#include <bits/stdc++.h>
long long gcd(long long a, long long b) {
  while (a && b) {
    if (a > b)
      a %= b;
    else
      b %= a;
  }
  return a + b;
}
long long lcm(long long a, long long b) { return (((a * b) / gcd(a, b))); }
int phi(int n) {
  int result = n;
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      while (n % i == 0) n /= i;
      result -= result / i;
    }
  }
  if (n > 1) result -= result / n;
  return result;
}
int catalan(unsigned int n) {
  int c[100];
  c[0] = c[1] = 1;
  for (int i = (int)(2); i < (int)(n); ++i) {
    for (int k = (int)(0); k < (int)(i); ++k) c[i] += c[k] * c[i - 1 - k];
  }
  return c[n];
}
using namespace std;
int n, a[1000100];
int dp[1000100];
int last = 0;
bool u[1000100];
int main() {
  {
    cin >> n;
    for (int i = 0; i < (int)(n); i++) scanf("%d", a + i);
    int mx = 0;
    for (int i = 0; i < n; ++i) {
      int ans = 1;
      long long x = a[i];
      if (x == 1) ans = max(ans, dp[1] + 1);
      for (long long j = 2; j * j <= x; ++j) {
        if (x % j) continue;
        while (x % j == 0) x /= j;
        ans = max(ans, dp[j] + 1);
      }
      if (x > 1) ans = max(ans, dp[x] + 1);
      x = a[i];
      if (x == 1) dp[1] = ans;
      for (long long j = 2; j * j <= x; ++j) {
        if (x % j) continue;
        while (x % j == 0) x /= j;
        dp[j] = ans;
      }
      if (x > 1) dp[x] = max(dp[x], ans);
    }
    for (int i = 0; i < (int)(1000100); i++) mx = max(mx, dp[i]);
    cout << mx << 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>
long long gcd(long long a, long long b) {
  while (a && b) {
    if (a > b)
      a %= b;
    else
      b %= a;
  }
  return a + b;
}
long long lcm(long long a, long long b) { return (((a * b) / gcd(a, b))); }
int phi(int n) {
  int result = n;
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      while (n % i == 0) n /= i;
      result -= result / i;
    }
  }
  if (n > 1) result -= result / n;
  return result;
}
int catalan(unsigned int n) {
  int c[100];
  c[0] = c[1] = 1;
  for (int i = (int)(2); i < (int)(n); ++i) {
    for (int k = (int)(0); k < (int)(i); ++k) c[i] += c[k] * c[i - 1 - k];
  }
  return c[n];
}
using namespace std;
int n, a[1000100];
int dp[1000100];
int last = 0;
bool u[1000100];
int main() {
  {
    cin >> n;
    for (int i = 0; i < (int)(n); i++) scanf("%d", a + i);
    int mx = 0;
    for (int i = 0; i < n; ++i) {
      int ans = 1;
      long long x = a[i];
      if (x == 1) ans = max(ans, dp[1] + 1);
      for (long long j = 2; j * j <= x; ++j) {
        if (x % j) continue;
        while (x % j == 0) x /= j;
        ans = max(ans, dp[j] + 1);
      }
      if (x > 1) ans = max(ans, dp[x] + 1);
      x = a[i];
      if (x == 1) dp[1] = ans;
      for (long long j = 2; j * j <= x; ++j) {
        if (x % j) continue;
        while (x % j == 0) x /= j;
        dp[j] = ans;
      }
      if (x > 1) dp[x] = max(dp[x], ans);
    }
    for (int i = 0; i < (int)(1000100); i++) mx = max(mx, dp[i]);
    cout << mx << endl;
  }
  return 0;
}

```