#include <bits/stdc++.h>
using namespace std;
const int ppr = 257;
const long long INF = 1e18;
const int inf = 1e9;
const int mod = 1e9 + 7;
const int N = 2e5 + 123;
const long double pi = 3.141592653589793238462643;
const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};
int n, x, dp[N], mx, ans;
vector<int> v;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    for (int j = 2; j * j <= x; j++)
      if (x % j == 0) {
        while (x % j == 0) x /= j;
        v.push_back(j);
        mx = max(mx, dp[j]);
      }
    if (x > 1) {
      v.push_back(x);
      mx = max(mx, dp[x]);
    }
    for (int j = 0; j < v.size(); j++) dp[v[j]] = mx + 1;
    ans = max(ans, mx + 1);
    v.clear();
    mx = 0;
  }
  cout << ans;
  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;
const int ppr = 257;
const long long INF = 1e18;
const int inf = 1e9;
const int mod = 1e9 + 7;
const int N = 2e5 + 123;
const long double pi = 3.141592653589793238462643;
const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};
int n, x, dp[N], mx, ans;
vector<int> v;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    for (int j = 2; j * j <= x; j++)
      if (x % j == 0) {
        while (x % j == 0) x /= j;
        v.push_back(j);
        mx = max(mx, dp[j]);
      }
    if (x > 1) {
      v.push_back(x);
      mx = max(mx, dp[x]);
    }
    for (int j = 0; j < v.size(); j++) dp[v[j]] = mx + 1;
    ans = max(ans, mx + 1);
    v.clear();
    mx = 0;
  }
  cout << ans;
  return 0;
}

```