#include <bits/stdc++.h>
using namespace std;
int n, A[100001], dp[100001], divi[100001];
void crib(void) {
  for (int i = 2; i * i <= 100001; i++) {
    if (divi[i]) continue;
    for (int j = i * 2; j < 100001; j += i) {
      divi[j] = i;
    }
  }
}
int main(void) {
  cin >> n;
  memset(dp, 0, sizeof(dp));
  memset(divi, 0, sizeof(divi));
  int rst = 0;
  crib();
  for (int i = 0; i < n; i++) cin >> A[i];
  for (int i = 0; i < n; i++) {
    int it = A[i];
    int m = 0;
    while (divi[it]) {
      if (dp[divi[it]] > m) m = dp[divi[it]];
      it /= divi[it];
    }
    if (dp[it] > m) m = dp[it];
    if (m + 1 > rst) rst = m + 1;
    it = A[i];
    while (divi[it]) {
      dp[divi[it]] = m + 1;
      it /= divi[it];
    }
    dp[it] = m + 1;
  }
  cout << rst << endl;
  return 0;
}

### Prompt

Your challenge is to write 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;
int n, A[100001], dp[100001], divi[100001];
void crib(void) {
  for (int i = 2; i * i <= 100001; i++) {
    if (divi[i]) continue;
    for (int j = i * 2; j < 100001; j += i) {
      divi[j] = i;
    }
  }
}
int main(void) {
  cin >> n;
  memset(dp, 0, sizeof(dp));
  memset(divi, 0, sizeof(divi));
  int rst = 0;
  crib();
  for (int i = 0; i < n; i++) cin >> A[i];
  for (int i = 0; i < n; i++) {
    int it = A[i];
    int m = 0;
    while (divi[it]) {
      if (dp[divi[it]] > m) m = dp[divi[it]];
      it /= divi[it];
    }
    if (dp[it] > m) m = dp[it];
    if (m + 1 > rst) rst = m + 1;
    it = A[i];
    while (divi[it]) {
      dp[divi[it]] = m + 1;
      it /= divi[it];
    }
    dp[it] = m + 1;
  }
  cout << rst << endl;
  return 0;
}

```