#include <bits/stdc++.h>
int GCD(int a, int b) {
  while (b) b ^= a ^= b ^= a %= b;
  return a;
}
int LCM(int a, int b) { return a * (b / GCD(a, b)); }
using namespace std;
int main() {
  int N;
  cin >> N;
  long long NN = 0;
  vector<long long> a(N, 0);
  for (int i = 0; i < a.size(); i++) {
    cin >> a[i];
    NN = max(NN, a[i]);
  }
  vector<bool> prime(NN + 1, true);
  for (long long i = 2; i <= sqrt(NN) + 1; i++) {
    if (prime[i])
      for (long long j = i * i; j <= NN + 1; j += i) {
        prime[j] = false;
      }
  }
  vector<long long> primes;
  for (int i = 2; i <= NN; i++) {
    if (prime[i]) primes.push_back(i);
  }
  vector<vector<int> > prime_to_num(NN + 1, vector<int>());
  vector<vector<int> > num_to_prime(NN + 1, vector<int>());
  for (int i = 0; i < N; i++) {
    int total = 0;
    long long init = a[i];
    for (int j = 0; j < primes.size(); j++) {
      if (a[i] % primes[j] == 0) {
        while (init != 1 && init % primes[j] == 0) {
          init /= primes[j];
        }
        prime_to_num[primes[j]].push_back(a[i]);
        num_to_prime[a[i]].push_back(primes[j]);
        total++;
      }
      if (primes[j] * primes[j] > a[i]) break;
    }
    if (init != 1) {
      if (init == a[i]) {
        num_to_prime[a[i]].push_back(a[i]);
        prime_to_num[a[i]].push_back(a[i]);
      } else {
        num_to_prime[a[i]].push_back(init);
        prime_to_num[init].push_back(a[i]);
      }
    }
  }
  vector<int> dp(NN + 1, 1);
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < num_to_prime[a[i]].size(); j++) {
      vector<int>::iterator it =
          lower_bound(((prime_to_num[num_to_prime[a[i]][j]]).begin()),
                      ((prime_to_num[num_to_prime[a[i]][j]]).end()), a[i]);
      if (it != prime_to_num[num_to_prime[a[i]][j]].end() &&
          it != prime_to_num[num_to_prime[a[i]][j]].begin()) {
        it--;
        dp[a[i]] = max(dp[a[i]], dp[*it] + 1);
      }
    }
  }
  int max_len = 0;
  for (int i = 0; i < NN + 1; i++) {
    max_len = max(dp[i], max_len);
  }
  cout << max_len << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution 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>
int GCD(int a, int b) {
  while (b) b ^= a ^= b ^= a %= b;
  return a;
}
int LCM(int a, int b) { return a * (b / GCD(a, b)); }
using namespace std;
int main() {
  int N;
  cin >> N;
  long long NN = 0;
  vector<long long> a(N, 0);
  for (int i = 0; i < a.size(); i++) {
    cin >> a[i];
    NN = max(NN, a[i]);
  }
  vector<bool> prime(NN + 1, true);
  for (long long i = 2; i <= sqrt(NN) + 1; i++) {
    if (prime[i])
      for (long long j = i * i; j <= NN + 1; j += i) {
        prime[j] = false;
      }
  }
  vector<long long> primes;
  for (int i = 2; i <= NN; i++) {
    if (prime[i]) primes.push_back(i);
  }
  vector<vector<int> > prime_to_num(NN + 1, vector<int>());
  vector<vector<int> > num_to_prime(NN + 1, vector<int>());
  for (int i = 0; i < N; i++) {
    int total = 0;
    long long init = a[i];
    for (int j = 0; j < primes.size(); j++) {
      if (a[i] % primes[j] == 0) {
        while (init != 1 && init % primes[j] == 0) {
          init /= primes[j];
        }
        prime_to_num[primes[j]].push_back(a[i]);
        num_to_prime[a[i]].push_back(primes[j]);
        total++;
      }
      if (primes[j] * primes[j] > a[i]) break;
    }
    if (init != 1) {
      if (init == a[i]) {
        num_to_prime[a[i]].push_back(a[i]);
        prime_to_num[a[i]].push_back(a[i]);
      } else {
        num_to_prime[a[i]].push_back(init);
        prime_to_num[init].push_back(a[i]);
      }
    }
  }
  vector<int> dp(NN + 1, 1);
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < num_to_prime[a[i]].size(); j++) {
      vector<int>::iterator it =
          lower_bound(((prime_to_num[num_to_prime[a[i]][j]]).begin()),
                      ((prime_to_num[num_to_prime[a[i]][j]]).end()), a[i]);
      if (it != prime_to_num[num_to_prime[a[i]][j]].end() &&
          it != prime_to_num[num_to_prime[a[i]][j]].begin()) {
        it--;
        dp[a[i]] = max(dp[a[i]], dp[*it] + 1);
      }
    }
  }
  int max_len = 0;
  for (int i = 0; i < NN + 1; i++) {
    max_len = max(dp[i], max_len);
  }
  cout << max_len << endl;
  return 0;
}

```