#include <bits/stdc++.h>
const int N = 100000 + 1;
int prime[N], vis[N], idx;
int d[N], n;
int use[N], idy;
void prepare() {
  idx = 0;
  memset(vis, 0, sizeof vis);
  for (int i = 2; i < N; ++i)
    if (!vis[i]) {
      prime[idx++] = i;
      if ((long long)i * i >= N) continue;
      for (int j = i * i; j < N; j += i) vis[j] = true;
    }
}
int main() {
  prepare();
  int n, a, b, mx;
  scanf("%d", &n);
  memset(d, 0, sizeof d);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a);
    if (1 == a) {
      ++d[1];
      continue;
    }
    b = a;
    mx = 0;
    idy = 0;
    for (int j = 0; j < idx; ++j) {
      if (prime[j] > a) break;
      if (a % prime[j] == 0) {
        use[idy++] = prime[j];
        if (d[prime[j]] > mx) mx = d[prime[j]];
        while (a % prime[j] == 0) a /= prime[j];
      }
    }
    if (a != 1) {
      use[idy++] = a;
      if (d[a] > mx) mx = d[a];
    }
    for (int j = 0; j < idy; ++j) d[use[j]] = mx + 1;
  }
  int ans = 0;
  for (int j = 0; j < N; ++j)
    if (d[j] > ans) ans = d[j];
  printf("%d\n", ans);
  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>
const int N = 100000 + 1;
int prime[N], vis[N], idx;
int d[N], n;
int use[N], idy;
void prepare() {
  idx = 0;
  memset(vis, 0, sizeof vis);
  for (int i = 2; i < N; ++i)
    if (!vis[i]) {
      prime[idx++] = i;
      if ((long long)i * i >= N) continue;
      for (int j = i * i; j < N; j += i) vis[j] = true;
    }
}
int main() {
  prepare();
  int n, a, b, mx;
  scanf("%d", &n);
  memset(d, 0, sizeof d);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a);
    if (1 == a) {
      ++d[1];
      continue;
    }
    b = a;
    mx = 0;
    idy = 0;
    for (int j = 0; j < idx; ++j) {
      if (prime[j] > a) break;
      if (a % prime[j] == 0) {
        use[idy++] = prime[j];
        if (d[prime[j]] > mx) mx = d[prime[j]];
        while (a % prime[j] == 0) a /= prime[j];
      }
    }
    if (a != 1) {
      use[idy++] = a;
      if (d[a] > mx) mx = d[a];
    }
    for (int j = 0; j < idy; ++j) d[use[j]] = mx + 1;
  }
  int ans = 0;
  for (int j = 0; j < N; ++j)
    if (d[j] > ans) ans = d[j];
  printf("%d\n", ans);
  return 0;
}

```