#include <bits/stdc++.h>
using namespace std;
template <class T>
inline T bigmod(T b, T p, T M) {
  long long int ret = 1;
  for (; p > 0; p >>= 1) {
    if (p & 1) ret = (ret * b) % M;
    b = (b * b) % M;
  }
  return (T)ret;
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T lcm(T a, T b) {
  return a * b / gcd<T>(a, b);
}
template <class T>
inline T power(T b, T p) {
  long long int x = 1;
  while (p--) x *= b;
  return x;
}
const long long int mod = 1e9 + 7;
const int siz = 202;
int vis[100110], diviCnt[100110];
vector<int> fac[100110];
void primeFactor() {
  for (int i = 2; i <= 100110; i++) {
    if (vis[i]) continue;
    for (int j = i; j <= 100110; j += i) {
      fac[j].push_back(i);
      vis[j] = 1;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  ;
  primeFactor();
  int n;
  cin >> n;
  int ans = 1;
  for (int i = 1; i <= n; i++) {
    int a;
    cin >> a;
    int have = 0;
    for (int f : fac[a]) have = max(have, diviCnt[f]);
    for (int f : fac[a]) diviCnt[f] = have + 1;
    ans = max(ans, have + 1);
  }
  cout << ans << 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>
using namespace std;
template <class T>
inline T bigmod(T b, T p, T M) {
  long long int ret = 1;
  for (; p > 0; p >>= 1) {
    if (p & 1) ret = (ret * b) % M;
    b = (b * b) % M;
  }
  return (T)ret;
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T lcm(T a, T b) {
  return a * b / gcd<T>(a, b);
}
template <class T>
inline T power(T b, T p) {
  long long int x = 1;
  while (p--) x *= b;
  return x;
}
const long long int mod = 1e9 + 7;
const int siz = 202;
int vis[100110], diviCnt[100110];
vector<int> fac[100110];
void primeFactor() {
  for (int i = 2; i <= 100110; i++) {
    if (vis[i]) continue;
    for (int j = i; j <= 100110; j += i) {
      fac[j].push_back(i);
      vis[j] = 1;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  ;
  primeFactor();
  int n;
  cin >> n;
  int ans = 1;
  for (int i = 1; i <= n; i++) {
    int a;
    cin >> a;
    int have = 0;
    for (int f : fac[a]) have = max(have, diviCnt[f]);
    for (int f : fac[a]) diviCnt[f] = have + 1;
    ans = max(ans, have + 1);
  }
  cout << ans << endl;
  return 0;
}

```