#include <bits/stdc++.h>
using namespace std;
int a[100006], f[100006];
int pre[100006], num;
bool p[100006];
void fun() {
  int i, j;
  int limt = sqrt(100006 + 0.1), num = 0;
  memset(p, 0, sizeof(p));
  pre[num++] = 2;
  for (i = 4; i <= 100006; i += 2) p[i] = 1;
  for (i = 3; i <= limt; i += 2) {
    if (p[i]) continue;
    pre[num++] = i;
    for (j = i * i; j <= 100006; j += i) p[j] = 1;
  }
  for (; i <= 100006; i++)
    if (p[i] == 0) pre[num++] = i;
}
int dp[100006];
int main() {
  int i, j, n;
  fun();
  while (scanf("%d", &n) != EOF) {
    for (i = 1; i <= n; i++) scanf("%d", &a[i]);
    memset(f, 0, sizeof(f));
    memset(dp, 0, sizeof(dp));
    if (n == 1) {
      printf("1\n");
    } else {
      for (i = 1; i <= n; i++) {
        int tmp = a[i];
        for (j = 0; pre[j] * pre[j] <= tmp; j++) {
          if (tmp % pre[j] == 0) {
            if (f[pre[j]] != 0) {
              dp[i] = max(dp[f[pre[j]]] + 1, dp[i]);
            } else
              dp[i] = max(1, dp[i]);
            f[pre[j]] = i;
            while (tmp % pre[j] == 0) {
              tmp = tmp / pre[j];
            }
          }
        }
        if (tmp > 1) {
          if (f[tmp] != 0) {
            dp[i] = max(dp[f[tmp]] + 1, dp[i]);
          } else
            dp[i] = max(dp[i], 1);
          f[tmp] = i;
        }
      }
      int ans = 0;
      for (i = 1; i <= n; i++) ans = max(dp[i], ans);
      printf("%d\n", ans);
    }
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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[100006], f[100006];
int pre[100006], num;
bool p[100006];
void fun() {
  int i, j;
  int limt = sqrt(100006 + 0.1), num = 0;
  memset(p, 0, sizeof(p));
  pre[num++] = 2;
  for (i = 4; i <= 100006; i += 2) p[i] = 1;
  for (i = 3; i <= limt; i += 2) {
    if (p[i]) continue;
    pre[num++] = i;
    for (j = i * i; j <= 100006; j += i) p[j] = 1;
  }
  for (; i <= 100006; i++)
    if (p[i] == 0) pre[num++] = i;
}
int dp[100006];
int main() {
  int i, j, n;
  fun();
  while (scanf("%d", &n) != EOF) {
    for (i = 1; i <= n; i++) scanf("%d", &a[i]);
    memset(f, 0, sizeof(f));
    memset(dp, 0, sizeof(dp));
    if (n == 1) {
      printf("1\n");
    } else {
      for (i = 1; i <= n; i++) {
        int tmp = a[i];
        for (j = 0; pre[j] * pre[j] <= tmp; j++) {
          if (tmp % pre[j] == 0) {
            if (f[pre[j]] != 0) {
              dp[i] = max(dp[f[pre[j]]] + 1, dp[i]);
            } else
              dp[i] = max(1, dp[i]);
            f[pre[j]] = i;
            while (tmp % pre[j] == 0) {
              tmp = tmp / pre[j];
            }
          }
        }
        if (tmp > 1) {
          if (f[tmp] != 0) {
            dp[i] = max(dp[f[tmp]] + 1, dp[i]);
          } else
            dp[i] = max(dp[i], 1);
          f[tmp] = i;
        }
      }
      int ans = 0;
      for (i = 1; i <= n; i++) ans = max(dp[i], ans);
      printf("%d\n", ans);
    }
  }
  return 0;
}

```