#include <bits/stdc++.h>
using namespace std;
const int maxlen = 600010;
int dp[maxlen], pre[maxlen], num[maxlen];
int Max(int x, int y) {
  if (x > y) return x;
  return y;
}
int main() {
  int n, ans = 1;
  memset(dp, 0, sizeof(dp));
  memset(pre, 0, sizeof(pre));
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &num[i]);
  }
  for (int i = 1; i <= n; i++) {
    int now = num[i];
    int end = (int)sqrt(num[i]);
    for (int j = 2; j <= end; j++) {
      if ((now % j) == 0) {
        while ((now % j) == 0) now /= j;
        dp[i] = Max(dp[i], dp[pre[j]] + 1);
        pre[j] = i;
      }
    }
    if (now > 1) {
      dp[i] = Max(dp[i], dp[pre[now]] + 1);
      pre[now] = i;
    }
    if (dp[i] > ans) ans = dp[i];
  }
  printf("%d\n", ans);
}

### 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 maxlen = 600010;
int dp[maxlen], pre[maxlen], num[maxlen];
int Max(int x, int y) {
  if (x > y) return x;
  return y;
}
int main() {
  int n, ans = 1;
  memset(dp, 0, sizeof(dp));
  memset(pre, 0, sizeof(pre));
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &num[i]);
  }
  for (int i = 1; i <= n; i++) {
    int now = num[i];
    int end = (int)sqrt(num[i]);
    for (int j = 2; j <= end; j++) {
      if ((now % j) == 0) {
        while ((now % j) == 0) now /= j;
        dp[i] = Max(dp[i], dp[pre[j]] + 1);
        pre[j] = i;
      }
    }
    if (now > 1) {
      dp[i] = Max(dp[i], dp[pre[now]] + 1);
      pre[now] = i;
    }
    if (dp[i] > ans) ans = dp[i];
  }
  printf("%d\n", ans);
}

```