#include <bits/stdc++.h>
using namespace std;
template <class T>
using min_heap = priority_queue<T, vector<T>, greater<T>>;
template <class T, class U>
inline void MAX(T &a, U b) {
  if (a < b) a = b;
}
template <class T, class U>
inline void MIN(T &a, U b) {
  if (a > b) a = b;
}
template <class T, class U>
inline bool ckmax(T &a, U b) {
  return b > a ? a = b, true : false;
}
template <class T, class U>
inline bool ckmin(T &a, U b) {
  return b < a ? a = b, true : false;
}
template <class T, class U>
inline bool INR(T i, U l, U r) {
  return l <= i && i <= r;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
  os << '{' << p.first << ',' << p.second << '}';
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &p) {
  os << "[";
  for (int i = 0; i < p.size(); i++)
    os << p[i] << (i == p.size() - 1 ? "" : ",");
  os << "]";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &p) {
  os << "{ ";
  for (T x : p) os << x << " ";
  os << "}";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const multiset<T> &p) {
  os << "{ ";
  for (T x : p) os << x << " ";
  os << "}";
  return os;
}
const int MOD = (int)(1e9) + 7;
const int INF = 0x3f3f3f3f;
const int NINF = -0x3f3f3f40;
const long long INF_L = 0x3f3f3f3f3f3f3f3f;
const long long NINF_L = -0x3f3f3f3f3f3f3f40;
const int MAXN = (int)(1e5 + 5);
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout.precision(20);
  vector<int> sieve(MAXN, 1);
  for (int i = 2; i * i < MAXN; i++) {
    if (sieve[i] == 1)
      for (int j = i * 2; j < MAXN; j += i) sieve[j] = i;
  }
  int n;
  cin >> n;
  vector<int> cmx(MAXN + 69);
  int tmp;
  int ans = 0;
  for (int i = 0; i < n; i++) {
    cin >> tmp;
    int cans = 0;
    set<int> fcts;
    while (sieve[tmp] != 1) {
      fcts.insert(sieve[tmp]);
      ckmax(cans, cmx[sieve[tmp]]);
      tmp /= sieve[tmp];
    }
    fcts.insert(tmp);
    ckmax(cans, cmx[tmp]);
    cans++;
    ckmax(ans, cans);
    for (auto &num : fcts) cmx[num] = cans;
  }
  cout << ans << '\n';
}

### Prompt

Develop a solution in cpp to the problem described below:

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>
using min_heap = priority_queue<T, vector<T>, greater<T>>;
template <class T, class U>
inline void MAX(T &a, U b) {
  if (a < b) a = b;
}
template <class T, class U>
inline void MIN(T &a, U b) {
  if (a > b) a = b;
}
template <class T, class U>
inline bool ckmax(T &a, U b) {
  return b > a ? a = b, true : false;
}
template <class T, class U>
inline bool ckmin(T &a, U b) {
  return b < a ? a = b, true : false;
}
template <class T, class U>
inline bool INR(T i, U l, U r) {
  return l <= i && i <= r;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
  os << '{' << p.first << ',' << p.second << '}';
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &p) {
  os << "[";
  for (int i = 0; i < p.size(); i++)
    os << p[i] << (i == p.size() - 1 ? "" : ",");
  os << "]";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &p) {
  os << "{ ";
  for (T x : p) os << x << " ";
  os << "}";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const multiset<T> &p) {
  os << "{ ";
  for (T x : p) os << x << " ";
  os << "}";
  return os;
}
const int MOD = (int)(1e9) + 7;
const int INF = 0x3f3f3f3f;
const int NINF = -0x3f3f3f40;
const long long INF_L = 0x3f3f3f3f3f3f3f3f;
const long long NINF_L = -0x3f3f3f3f3f3f3f40;
const int MAXN = (int)(1e5 + 5);
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout.precision(20);
  vector<int> sieve(MAXN, 1);
  for (int i = 2; i * i < MAXN; i++) {
    if (sieve[i] == 1)
      for (int j = i * 2; j < MAXN; j += i) sieve[j] = i;
  }
  int n;
  cin >> n;
  vector<int> cmx(MAXN + 69);
  int tmp;
  int ans = 0;
  for (int i = 0; i < n; i++) {
    cin >> tmp;
    int cans = 0;
    set<int> fcts;
    while (sieve[tmp] != 1) {
      fcts.insert(sieve[tmp]);
      ckmax(cans, cmx[sieve[tmp]]);
      tmp /= sieve[tmp];
    }
    fcts.insert(tmp);
    ckmax(cans, cmx[tmp]);
    cans++;
    ckmax(ans, cans);
    for (auto &num : fcts) cmx[num] = cans;
  }
  cout << ans << '\n';
}

```