#include <bits/stdc++.h>
using namespace std;
long long dp[200005];
void get_all_primes(long long upto_primes, vector<long long> &primes) {
  primes.clear();
  vector<bool> mark(upto_primes + 2, true);
  for (long long i = 3; i * i <= upto_primes; i += 2) {
    if (mark[i]) {
      for (long long j = i * i; j <= upto_primes; j += 2 * i) {
        mark[j] = false;
      }
    }
  }
  primes.push_back(2);
  for (long long i = 3; i <= upto_primes; i += 2) {
    if (mark[i]) {
      primes.push_back(i);
    }
  };
}
void solve() {
  long long n;
  cin >> n;
  if (n == 1) {
    cout << 1 << '\n';
    return;
  }
  vector<long long> primes;
  get_all_primes(100000, primes);
  auto *pDiv = new vector<long long>[100001];
  for (auto i : primes) {
    for (long long j = i; j <= 100000; j += i) {
      pDiv[j].push_back(i);
    }
  }
  long long a[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  map<long long, long long> have;
  for (long long i = a[0] == 1; i < n; i++) {
    long long maxi = 0;
    for (auto p : pDiv[a[i]]) {
      maxi = max(maxi, have[p]);
    }
    for (auto p : pDiv[a[i]]) {
      have[p] = maxi + 1;
    }
  }
  long long maxi = 0;
  for (auto i : have) {
    maxi = max(maxi, i.second);
  }
  cout << maxi << '\n';
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  long long t = 1;
  while (t--) solve();
  return 0;
}

### Prompt

Your task is to create 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>
using namespace std;
long long dp[200005];
void get_all_primes(long long upto_primes, vector<long long> &primes) {
  primes.clear();
  vector<bool> mark(upto_primes + 2, true);
  for (long long i = 3; i * i <= upto_primes; i += 2) {
    if (mark[i]) {
      for (long long j = i * i; j <= upto_primes; j += 2 * i) {
        mark[j] = false;
      }
    }
  }
  primes.push_back(2);
  for (long long i = 3; i <= upto_primes; i += 2) {
    if (mark[i]) {
      primes.push_back(i);
    }
  };
}
void solve() {
  long long n;
  cin >> n;
  if (n == 1) {
    cout << 1 << '\n';
    return;
  }
  vector<long long> primes;
  get_all_primes(100000, primes);
  auto *pDiv = new vector<long long>[100001];
  for (auto i : primes) {
    for (long long j = i; j <= 100000; j += i) {
      pDiv[j].push_back(i);
    }
  }
  long long a[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  map<long long, long long> have;
  for (long long i = a[0] == 1; i < n; i++) {
    long long maxi = 0;
    for (auto p : pDiv[a[i]]) {
      maxi = max(maxi, have[p]);
    }
    for (auto p : pDiv[a[i]]) {
      have[p] = maxi + 1;
    }
  }
  long long maxi = 0;
  for (auto i : have) {
    maxi = max(maxi, i.second);
  }
  cout << maxi << '\n';
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  long long t = 1;
  while (t--) solve();
  return 0;
}

```