#include <bits/stdc++.h>
using namespace std;
template <typename T>
std::ostream& operator<<(std::ostream& os, const std::vector<T>& vector) {
  for (size_t i = 0; i < vector.size(); ++i) {
    os << vector[i] << " ";
  }
  os << std::endl;
  return os;
}
int gcd(int a, int b) {
  if (b == 0) {
    return a;
  } else {
    return gcd(b, a % b);
  }
}
void Print(int* begin, int n) {
  for (int i = 0; i < n; ++i) {
    printf("%d ", *(begin + i));
  }
  printf("\n");
}
const int max_size = 100010;
int n;
int a[max_size], d[max_size], dp[max_size];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  int max_len = 1;
  for (int i = 1; i <= n; ++i) {
    dp[a[i]] = 0;
    for (int j = 2; j <= sqrt(a[i]); ++j) {
      if (a[i] % j == 0) {
        dp[a[i]] = max(dp[a[i]], d[j]);
        dp[a[i]] = max(dp[a[i]], d[a[i] / j]);
      }
    }
    ++dp[a[i]];
    max_len = max(max_len, dp[a[i]]);
    for (int j = 2; j <= sqrt(a[i]); ++j) {
      if (a[i] % j == 0) {
        d[j] = dp[a[i]];
        d[a[i] / j] = dp[a[i]];
      }
    }
    d[a[i]] = dp[a[i]];
  }
  printf("%d\n", max_len);
  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;
template <typename T>
std::ostream& operator<<(std::ostream& os, const std::vector<T>& vector) {
  for (size_t i = 0; i < vector.size(); ++i) {
    os << vector[i] << " ";
  }
  os << std::endl;
  return os;
}
int gcd(int a, int b) {
  if (b == 0) {
    return a;
  } else {
    return gcd(b, a % b);
  }
}
void Print(int* begin, int n) {
  for (int i = 0; i < n; ++i) {
    printf("%d ", *(begin + i));
  }
  printf("\n");
}
const int max_size = 100010;
int n;
int a[max_size], d[max_size], dp[max_size];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  int max_len = 1;
  for (int i = 1; i <= n; ++i) {
    dp[a[i]] = 0;
    for (int j = 2; j <= sqrt(a[i]); ++j) {
      if (a[i] % j == 0) {
        dp[a[i]] = max(dp[a[i]], d[j]);
        dp[a[i]] = max(dp[a[i]], d[a[i] / j]);
      }
    }
    ++dp[a[i]];
    max_len = max(max_len, dp[a[i]]);
    for (int j = 2; j <= sqrt(a[i]); ++j) {
      if (a[i] % j == 0) {
        d[j] = dp[a[i]];
        d[a[i] / j] = dp[a[i]];
      }
    }
    d[a[i]] = dp[a[i]];
  }
  printf("%d\n", max_len);
  return 0;
}

```