#include <bits/stdc++.h>
using namespace std;
long double pi = acos(-1.0);
long long int power(long long int x, long long int y) {
  long long int temp;
  if (y == 0) return 1;
  temp = power(x, y / 2);
  if (y % 2 == 0)
    return (temp * temp);
  else {
    if (y > 0)
      return (((x * temp)) * temp);
    else
      return (temp * temp) / x;
  }
}
long long int mul_inv(long long int n, long long int p, long long int m) {
  if (p == 1) {
    return n % m;
  } else if (p % 2 == 0) {
    long long int t = mul_inv(n, p / 2, m);
    return ((t) * (t)) % m;
  } else {
    long long int t = mul_inv(n, p / 2, m);
    return ((((n % m) * t) % m) * t) % m;
  }
}
bool isprime(long long int n) {
  if (n == 2) {
    return true;
  }
  if (n % 2 == 0) {
    return false;
  }
  for (long long int i = 3; i <= sqrt(n); i += 2) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<vector<int> > pf(100005);
  int temp[100005] = {0};
  vector<int> prime(100005);
  for (int i = 2; i <= 100005; i++) {
    if (temp[i] == 0) {
      pf[i].push_back(i);
      for (int j = 2 * i; j <= 100005; j += i) {
        pf[j].push_back(i);
        temp[j] = 1;
      }
    }
  }
  int a[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  int ans = 0;
  for (int i = 0; i < n; i++) {
    int cur = 0;
    for (int j = 0; j < pf[a[i]].size(); j++) {
      cur = max(cur, prime[pf[a[i]][j]]);
    }
    cur++;
    ans = max(ans, cur);
    for (int j = 0; j < pf[a[i]].size(); j++) {
      prime[pf[a[i]][j]] = cur;
    }
  }
  cout << ans;
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
long double pi = acos(-1.0);
long long int power(long long int x, long long int y) {
  long long int temp;
  if (y == 0) return 1;
  temp = power(x, y / 2);
  if (y % 2 == 0)
    return (temp * temp);
  else {
    if (y > 0)
      return (((x * temp)) * temp);
    else
      return (temp * temp) / x;
  }
}
long long int mul_inv(long long int n, long long int p, long long int m) {
  if (p == 1) {
    return n % m;
  } else if (p % 2 == 0) {
    long long int t = mul_inv(n, p / 2, m);
    return ((t) * (t)) % m;
  } else {
    long long int t = mul_inv(n, p / 2, m);
    return ((((n % m) * t) % m) * t) % m;
  }
}
bool isprime(long long int n) {
  if (n == 2) {
    return true;
  }
  if (n % 2 == 0) {
    return false;
  }
  for (long long int i = 3; i <= sqrt(n); i += 2) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<vector<int> > pf(100005);
  int temp[100005] = {0};
  vector<int> prime(100005);
  for (int i = 2; i <= 100005; i++) {
    if (temp[i] == 0) {
      pf[i].push_back(i);
      for (int j = 2 * i; j <= 100005; j += i) {
        pf[j].push_back(i);
        temp[j] = 1;
      }
    }
  }
  int a[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  int ans = 0;
  for (int i = 0; i < n; i++) {
    int cur = 0;
    for (int j = 0; j < pf[a[i]].size(); j++) {
      cur = max(cur, prime[pf[a[i]][j]]);
    }
    cur++;
    ans = max(ans, cur);
    for (int j = 0; j < pf[a[i]].size(); j++) {
      prime[pf[a[i]][j]] = cur;
    }
  }
  cout << ans;
  return 0;
}

```