#include <bits/stdc++.h>
using namespace std;
bool prime(int m) {
  int u = sqrt(m);
  for (int i = 2; i <= u + 1; i++) {
    if (m % i == 0) {
      return false;
    }
  }
  return true;
}
long long max(long long a, long long b) {
  if (a > b) {
    return a;
  } else {
    return b;
  }
}
long long min(long long a, long long b) {
  if (a < b) {
    return a;
  } else {
    return b;
  }
}
bool file = false;
int main() {
  if (file) {
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
  }
  int n;
  cin >> n;
  int a[n + 1];
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  int b[100001], l[100001], v[100001];
  for (int i = 1; i <= 100001; i++) {
    b[i] = -1;
    l[i] = 0;
    v[i] = 0;
  }
  b[0] = -99999;
  for (int i = 1; i <= n; i++) {
    int c = a[i];
    b[c] = max(b[c], 1);
    if (c != 1) {
      for (int j = 2; j <= sqrt(c); j++) {
        if (c % j == 0) {
          if (b[c] < l[j] + 1) {
            b[c] = l[j] + 1;
          }
          if (b[c] < l[c / j] + 1) {
            b[c] = l[c / j] + 1;
          }
        }
      }
      for (int j = 2; j <= sqrt(c); j++) {
        if ((c % j == 0) && (l[j] < b[c])) {
          l[j] = b[c];
        }
        if ((c % j == 0) && (l[c / j] < b[c])) {
          l[c / j] = b[c];
        }
      }
      l[c] = b[c];
    }
  }
  int maxval = 0;
  for (int i = 2; i <= 100000; i++) {
    if (b[i] > maxval) {
      maxval = b[i];
    }
  }
  cout << max(1, maxval);
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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(int m) {
  int u = sqrt(m);
  for (int i = 2; i <= u + 1; i++) {
    if (m % i == 0) {
      return false;
    }
  }
  return true;
}
long long max(long long a, long long b) {
  if (a > b) {
    return a;
  } else {
    return b;
  }
}
long long min(long long a, long long b) {
  if (a < b) {
    return a;
  } else {
    return b;
  }
}
bool file = false;
int main() {
  if (file) {
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
  }
  int n;
  cin >> n;
  int a[n + 1];
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  int b[100001], l[100001], v[100001];
  for (int i = 1; i <= 100001; i++) {
    b[i] = -1;
    l[i] = 0;
    v[i] = 0;
  }
  b[0] = -99999;
  for (int i = 1; i <= n; i++) {
    int c = a[i];
    b[c] = max(b[c], 1);
    if (c != 1) {
      for (int j = 2; j <= sqrt(c); j++) {
        if (c % j == 0) {
          if (b[c] < l[j] + 1) {
            b[c] = l[j] + 1;
          }
          if (b[c] < l[c / j] + 1) {
            b[c] = l[c / j] + 1;
          }
        }
      }
      for (int j = 2; j <= sqrt(c); j++) {
        if ((c % j == 0) && (l[j] < b[c])) {
          l[j] = b[c];
        }
        if ((c % j == 0) && (l[c / j] < b[c])) {
          l[c / j] = b[c];
        }
      }
      l[c] = b[c];
    }
  }
  int maxval = 0;
  for (int i = 2; i <= 100000; i++) {
    if (b[i] > maxval) {
      maxval = b[i];
    }
  }
  cout << max(1, maxval);
  return 0;
}

```