#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 1;
int n, a[N], dp[N];
vector<int> p, d[N], nxt[N];
bitset<N> bs;
void add(int at) {
  int x = a[at];
  for (size_t i = 0; i < p.size() && p[i] * p[i] <= x; ++i)
    if (x % p[i] == 0) {
      d[at].push_back(p[i]);
      nxt[p[i]].push_back(at);
      while (x % p[i] == 0) x /= p[i];
    }
  if (x != 1) {
    d[at].push_back(x);
    nxt[x].push_back(at);
  }
}
int calc(int at) {
  int &res = dp[at];
  if (res != -1) return res;
  res = 0;
  for (size_t i = 0; i < d[at].size(); ++i) {
    int v = d[at][i];
    int k = upper_bound(nxt[v].begin(), nxt[v].end(), at) - nxt[v].begin();
    if (k != (int)nxt[v].size()) res = max(res, calc(nxt[v][k]) + 1);
  }
  return res;
}
int main() {
  bs[0] = bs[1] = 1;
  for (int i = 2; i * i < N; ++i)
    if (!bs[i])
      for (int j = i * i; j < N; j += i) bs[j] = 1;
  for (int i = 0; i < N; ++i)
    if (!bs[i]) p.push_back(i);
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    add(i);
    dp[i] = -1;
  }
  int res = 0;
  for (int i = 0; i < n; ++i) res = max(res, calc(i) + 1);
  printf("%d\n", res);
  return 0;
}

### Prompt

Your task is to create 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 N = 1e5 + 1;
int n, a[N], dp[N];
vector<int> p, d[N], nxt[N];
bitset<N> bs;
void add(int at) {
  int x = a[at];
  for (size_t i = 0; i < p.size() && p[i] * p[i] <= x; ++i)
    if (x % p[i] == 0) {
      d[at].push_back(p[i]);
      nxt[p[i]].push_back(at);
      while (x % p[i] == 0) x /= p[i];
    }
  if (x != 1) {
    d[at].push_back(x);
    nxt[x].push_back(at);
  }
}
int calc(int at) {
  int &res = dp[at];
  if (res != -1) return res;
  res = 0;
  for (size_t i = 0; i < d[at].size(); ++i) {
    int v = d[at][i];
    int k = upper_bound(nxt[v].begin(), nxt[v].end(), at) - nxt[v].begin();
    if (k != (int)nxt[v].size()) res = max(res, calc(nxt[v][k]) + 1);
  }
  return res;
}
int main() {
  bs[0] = bs[1] = 1;
  for (int i = 2; i * i < N; ++i)
    if (!bs[i])
      for (int j = i * i; j < N; j += i) bs[j] = 1;
  for (int i = 0; i < N; ++i)
    if (!bs[i]) p.push_back(i);
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    add(i);
    dp[i] = -1;
  }
  int res = 0;
  for (int i = 0; i < n; ++i) res = max(res, calc(i) + 1);
  printf("%d\n", res);
  return 0;
}

```