#include <bits/stdc++.h>
using namespace std;
bool prime[1000001];
int spf[1000001];
vector<int> primes;
void sieve() {
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= 1000000; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= 1000000; i += p) {
        prime[i] = false;
        spf[i] = p;
      }
    }
  }
  for (int p = 2; p <= 1000000; p++)
    if (prime[p]) primes.push_back(p);
}
int dp[200005];
int main() {
  int n;
  cin >> n;
  int a[n];
  for (long long int i = 0; i < n; i++) cin >> a[i];
  map<int, int> m;
  for (long long int i = 0; i < n; i++) {
    vector<int> factors;
    int x = a[i];
    int sq = sqrt(a[i]) + 1;
    for (long long int j = 2; j < sq + 1; j++) {
      if (x % j == 0) {
        factors.push_back(j);
        while (x % j == 0) x /= j;
      }
    }
    if (x > 2) factors.push_back(x);
    for (long long int j = 0; j < factors.size(); j++) {
      if (m[factors[j]] > 0)
        dp[i] = max(dp[i], m[factors[j]] + 1);
      else {
        dp[i] = max(dp[i], 1);
      }
    }
    for (long long int j = 0; j < factors.size(); j++) {
      m[factors[j]] = dp[i];
    }
  }
  int ans = *max_element(dp, dp + n);
  if (ans == 0) ans++;
  cout << ans;
}

### 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>
using namespace std;
bool prime[1000001];
int spf[1000001];
vector<int> primes;
void sieve() {
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= 1000000; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= 1000000; i += p) {
        prime[i] = false;
        spf[i] = p;
      }
    }
  }
  for (int p = 2; p <= 1000000; p++)
    if (prime[p]) primes.push_back(p);
}
int dp[200005];
int main() {
  int n;
  cin >> n;
  int a[n];
  for (long long int i = 0; i < n; i++) cin >> a[i];
  map<int, int> m;
  for (long long int i = 0; i < n; i++) {
    vector<int> factors;
    int x = a[i];
    int sq = sqrt(a[i]) + 1;
    for (long long int j = 2; j < sq + 1; j++) {
      if (x % j == 0) {
        factors.push_back(j);
        while (x % j == 0) x /= j;
      }
    }
    if (x > 2) factors.push_back(x);
    for (long long int j = 0; j < factors.size(); j++) {
      if (m[factors[j]] > 0)
        dp[i] = max(dp[i], m[factors[j]] + 1);
      else {
        dp[i] = max(dp[i], 1);
      }
    }
    for (long long int j = 0; j < factors.size(); j++) {
      m[factors[j]] = dp[i];
    }
  }
  int ans = *max_element(dp, dp + n);
  if (ans == 0) ans++;
  cout << ans;
}

```