#include <bits/stdc++.h>
using namespace std;
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> a) {
  out << a.first << " " << a.second;
  return out;
}
template <typename T, typename T1>
T maxs(T& a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T mins(T& a, T1 b) {
  if (b < a) a = b;
  return a;
}
long long solve() {
  long long n;
  cin >> n;
  vector<long long> a(n + 1, 0);
  for (long long i = 1; i < n + 1; i++) cin >> a[i];
  vector<long long> dp(n + 1, 0);
  vector<long long> dp2(1e5 + 1, 0);
  long long mx = 1;
  for (long long i = 1; i < n + 1; i++) {
    long long temp = a[i];
    dp[i] = 1;
    for (long long j = 2; j <= sqrt(temp); j++) {
      if (temp % j == 0) {
        while (temp % j == 0) temp /= j;
        maxs(dp[i], 1 + dp2[j]);
      }
    }
    if (temp > 1) maxs(dp[i], 1 + dp2[temp]);
    temp = a[i];
    for (long long j = 2; j <= sqrt(temp); j++) {
      if (temp % j == 0) {
        while (temp % j == 0) temp /= j;
        maxs(dp2[j], dp[i]);
      }
    }
    if (temp > 1) maxs(dp2[temp], dp[i]);
    maxs(mx, dp[i]);
  }
  cout << mx << "\n";
  return 0;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### 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;
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> a) {
  out << a.first << " " << a.second;
  return out;
}
template <typename T, typename T1>
T maxs(T& a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T mins(T& a, T1 b) {
  if (b < a) a = b;
  return a;
}
long long solve() {
  long long n;
  cin >> n;
  vector<long long> a(n + 1, 0);
  for (long long i = 1; i < n + 1; i++) cin >> a[i];
  vector<long long> dp(n + 1, 0);
  vector<long long> dp2(1e5 + 1, 0);
  long long mx = 1;
  for (long long i = 1; i < n + 1; i++) {
    long long temp = a[i];
    dp[i] = 1;
    for (long long j = 2; j <= sqrt(temp); j++) {
      if (temp % j == 0) {
        while (temp % j == 0) temp /= j;
        maxs(dp[i], 1 + dp2[j]);
      }
    }
    if (temp > 1) maxs(dp[i], 1 + dp2[temp]);
    temp = a[i];
    for (long long j = 2; j <= sqrt(temp); j++) {
      if (temp % j == 0) {
        while (temp % j == 0) temp /= j;
        maxs(dp2[j], dp[i]);
      }
    }
    if (temp > 1) maxs(dp2[temp], dp[i]);
    maxs(mx, dp[i]);
  }
  cout << mx << "\n";
  return 0;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```