#include <bits/stdc++.h>
using namespace std;
int memo[100003], A[100003];
int F[100003], last[100003];
void criba() {
  for (int i = 2; i * i < 100003; i++) {
    if (!F[i])
      for (int j = i * i; j < 100003; j += i) F[j] = i;
  }
  for (int i = 2; i < 100003; i++)
    if (F[i] == 0) F[i] = i;
}
int main() {
  int n;
  criba();
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &A[i]);
  memset(memo, 0, sizeof memo);
  memset(last, -1, sizeof last);
  int ans = 0;
  for (int i = 0; i < n; i++) {
    memo[i] = 1;
    while (A[i] > 1) {
      int f = F[A[i]];
      if (last[f] >= 0 && last[f] != i)
        memo[i] = max(memo[i], 1 + memo[last[f]]);
      last[f] = i;
      A[i] /= f;
    }
    ans = max(ans, memo[i]);
  }
  printf("%d\n", ans);
}

### Prompt

Develop a solution in CPP 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 memo[100003], A[100003];
int F[100003], last[100003];
void criba() {
  for (int i = 2; i * i < 100003; i++) {
    if (!F[i])
      for (int j = i * i; j < 100003; j += i) F[j] = i;
  }
  for (int i = 2; i < 100003; i++)
    if (F[i] == 0) F[i] = i;
}
int main() {
  int n;
  criba();
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &A[i]);
  memset(memo, 0, sizeof memo);
  memset(last, -1, sizeof last);
  int ans = 0;
  for (int i = 0; i < n; i++) {
    memo[i] = 1;
    while (A[i] > 1) {
      int f = F[A[i]];
      if (last[f] >= 0 && last[f] != i)
        memo[i] = max(memo[i], 1 + memo[last[f]]);
      last[f] = i;
      A[i] /= f;
    }
    ans = max(ans, memo[i]);
  }
  printf("%d\n", ans);
}

```