#include <bits/stdc++.h>
using namespace std;
const long double PI = 3.141592653589793238L;
void init() {
  cin.tie(0);
  ios::sync_with_stdio(0);
}
const int N = 1e5 + 5, M = 1e6, OO = 0x3f3f3f3f;
const long long llOO = 1e18;
int n, a[N], dp[N], dpF[N];
int solve() {
  for (int i = 0; i < n; i++) {
    vector<int> divisors;
    for (int j = 1; j * j <= a[i]; j++) {
      if (a[i] % j) continue;
      if (j != 1) divisors.push_back(j);
      if (a[i] / j != j) divisors.push_back(a[i] / j);
    }
    int best = 1;
    for (int j = 0; j < divisors.size(); j++)
      best = max(best, 1 + dpF[divisors[j]]);
    dp[i] = best;
    for (int j = 0; j < divisors.size(); j++) dpF[divisors[j]] = best;
  }
  return *max_element(dp, dp + n);
}
int main() {
  init();
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  cout << solve();
}

### 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 long double PI = 3.141592653589793238L;
void init() {
  cin.tie(0);
  ios::sync_with_stdio(0);
}
const int N = 1e5 + 5, M = 1e6, OO = 0x3f3f3f3f;
const long long llOO = 1e18;
int n, a[N], dp[N], dpF[N];
int solve() {
  for (int i = 0; i < n; i++) {
    vector<int> divisors;
    for (int j = 1; j * j <= a[i]; j++) {
      if (a[i] % j) continue;
      if (j != 1) divisors.push_back(j);
      if (a[i] / j != j) divisors.push_back(a[i] / j);
    }
    int best = 1;
    for (int j = 0; j < divisors.size(); j++)
      best = max(best, 1 + dpF[divisors[j]]);
    dp[i] = best;
    for (int j = 0; j < divisors.size(); j++) dpF[divisors[j]] = best;
  }
  return *max_element(dp, dp + n);
}
int main() {
  init();
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  cout << solve();
}

```