#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
const int M = 1e9 + 7;
int prime[N];
int ans[N];
int nums[N];
vector<int> factor[N];
void SIEVE() {
  for (int i = 2; i < N; i++) {
    if (prime[i] != 0) continue;
    for (int j = i; j < N; j += i) {
      prime[j] = i;
    }
  }
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  SIEVE();
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> nums[i], ans[i] = 1;
  for (int i = 0; i < n; i++) {
    int x = nums[i];
    while (x > 1) {
      int y = prime[x];
      int z = factor[y].size() - 1;
      if (z >= 0) ans[i] = max(ans[i], ans[factor[y][z]] + 1);
      while (x % y == 0) x = x / y;
      factor[y].push_back(i);
    }
  }
  cout << *max_element(ans, ans + n);
  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;
const int N = 1e5 + 5;
const int M = 1e9 + 7;
int prime[N];
int ans[N];
int nums[N];
vector<int> factor[N];
void SIEVE() {
  for (int i = 2; i < N; i++) {
    if (prime[i] != 0) continue;
    for (int j = i; j < N; j += i) {
      prime[j] = i;
    }
  }
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  SIEVE();
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> nums[i], ans[i] = 1;
  for (int i = 0; i < n; i++) {
    int x = nums[i];
    while (x > 1) {
      int y = prime[x];
      int z = factor[y].size() - 1;
      if (z >= 0) ans[i] = max(ans[i], ans[factor[y][z]] + 1);
      while (x % y == 0) x = x / y;
      factor[y].push_back(i);
    }
  }
  cout << *max_element(ans, ans + n);
  return 0;
}

```