#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
bool pf[maxn];
int dp[maxn];
vector<int> pri;
vector<int> v[maxn];
int n;
int a[maxn];
bool pr(int u) {
  for (int i = 2; i <= sqrt(u); i++) {
    if (u % i == 0) {
      return 0;
    }
  }
  return 1;
}
inline void input();
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  input();
}
inline void input() {
  for (int i = 2; i <= sqrt(maxn); i++) {
    if (pr(i)) {
      pri.emplace_back(i);
    }
  }
  for (int i = 0; i < pri.size(); i++) {
    int p = pri[i];
    int o = 2;
    while (1ll * p * o < maxn) {
      pf[p * o] = 1;
      o++;
    }
  }
  pri.clear();
  for (int i = 2; i < maxn; i++) {
    if (pf[i] == 0) {
      pri.push_back(i);
    }
  }
  for (int i = 0; i < pri.size(); i++) {
    int o = pri[i];
    int k = 1;
    while (1ll * o * k < maxn) {
      v[o * k].emplace_back(o);
      k++;
    }
  }
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n; i++) {
    int r = 0;
    for (int u = 0; u < v[a[i]].size(); u++) {
      r = max(r, dp[v[a[i]][u]]);
    }
    for (int u = 0; u < v[a[i]].size(); u++) {
      dp[v[a[i]][u]] = r + 1;
    }
  }
  int res = 0;
  for (int i = 0; i < pri.size(); i++) {
    res = max(res, dp[pri[i]]);
  }
  cout << max(res, 1);
}

### Prompt

Your challenge is to write a cpp solution 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;
const int maxn = 1e5 + 10;
bool pf[maxn];
int dp[maxn];
vector<int> pri;
vector<int> v[maxn];
int n;
int a[maxn];
bool pr(int u) {
  for (int i = 2; i <= sqrt(u); i++) {
    if (u % i == 0) {
      return 0;
    }
  }
  return 1;
}
inline void input();
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  input();
}
inline void input() {
  for (int i = 2; i <= sqrt(maxn); i++) {
    if (pr(i)) {
      pri.emplace_back(i);
    }
  }
  for (int i = 0; i < pri.size(); i++) {
    int p = pri[i];
    int o = 2;
    while (1ll * p * o < maxn) {
      pf[p * o] = 1;
      o++;
    }
  }
  pri.clear();
  for (int i = 2; i < maxn; i++) {
    if (pf[i] == 0) {
      pri.push_back(i);
    }
  }
  for (int i = 0; i < pri.size(); i++) {
    int o = pri[i];
    int k = 1;
    while (1ll * o * k < maxn) {
      v[o * k].emplace_back(o);
      k++;
    }
  }
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n; i++) {
    int r = 0;
    for (int u = 0; u < v[a[i]].size(); u++) {
      r = max(r, dp[v[a[i]][u]]);
    }
    for (int u = 0; u < v[a[i]].size(); u++) {
      dp[v[a[i]][u]] = r + 1;
    }
  }
  int res = 0;
  for (int i = 0; i < pri.size(); i++) {
    res = max(res, dp[pri[i]]);
  }
  cout << max(res, 1);
}

```