#include <bits/stdc++.h>
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  std::cerr << name << " = " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  std::cerr.write(names, comma - names) << " = " << arg1 << " |";
  __f(comma + 1, args...);
}
const double pi = acos(-1.0);
using namespace std;
void MOD(long long& x) {
  if (x >= 1000000007) x -= 1000000007;
  if (x < 0) x += 1000000007;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
void solve() {
  int n;
  cin >> n;
  int arr[n];
  for (int i = 0; i < n; i++) cin >> arr[i];
  vector<int> facts[100005];
  for (int i = 2; i < 100005; i++) {
    for (int j = i; j < 100005; j += i) {
      facts[j].push_back(i);
    }
  }
  int dp[100005];
  for (int i = 0; i < 100005; i++) dp[i] = 0;
  int ans = 1;
  for (int i = 0; i < n; i++) {
    int val = 0;
    for (auto fact : facts[arr[i]]) {
      val = max(val, dp[fact]);
    }
    val++;
    for (auto fact : facts[arr[i]]) {
      dp[fact] = val;
    }
    ans = max(ans, val);
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Your challenge is to write 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>
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  std::cerr << name << " = " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  std::cerr.write(names, comma - names) << " = " << arg1 << " |";
  __f(comma + 1, args...);
}
const double pi = acos(-1.0);
using namespace std;
void MOD(long long& x) {
  if (x >= 1000000007) x -= 1000000007;
  if (x < 0) x += 1000000007;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
void solve() {
  int n;
  cin >> n;
  int arr[n];
  for (int i = 0; i < n; i++) cin >> arr[i];
  vector<int> facts[100005];
  for (int i = 2; i < 100005; i++) {
    for (int j = i; j < 100005; j += i) {
      facts[j].push_back(i);
    }
  }
  int dp[100005];
  for (int i = 0; i < 100005; i++) dp[i] = 0;
  int ans = 1;
  for (int i = 0; i < n; i++) {
    int val = 0;
    for (auto fact : facts[arr[i]]) {
      val = max(val, dp[fact]);
    }
    val++;
    for (auto fact : facts[arr[i]]) {
      dp[fact] = val;
    }
    ans = max(ans, val);
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```