#include <bits/stdc++.h>
using namespace std;
int a[100010], f[100010], c[100010];
bool noprime[100010];
vector<int> prime;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
void Prime() {
  for (int i = 2; i <= 100000; i++) {
    if (!noprime[i]) {
      prime.push_back(i);
    }
    for (int j = 0; j < prime.size() && prime[j] * i <= 100000; j++) {
      noprime[prime[j] * i] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
int main() {
  Prime();
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    int tmp = a[i];
    f[i] = 1;
    for (int j = 0; j < prime.size(); j++) {
      int p = prime[j];
      if (p * p > tmp) break;
      if (tmp % p == 0) {
        f[i] = max(f[i], c[p] + 1);
        while (tmp % p == 0) tmp /= p;
      }
    }
    if (tmp > 1) {
      f[i] = max(f[i], c[tmp] + 1);
    }
    for (int j = 0; j < prime.size(); j++) {
      int p = prime[j];
      if (p * p > a[i]) break;
      if (a[i] % p == 0) {
        c[p] = f[i];
        while (a[i] % p == 0) a[i] /= p;
      }
    }
    if (a[i] > 1) {
      c[a[i]] = f[i];
    }
  }
  int maxn = 1;
  for (int i = 0; i < n; i++)
    if (f[i] > maxn) maxn = f[i];
  cout << maxn << endl;
}

### 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;
int a[100010], f[100010], c[100010];
bool noprime[100010];
vector<int> prime;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
void Prime() {
  for (int i = 2; i <= 100000; i++) {
    if (!noprime[i]) {
      prime.push_back(i);
    }
    for (int j = 0; j < prime.size() && prime[j] * i <= 100000; j++) {
      noprime[prime[j] * i] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
int main() {
  Prime();
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    int tmp = a[i];
    f[i] = 1;
    for (int j = 0; j < prime.size(); j++) {
      int p = prime[j];
      if (p * p > tmp) break;
      if (tmp % p == 0) {
        f[i] = max(f[i], c[p] + 1);
        while (tmp % p == 0) tmp /= p;
      }
    }
    if (tmp > 1) {
      f[i] = max(f[i], c[tmp] + 1);
    }
    for (int j = 0; j < prime.size(); j++) {
      int p = prime[j];
      if (p * p > a[i]) break;
      if (a[i] % p == 0) {
        c[p] = f[i];
        while (a[i] % p == 0) a[i] /= p;
      }
    }
    if (a[i] > 1) {
      c[a[i]] = f[i];
    }
  }
  int maxn = 1;
  for (int i = 0; i < n; i++)
    if (f[i] > maxn) maxn = f[i];
  cout << maxn << endl;
}

```