#include <bits/stdc++.h>
using namespace std;
long long int a[200000], p[200000];
vector<long long int> v;
int main() {
  long long int n, m, i, j, k, l, ans = 0, x, y;
  scanf("%lld", &n);
  for (i = 2; i <= n; i++) {
    if (p[i] == 0) {
      for (j = i * i; j <= n; j += i) p[j] = 1;
    }
  }
  for (i = 0; i < n; i++) {
    scanf("%lld", &x);
    l = 1;
    v.clear();
    for (j = 2; j <= sqrt(x); j++) {
      if (x % j == 0) {
        if (p[j] == 0) {
          a[j]++;
          v.push_back(j);
        }
        if (j != (x / j) && p[x / j] == 0) {
          a[x / j]++;
          v.push_back(x / j);
        }
        l = max(l, max(a[j], a[x / j]));
      }
    }
    for (j = 0; j < v.size(); j++) a[v[j]] = l;
    if (p[x] == 0) a[x] = l;
    ans = max(ans, l);
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Your challenge is to write 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>
using namespace std;
long long int a[200000], p[200000];
vector<long long int> v;
int main() {
  long long int n, m, i, j, k, l, ans = 0, x, y;
  scanf("%lld", &n);
  for (i = 2; i <= n; i++) {
    if (p[i] == 0) {
      for (j = i * i; j <= n; j += i) p[j] = 1;
    }
  }
  for (i = 0; i < n; i++) {
    scanf("%lld", &x);
    l = 1;
    v.clear();
    for (j = 2; j <= sqrt(x); j++) {
      if (x % j == 0) {
        if (p[j] == 0) {
          a[j]++;
          v.push_back(j);
        }
        if (j != (x / j) && p[x / j] == 0) {
          a[x / j]++;
          v.push_back(x / j);
        }
        l = max(l, max(a[j], a[x / j]));
      }
    }
    for (j = 0; j < v.size(); j++) a[v[j]] = l;
    if (p[x] == 0) a[x] = l;
    ans = max(ans, l);
  }
  printf("%lld\n", ans);
  return 0;
}

```