#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int n, num, ans, a[maxn], fp[maxn], dp[maxn], f[maxn], pr[maxn];
vector<int> p;
void pre() {
  int N = 1e5;
  fp[1] = -1;
  for (int i = 2; i <= N; i++) {
    if (!fp[i]) pr[num++] = i;
    for (int j = 0; j < num; j++) {
      int k = i * pr[j];
      if (k > N) break;
      fp[k] = j;
      if (i % pr[j] == 0) break;
    }
  }
  return;
}
int main() {
  pre();
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  ans = 1;
  for (int i = 1; i <= n; i++) {
    int x = a[i];
    p.clear();
    for (int j = 0; j < num; j++) {
      if (x == 1 || pr[j] * pr[j] > x) break;
      if (x % pr[j] == 0) {
        dp[i] = max(dp[i], f[pr[j]] + 1);
        p.push_back(pr[j]);
        while (x % pr[j] == 0) x /= pr[j];
      }
    }
    if (x > 1) {
      dp[i] = max(dp[i], f[x] + 1);
      p.push_back(x);
    }
    for (int j = 0; j < p.size(); j++) f[p[j]] = max(f[p[j]], dp[i]);
    ans = max(ans, dp[i]);
  }
  printf("%d\n", ans);
  return 0;
}

### 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;
const int maxn = 1e5 + 10;
int n, num, ans, a[maxn], fp[maxn], dp[maxn], f[maxn], pr[maxn];
vector<int> p;
void pre() {
  int N = 1e5;
  fp[1] = -1;
  for (int i = 2; i <= N; i++) {
    if (!fp[i]) pr[num++] = i;
    for (int j = 0; j < num; j++) {
      int k = i * pr[j];
      if (k > N) break;
      fp[k] = j;
      if (i % pr[j] == 0) break;
    }
  }
  return;
}
int main() {
  pre();
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  ans = 1;
  for (int i = 1; i <= n; i++) {
    int x = a[i];
    p.clear();
    for (int j = 0; j < num; j++) {
      if (x == 1 || pr[j] * pr[j] > x) break;
      if (x % pr[j] == 0) {
        dp[i] = max(dp[i], f[pr[j]] + 1);
        p.push_back(pr[j]);
        while (x % pr[j] == 0) x /= pr[j];
      }
    }
    if (x > 1) {
      dp[i] = max(dp[i], f[x] + 1);
      p.push_back(x);
    }
    for (int j = 0; j < p.size(); j++) f[p[j]] = max(f[p[j]], dp[i]);
    ans = max(ans, dp[i]);
  }
  printf("%d\n", ans);
  return 0;
}

```