) net[j] = 1;
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    int mx = 0;
    for (int j = 0; j < pr.size(); j++) {
      if (a[i] % pr[j] == 0) {
        mx = max(mx, pq[j]);
      }
    }
    for (int j = 0; j < pr.size(); j++) {
      if (a[i] % pr[j] == 0) {
        pq[j] = max(mx + 1, pq[j]);
      }
    }
  }
  int mx = 1;
  for (int i = 0; i < pr.size(); i++) mx = max(mx, pq[i]);
  printf("%d", mx);
}

```