#include <bits/stdc++.h>
using namespace std;
int notprime[100005];
vector<int> divisors[100005];
int dp[100005];
int arr[100005];
map<int, int> last;
int main() {
  notprime[0] = 1;
  notprime[1] = 1;
  for (int g = 2; g * g <= 100000; g++) {
    if (notprime[g]) continue;
    for (int y = 2 * g; y <= 100000; y += g) {
      notprime[y] = 1;
    }
  }
  for (int g = 2; g <= 100000; g++) {
    if (notprime[g]) continue;
    for (int y = g; y <= 100000; y += g) {
      divisors[y].push_back(g);
    }
  }
  int a;
  cin >> a;
  for (int g = 1; g <= a; g++) cin >> arr[g];
  int ans = 0;
  for (int g = 1; g <= a; g++) {
    int z = 0;
    for (int y = 0; y < divisors[arr[g]].size(); y++) {
      z = max(z, 1 + dp[last[divisors[arr[g]][y]]]);
    }
    dp[g] = max(z, 1);
    ans = max(ans, dp[g]);
    for (int y = 0; y < divisors[arr[g]].size(); y++) {
      last[divisors[arr[g]][y]] = g;
    }
  }
  cout << ans;
  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;
int notprime[100005];
vector<int> divisors[100005];
int dp[100005];
int arr[100005];
map<int, int> last;
int main() {
  notprime[0] = 1;
  notprime[1] = 1;
  for (int g = 2; g * g <= 100000; g++) {
    if (notprime[g]) continue;
    for (int y = 2 * g; y <= 100000; y += g) {
      notprime[y] = 1;
    }
  }
  for (int g = 2; g <= 100000; g++) {
    if (notprime[g]) continue;
    for (int y = g; y <= 100000; y += g) {
      divisors[y].push_back(g);
    }
  }
  int a;
  cin >> a;
  for (int g = 1; g <= a; g++) cin >> arr[g];
  int ans = 0;
  for (int g = 1; g <= a; g++) {
    int z = 0;
    for (int y = 0; y < divisors[arr[g]].size(); y++) {
      z = max(z, 1 + dp[last[divisors[arr[g]][y]]]);
    }
    dp[g] = max(z, 1);
    ans = max(ans, dp[g]);
    for (int y = 0; y < divisors[arr[g]].size(); y++) {
      last[divisors[arr[g]][y]] = g;
    }
  }
  cout << ans;
  return 0;
}

```