#include <bits/stdc++.h>
const int maxn = 1500000;
inline int max(int a, int b) { return a > b ? a : b; }
int a, n, m;
int k[maxn];
int main() {
  scanf("%d", &n);
  m = 1;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a);
    k[a] = 1;
  }
  int tmp = a;
  for (int i = 1; i <= tmp; ++i)
    if (k[i])
      for (int a = 2 * i; a <= tmp; a += i)
        if (k[a]) {
          k[a] = max(k[a], k[i] + 1);
          m = max(m, k[a]);
        }
  printf("%d\n", m);
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

As you must know, the maximum clique problem in an arbitrary graph is NP-hard. Nevertheless, for some graphs of specific kinds it can be solved effectively.

Just in case, let us remind you that a clique in a non-directed graph is a subset of the vertices of a graph, such that any two vertices of this subset are connected by an edge. In particular, an empty set of vertexes and a set consisting of a single vertex, are cliques.

Let's define a divisibility graph for a set of positive integers A = {a1, a2, ..., an} as follows. The vertices of the given graph are numbers from set A, and two numbers ai and aj (i ≠ j) are connected by an edge if and only if either ai is divisible by aj, or aj is divisible by ai.

You are given a set of non-negative integers A. Determine the size of a maximum clique in a divisibility graph for set A.

Input

The first line contains integer n (1 ≤ n ≤ 106), that sets the size of set A.

The second line contains n distinct positive integers a1, a2, ..., an (1 ≤ ai ≤ 106) — elements of subset A. The numbers in the line follow in the ascending order.

Output

Print a single number — the maximum size of a clique in a divisibility graph for set A.

Examples

Input

8
3 4 6 8 10 18 21 24


Output

3

Note

In the first sample test a clique of size 3 is, for example, a subset of vertexes {3, 6, 18}. A clique of a larger size doesn't exist in this graph.

### Response

```cpp
#include <bits/stdc++.h>
const int maxn = 1500000;
inline int max(int a, int b) { return a > b ? a : b; }
int a, n, m;
int k[maxn];
int main() {
  scanf("%d", &n);
  m = 1;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a);
    k[a] = 1;
  }
  int tmp = a;
  for (int i = 1; i <= tmp; ++i)
    if (k[i])
      for (int a = 2 * i; a <= tmp; a += i)
        if (k[a]) {
          k[a] = max(k[a], k[i] + 1);
          m = max(m, k[a]);
        }
  printf("%d\n", m);
  return 0;
}

```