#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:20000000")
using namespace std;
int ri() {
  int x;
  scanf("%d", &x);
  return x;
}
long long rll() {
  long long x;
  scanf("%lld", &x);
  return x;
}
set<int> st[105000];
int dp[105000];
void solve() {
  int n = ri();
  for (int i = (int)(1); i <= (int)(n); i++) {
    int x = ri();
    for (int j = 2; j * j <= x; j++)
      while (x % j == 0) st[i].insert(j), x /= j;
    if (x != 1) st[i].insert(x);
  }
  int res = 1;
  map<int, int> best;
  for (int i = (int)(1); i <= (int)(n); i++) {
    dp[i] = 1;
    set<int>::iterator it;
    int q = 0;
    for (it = st[i].begin(); it != st[i].end(); it++) {
      if (best.find(*it) != best.end()) q = max(best[*it], q);
    }
    dp[i] = q + 1;
    for (it = st[i].begin(); it != st[i].end(); it++)
      best[*it] = max(best[*it], q + 1);
    res = max(res, dp[i]);
  }
  cout << res << endl;
}
int main() {
  solve();
  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>
#pragma comment(linker, "/STACK:20000000")
using namespace std;
int ri() {
  int x;
  scanf("%d", &x);
  return x;
}
long long rll() {
  long long x;
  scanf("%lld", &x);
  return x;
}
set<int> st[105000];
int dp[105000];
void solve() {
  int n = ri();
  for (int i = (int)(1); i <= (int)(n); i++) {
    int x = ri();
    for (int j = 2; j * j <= x; j++)
      while (x % j == 0) st[i].insert(j), x /= j;
    if (x != 1) st[i].insert(x);
  }
  int res = 1;
  map<int, int> best;
  for (int i = (int)(1); i <= (int)(n); i++) {
    dp[i] = 1;
    set<int>::iterator it;
    int q = 0;
    for (it = st[i].begin(); it != st[i].end(); it++) {
      if (best.find(*it) != best.end()) q = max(best[*it], q);
    }
    dp[i] = q + 1;
    for (it = st[i].begin(); it != st[i].end(); it++)
      best[*it] = max(best[*it], q + 1);
    res = max(res, dp[i]);
  }
  cout << res << endl;
}
int main() {
  solve();
  return 0;
}

```