#include <bits/stdc++.h>
using namespace std;
const int M1 = 111111;
int N;
int f[M1];
int d[M1];
int main() {
  scanf("%d", &N);
  int res = 1;
  for (int i = 0; i != N; ++i) {
    int tmp;
    int dlen = 0, mx = 0;
    scanf("%d", &tmp);
    for (int j = 2; j * j <= tmp; ++j) {
      if (0 == tmp % j) {
        d[dlen++] = j;
        if (f[j] + 1 > mx) mx = f[j] + 1;
        while (0 == tmp % j) tmp /= j;
      }
    }
    if (tmp > 1) {
      d[dlen++] = tmp;
      if (f[tmp] + 1 > mx) mx = f[tmp] + 1;
    }
    for (int j = 0; j != dlen; ++j) f[d[j]] = mx;
    if (res < mx) res = mx;
  }
  printf("%d\n", res);
  return 0;
}

### Prompt

Please create a solution in Cpp 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 M1 = 111111;
int N;
int f[M1];
int d[M1];
int main() {
  scanf("%d", &N);
  int res = 1;
  for (int i = 0; i != N; ++i) {
    int tmp;
    int dlen = 0, mx = 0;
    scanf("%d", &tmp);
    for (int j = 2; j * j <= tmp; ++j) {
      if (0 == tmp % j) {
        d[dlen++] = j;
        if (f[j] + 1 > mx) mx = f[j] + 1;
        while (0 == tmp % j) tmp /= j;
      }
    }
    if (tmp > 1) {
      d[dlen++] = tmp;
      if (f[tmp] + 1 > mx) mx = f[tmp] + 1;
    }
    for (int j = 0; j != dlen; ++j) f[d[j]] = mx;
    if (res < mx) res = mx;
  }
  printf("%d\n", res);
  return 0;
}

```