#include <bits/stdc++.h>
using namespace std;
int n, a[100100];
int cnt, prime[100100];
int dp[100100];
map<int, int> mm;
int main() {
  cnt = 0;
  for (int i = 2; i <= 100000; i++) {
    bool flag = true;
    for (int j = 0; j < cnt && ((prime[j]) * (prime[j])) <= i && flag; j++)
      if (i % prime[j] == 0) flag = false;
    if (flag) prime[cnt++] = i;
  }
  for (int i = 0; i < cnt; i++) mm[prime[i]] = i;
  scanf("%d", &n);
  memset(dp, 0, sizeof(dp));
  for (int i = 0; i < n; i++) {
    int x, y;
    scanf("%d", &x);
    y = x;
    int now = 0;
    for (int j = 0; j < cnt && ((prime[j]) * (prime[j])) <= x; j++) {
      if (x % prime[j] == 0) now = max(now, dp[j]);
      while (x % prime[j] == 0) x /= prime[j];
    }
    if (x != 1) now = max(now, dp[mm[x]]);
    x = y;
    for (int j = 0; j < cnt && ((prime[j]) * (prime[j])) <= x; j++) {
      if (x % prime[j] == 0) dp[j] = now + 1;
      while (x % prime[j] == 0) x /= prime[j];
    }
    if (x != 1) dp[mm[x]] = now + 1;
  }
  int ans = 1;
  for (int i = 0; i < n; i++) ans = max(ans, dp[i]);
  printf("%d\n", ans);
  return 0;
}

### 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 n, a[100100];
int cnt, prime[100100];
int dp[100100];
map<int, int> mm;
int main() {
  cnt = 0;
  for (int i = 2; i <= 100000; i++) {
    bool flag = true;
    for (int j = 0; j < cnt && ((prime[j]) * (prime[j])) <= i && flag; j++)
      if (i % prime[j] == 0) flag = false;
    if (flag) prime[cnt++] = i;
  }
  for (int i = 0; i < cnt; i++) mm[prime[i]] = i;
  scanf("%d", &n);
  memset(dp, 0, sizeof(dp));
  for (int i = 0; i < n; i++) {
    int x, y;
    scanf("%d", &x);
    y = x;
    int now = 0;
    for (int j = 0; j < cnt && ((prime[j]) * (prime[j])) <= x; j++) {
      if (x % prime[j] == 0) now = max(now, dp[j]);
      while (x % prime[j] == 0) x /= prime[j];
    }
    if (x != 1) now = max(now, dp[mm[x]]);
    x = y;
    for (int j = 0; j < cnt && ((prime[j]) * (prime[j])) <= x; j++) {
      if (x % prime[j] == 0) dp[j] = now + 1;
      while (x % prime[j] == 0) x /= prime[j];
    }
    if (x != 1) dp[mm[x]] = now + 1;
  }
  int ans = 1;
  for (int i = 0; i < n; i++) ans = max(ans, dp[i]);
  printf("%d\n", ans);
  return 0;
}

```