#include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  std::cerr << name << " : " << arg1 << 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...);
}
long long hell = round(1e18);
vector<long long> p(100001, 1);
vector<long long> pr;
void seive(long long n) {
  p[0] = p[1] = 0;
  for (long long i = 2; i * i <= n; i++) {
    if (p[i] == 0) continue;
    for (long long j = i * i; j <= n; j += i) {
      p[j] = 0;
    }
  }
  for (long long i = 2; i < n + 1; i++)
    if (p[i]) {
      pr.push_back(i);
    }
}
signed solve() {
  long long n;
  cin >> n;
  vector<long long> v(n);
  for (long long i = 0; i < n; i++) cin >> v[i];
  map<long long, long long> m;
  for (long long i = 0; i < n; i++) {
    long long mx = 0;
    for (long long j = 1; j * j <= v[i]; j++) {
      if (v[i] % j == 0) {
        if (j != 1) mx = max(mx, m[j]);
        mx = max(mx, m[v[i] / j]);
      }
    }
    for (long long j = 1; j * j <= v[i]; j++) {
      if (v[i] % j == 0) {
        if (j != 1) m[j] = mx + 1;
        m[v[i] / j] = mx + 1;
      }
    }
  }
  long long mx = 1;
  for (auto c = (m).begin(); c != (m).end(); c++) mx = max(mx, c->second);
  cout << mx;
  return 0;
}
signed main() {
  long long t = 1;
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  while (t--) {
    solve();
  }
  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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  std::cerr << name << " : " << arg1 << 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...);
}
long long hell = round(1e18);
vector<long long> p(100001, 1);
vector<long long> pr;
void seive(long long n) {
  p[0] = p[1] = 0;
  for (long long i = 2; i * i <= n; i++) {
    if (p[i] == 0) continue;
    for (long long j = i * i; j <= n; j += i) {
      p[j] = 0;
    }
  }
  for (long long i = 2; i < n + 1; i++)
    if (p[i]) {
      pr.push_back(i);
    }
}
signed solve() {
  long long n;
  cin >> n;
  vector<long long> v(n);
  for (long long i = 0; i < n; i++) cin >> v[i];
  map<long long, long long> m;
  for (long long i = 0; i < n; i++) {
    long long mx = 0;
    for (long long j = 1; j * j <= v[i]; j++) {
      if (v[i] % j == 0) {
        if (j != 1) mx = max(mx, m[j]);
        mx = max(mx, m[v[i] / j]);
      }
    }
    for (long long j = 1; j * j <= v[i]; j++) {
      if (v[i] % j == 0) {
        if (j != 1) m[j] = mx + 1;
        m[v[i] / j] = mx + 1;
      }
    }
  }
  long long mx = 1;
  for (auto c = (m).begin(); c != (m).end(); c++) mx = max(mx, c->second);
  cout << mx;
  return 0;
}
signed main() {
  long long t = 1;
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  while (t--) {
    solve();
  }
  return 0;
}

```