#include <bits/stdc++.h>
using namespace std;
int n;
vector<int> f(100001), a, dp, mp(100001);
vector<vector<int> > af;
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  a.resize(n);
  af.resize(n);
  dp.resize(n);
  for (int i = (0); i < (n); i++) cin >> a[i];
  f[0] = f[1] = 1;
  for (int i = (2); i < (100001); i++) {
    if (!f[i]) {
      f[i] = i;
      for (long long j = ((long long)i) * i; j < 100001; j += i) {
        f[j] = i;
      }
    }
  }
  for (int i = (0); i < (n); i++) {
    int t = a[i];
    while (t > 1) {
      if ((!af[i].size()) || af[i][af[i].size() - 1] != f[t])
        af[i].push_back(f[t]);
      t /= f[t];
    }
  }
  for (int i = (0); i < (100001); i++) mp[i] = 0;
  dp[0] = 1;
  for (int i = (0); i < (af[0].size()); i++) {
    mp[af[0][i]] = 1;
  }
  for (int i = (1); i < (n); i++) {
    int mx = 0;
    for (int j = (0); j < (af[i].size()); j++) {
      mx = max(mx, mp[af[i][j]]);
    }
    mx += 1;
    dp[i] = mx;
    for (int j = (0); j < (af[i].size()); j++) {
      mp[af[i][j]] = mx;
    }
  }
  int mx = 0;
  for (int i = (0); i < (n); i++) mx = max(mx, dp[i]);
  cout << mx;
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
int n;
vector<int> f(100001), a, dp, mp(100001);
vector<vector<int> > af;
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  a.resize(n);
  af.resize(n);
  dp.resize(n);
  for (int i = (0); i < (n); i++) cin >> a[i];
  f[0] = f[1] = 1;
  for (int i = (2); i < (100001); i++) {
    if (!f[i]) {
      f[i] = i;
      for (long long j = ((long long)i) * i; j < 100001; j += i) {
        f[j] = i;
      }
    }
  }
  for (int i = (0); i < (n); i++) {
    int t = a[i];
    while (t > 1) {
      if ((!af[i].size()) || af[i][af[i].size() - 1] != f[t])
        af[i].push_back(f[t]);
      t /= f[t];
    }
  }
  for (int i = (0); i < (100001); i++) mp[i] = 0;
  dp[0] = 1;
  for (int i = (0); i < (af[0].size()); i++) {
    mp[af[0][i]] = 1;
  }
  for (int i = (1); i < (n); i++) {
    int mx = 0;
    for (int j = (0); j < (af[i].size()); j++) {
      mx = max(mx, mp[af[i][j]]);
    }
    mx += 1;
    dp[i] = mx;
    for (int j = (0); j < (af[i].size()); j++) {
      mp[af[i][j]] = mx;
    }
  }
  int mx = 0;
  for (int i = (0); i < (n); i++) mx = max(mx, dp[i]);
  cout << mx;
  return 0;
}

```