#include <bits/stdc++.h>
using namespace std;
const int mxs = 101100;
int n;
int a[mxs];
int prime[mxs];
int len;
bool use[mxs];
int dp[mxs];
int dv[mxs];
int ln;
int main() {
  for (int i = 2; i < 1e5; i++) {
    if (!use[i]) {
      for (int j = i; j < 1e5; j += i) use[j] = true;
      prime[len++] = i;
    }
  }
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  int ans = 1;
  for (int i = 0; i < n; i++) {
    ln = 0;
    int w = a[i];
    for (int j = 2; j * j <= w; j++) {
      if (w % j == 0) {
        while (w % j == 0) w /= j;
        dv[ln++] = j;
      }
    }
    if (w != 1) dv[ln++] = w;
    int mx = 0;
    for (int j = 0; j < ln; j++) {
      mx = max(dp[dv[j]] + 1, mx);
    }
    for (int j = 0; j < ln; j++) dp[dv[j]] = mx;
    ans = max(mx, ans);
  }
  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 mxs = 101100;
int n;
int a[mxs];
int prime[mxs];
int len;
bool use[mxs];
int dp[mxs];
int dv[mxs];
int ln;
int main() {
  for (int i = 2; i < 1e5; i++) {
    if (!use[i]) {
      for (int j = i; j < 1e5; j += i) use[j] = true;
      prime[len++] = i;
    }
  }
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  int ans = 1;
  for (int i = 0; i < n; i++) {
    ln = 0;
    int w = a[i];
    for (int j = 2; j * j <= w; j++) {
      if (w % j == 0) {
        while (w % j == 0) w /= j;
        dv[ln++] = j;
      }
    }
    if (w != 1) dv[ln++] = w;
    int mx = 0;
    for (int j = 0; j < ln; j++) {
      mx = max(dp[dv[j]] + 1, mx);
    }
    for (int j = 0; j < ln; j++) dp[dv[j]] = mx;
    ans = max(mx, ans);
  }
  printf("%d\n", ans);
}

```