#include <bits/stdc++.h>
using namespace std;
using namespace std;
unsigned long long mod_pow(unsigned long long num, unsigned long long pow,
                           unsigned long long mod) {
  unsigned long long test, n = num;
  for (test = 1; pow; pow >>= 1) {
    if (pow & 1) test = ((test % mod) * (n % mod)) % mod;
    n = ((n % mod) * (n % mod)) % mod;
  }
  return test;
}
long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
vector<int> divsior[100005];
int dp[100005];
int main() {
  int x = 0;
  for (int i = 2; i < 100005; ++i) {
    if (divsior[i].size() == 0) {
      for (int j = i + i; j < 100005; j += i) {
        divsior[j].push_back(i);
        x++;
      }
    }
  }
  int n;
  scanf("%d", &n);
  std::vector<int> v;
  for (int i = 0; i < n; ++i) {
    int x;
    scanf("%d", &x);
    v.push_back(x);
    dp[x] = 1;
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < divsior[v[i]].size(); ++j) {
      int d = divsior[v[i]][j];
      dp[v[i]] = max(dp[v[i]], 1 + dp[d]);
    }
    for (int j = 0; j < divsior[v[i]].size(); ++j) {
      int d = divsior[v[i]][j];
      dp[d] = max(dp[d], dp[v[i]]);
    }
  }
  int ans = -1;
  for (int i = 0; i < 100005; ++i) {
    ans = max(ans, dp[i]);
  }
  printf("%d\n", ans);
}

### 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;
using namespace std;
unsigned long long mod_pow(unsigned long long num, unsigned long long pow,
                           unsigned long long mod) {
  unsigned long long test, n = num;
  for (test = 1; pow; pow >>= 1) {
    if (pow & 1) test = ((test % mod) * (n % mod)) % mod;
    n = ((n % mod) * (n % mod)) % mod;
  }
  return test;
}
long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
vector<int> divsior[100005];
int dp[100005];
int main() {
  int x = 0;
  for (int i = 2; i < 100005; ++i) {
    if (divsior[i].size() == 0) {
      for (int j = i + i; j < 100005; j += i) {
        divsior[j].push_back(i);
        x++;
      }
    }
  }
  int n;
  scanf("%d", &n);
  std::vector<int> v;
  for (int i = 0; i < n; ++i) {
    int x;
    scanf("%d", &x);
    v.push_back(x);
    dp[x] = 1;
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < divsior[v[i]].size(); ++j) {
      int d = divsior[v[i]][j];
      dp[v[i]] = max(dp[v[i]], 1 + dp[d]);
    }
    for (int j = 0; j < divsior[v[i]].size(); ++j) {
      int d = divsior[v[i]][j];
      dp[d] = max(dp[d], dp[v[i]]);
    }
  }
  int ans = -1;
  for (int i = 0; i < 100005; ++i) {
    ans = max(ans, dp[i]);
  }
  printf("%d\n", ans);
}

```