#include <bits/stdc++.h>
using namespace std;
int a[100000 + 10];
int p[100000 + 10];
int b[100000];
int po[100000 + 10];
int dp[100000 + 10];
int num;
void prime() {
  memset(po, 0, sizeof(po));
  memset(dp, 0, sizeof(dp));
  memset(p, 0, sizeof(p));
  int i = 0, sum = 0, x = 0;
  p[0] = 1;
  p[1] = 1;
  for (i = 2; i <= 100000; i++) {
    sum = i;
    x = 2;
    if (p[sum] == 1) continue;
    sum = i * x;
    while (sum <= 100000) {
      p[sum] = 1;
      x++;
      sum = i * x;
    }
  }
  num = 0;
  for (i = 1; i <= 100000; i++) {
    if (!p[i]) {
      b[++num] = i;
    }
  }
}
int max(int a, int b) { return a > b ? a : b; }
int solve(int x, int p) {
  int tmp = x, sum = 0;
  int i, tru;
  for (i = 1; i <= num; i++) {
    if (tmp == 1) break;
    if (b[i] > 316) {
      sum = max(sum, dp[po[tmp]] + 1);
      po[tmp] = p;
      break;
    }
    tru = 0;
    while (tmp % b[i] == 0) {
      tru = 1;
      tmp /= b[i];
      sum = max(sum, dp[po[b[i]]] + 1);
    }
    if (tru) {
      po[b[i]] = p;
    }
  }
  return sum;
}
int main() {
  prime();
  int n, i;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  int ans = 1;
  for (i = n; i >= 1; i--) {
    dp[i] = solve(a[i], i);
    if (dp[i] > ans) ans = dp[i];
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Generate 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 a[100000 + 10];
int p[100000 + 10];
int b[100000];
int po[100000 + 10];
int dp[100000 + 10];
int num;
void prime() {
  memset(po, 0, sizeof(po));
  memset(dp, 0, sizeof(dp));
  memset(p, 0, sizeof(p));
  int i = 0, sum = 0, x = 0;
  p[0] = 1;
  p[1] = 1;
  for (i = 2; i <= 100000; i++) {
    sum = i;
    x = 2;
    if (p[sum] == 1) continue;
    sum = i * x;
    while (sum <= 100000) {
      p[sum] = 1;
      x++;
      sum = i * x;
    }
  }
  num = 0;
  for (i = 1; i <= 100000; i++) {
    if (!p[i]) {
      b[++num] = i;
    }
  }
}
int max(int a, int b) { return a > b ? a : b; }
int solve(int x, int p) {
  int tmp = x, sum = 0;
  int i, tru;
  for (i = 1; i <= num; i++) {
    if (tmp == 1) break;
    if (b[i] > 316) {
      sum = max(sum, dp[po[tmp]] + 1);
      po[tmp] = p;
      break;
    }
    tru = 0;
    while (tmp % b[i] == 0) {
      tru = 1;
      tmp /= b[i];
      sum = max(sum, dp[po[b[i]]] + 1);
    }
    if (tru) {
      po[b[i]] = p;
    }
  }
  return sum;
}
int main() {
  prime();
  int n, i;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  int ans = 1;
  for (i = n; i >= 1; i--) {
    dp[i] = solve(a[i], i);
    if (dp[i] > ans) ans = dp[i];
  }
  printf("%d\n", ans);
  return 0;
}

```