#include <bits/stdc++.h>
using namespace std;
bool sieve[100000];
int primes[100000];
int dp[100000] = {0};
int divis[100000] = {0};
int a[100000];
int main() {
  long long j, x, n;
  cin >> n;
  int m = 0;
  for (long long i = 2; i < 100000; ++i) {
    if (!sieve[i]) {
      for (long long j = i * i; j < 100000; j += i) sieve[j] = true;
      primes[m++] = i;
    }
  }
  for (j = 0; j < n; ++j) cin >> a[j];
  for (int i = 0; i < n; ++i) {
    x = a[i];
    for (int k = 0; primes[k] * primes[k] <= x; ++k) {
      if (x % primes[k] == 0) {
        j = primes[k];
        dp[i] = max(dp[i], divis[j] + 1);
        while (x % j == 0) x /= j;
      }
    }
    if (x > 1) {
      dp[i] = max(dp[i], divis[x] + 1);
    }
    x = a[i];
    for (int k = 0; primes[k] * primes[k] <= x; ++k) {
      if (x % primes[k] == 0) {
        j = primes[k];
        while (x % j == 0) x /= j;
        divis[j] = dp[i];
      }
    }
    if (x > 1) {
      divis[x] = dp[i];
    }
  }
  int ans = *max_element(dp, dp + n);
  if (ans == 0) ++ans;
  cout << ans << endl;
  return 0;
}

### Prompt

Please create a solution in CPP 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 sieve[100000];
int primes[100000];
int dp[100000] = {0};
int divis[100000] = {0};
int a[100000];
int main() {
  long long j, x, n;
  cin >> n;
  int m = 0;
  for (long long i = 2; i < 100000; ++i) {
    if (!sieve[i]) {
      for (long long j = i * i; j < 100000; j += i) sieve[j] = true;
      primes[m++] = i;
    }
  }
  for (j = 0; j < n; ++j) cin >> a[j];
  for (int i = 0; i < n; ++i) {
    x = a[i];
    for (int k = 0; primes[k] * primes[k] <= x; ++k) {
      if (x % primes[k] == 0) {
        j = primes[k];
        dp[i] = max(dp[i], divis[j] + 1);
        while (x % j == 0) x /= j;
      }
    }
    if (x > 1) {
      dp[i] = max(dp[i], divis[x] + 1);
    }
    x = a[i];
    for (int k = 0; primes[k] * primes[k] <= x; ++k) {
      if (x % primes[k] == 0) {
        j = primes[k];
        while (x % j == 0) x /= j;
        divis[j] = dp[i];
      }
    }
    if (x > 1) {
      divis[x] = dp[i];
    }
  }
  int ans = *max_element(dp, dp + n);
  if (ans == 0) ++ans;
  cout << ans << endl;
  return 0;
}

```