#include <bits/stdc++.h>
using namespace std;
const int max_n = 100005;
int n;
int a[max_n];
bool p[max_n];
int dp[max_n];
int mx[max_n];
int pm[max_n];
int np = 0;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  if (n == 1)
    printf("1");
  else {
    fill(p, p + max_n, 1);
    for (int i = 2; i * i <= max_n; i++)
      if (p[i])
        for (int j = i; i * j <= max_n; j++) p[i * j] = 0;
    p[1] = 0;
    for (int i = 2; i <= max_n; i++)
      if (p[i]) {
        pm[np++] = i;
      }
    dp[0] = 0;
    for (int i = 1; i <= n; i++) {
      int mp = 0, mj = 0;
      for (int j = 1; j * j <= a[i - 1]; j++) {
        if (p[j] and a[i - 1] % j == 0) {
          if (mp <= mx[j]) {
            mp = mx[j];
            mj = j;
          }
        }
        if (a[i - 1] % j == 0) {
          int p2 = a[i - 1] / j;
          if (p2 != j and p[p2]) {
            if (mp <= mx[p2]) {
              mp = mx[p2];
              mj = p2;
            }
          }
        }
      }
      for (int j = 1; j * j <= a[i - 1]; j++) {
        if (p[j] and a[i - 1] % j == 0) mx[j] = mp + 1;
        if (a[i - 1] % j == 0) {
          int p2 = a[i - 1] / j;
          if (p2 != j and p[p2]) mx[p2] = mp + 1;
        }
      }
    }
    int res = 0;
    for (int i = 2; i <= max_n; i++) res = max(res, mx[i]);
    printf("%d", res);
  }
  return 0;
}

### Prompt

Generate 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;
const int max_n = 100005;
int n;
int a[max_n];
bool p[max_n];
int dp[max_n];
int mx[max_n];
int pm[max_n];
int np = 0;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  if (n == 1)
    printf("1");
  else {
    fill(p, p + max_n, 1);
    for (int i = 2; i * i <= max_n; i++)
      if (p[i])
        for (int j = i; i * j <= max_n; j++) p[i * j] = 0;
    p[1] = 0;
    for (int i = 2; i <= max_n; i++)
      if (p[i]) {
        pm[np++] = i;
      }
    dp[0] = 0;
    for (int i = 1; i <= n; i++) {
      int mp = 0, mj = 0;
      for (int j = 1; j * j <= a[i - 1]; j++) {
        if (p[j] and a[i - 1] % j == 0) {
          if (mp <= mx[j]) {
            mp = mx[j];
            mj = j;
          }
        }
        if (a[i - 1] % j == 0) {
          int p2 = a[i - 1] / j;
          if (p2 != j and p[p2]) {
            if (mp <= mx[p2]) {
              mp = mx[p2];
              mj = p2;
            }
          }
        }
      }
      for (int j = 1; j * j <= a[i - 1]; j++) {
        if (p[j] and a[i - 1] % j == 0) mx[j] = mp + 1;
        if (a[i - 1] % j == 0) {
          int p2 = a[i - 1] / j;
          if (p2 != j and p[p2]) mx[p2] = mp + 1;
        }
      }
    }
    int res = 0;
    for (int i = 2; i <= max_n; i++) res = max(res, mx[i]);
    printf("%d", res);
  }
  return 0;
}

```