#include <bits/stdc++.h>
using namespace std;
int dp[100002], pos[100002];
int n, num[100002];
int prime[100002];
bool isprime[100002];
void getPrime() {
  for (int i = 0; i <= n + 2; i += 2) {
    isprime[i + 1] = true;
    isprime[i] = 0;
  }
  prime[prime[0] = 1] = 2;
  for (int i = 3;; i += 2)
    if (isprime[i]) {
      int j = i * i, k = i << 1;
      if (j > 100002) break;
      while (j < 100002) {
        isprime[j] = 0;
        j += k;
      }
    }
  for (int i = 3; i <= (n + 3); i += 2)
    if (isprime[i]) prime[++prime[0]] = i;
}
void solve() {
  memset(pos, -1, 4 * (n + 2));
  int ans = 0;
  for (int i = 1; i <= n; ++i) {
    dp[i] = 1;
    for (int j = 1; j <= prime[0] && prime[j] * prime[j] <= num[i]; ++j)
      if (num[i] % prime[j] == 0) {
        if (pos[prime[j]] != -1) {
          dp[i] = max(dp[i], dp[pos[prime[j]]] + 1);
        }
        pos[prime[j]] = i;
        while (num[i] % prime[j] == 0) num[i] /= prime[j];
      }
    if (num[i] > 1) {
      if (pos[num[i]] != -1) dp[i] = max(dp[i], dp[pos[num[i]]] + 1);
      pos[num[i]] = i;
    }
    if (ans < dp[i]) ans = dp[i];
  }
  printf("%d\n", ans);
}
int main(int argc, char** argv) {
  scanf("%d", &n);
  getPrime();
  for (int i = 1; i <= n; ++i) scanf("%d", num + i);
  solve();
}

### Prompt

Create a solution in CPP for 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 dp[100002], pos[100002];
int n, num[100002];
int prime[100002];
bool isprime[100002];
void getPrime() {
  for (int i = 0; i <= n + 2; i += 2) {
    isprime[i + 1] = true;
    isprime[i] = 0;
  }
  prime[prime[0] = 1] = 2;
  for (int i = 3;; i += 2)
    if (isprime[i]) {
      int j = i * i, k = i << 1;
      if (j > 100002) break;
      while (j < 100002) {
        isprime[j] = 0;
        j += k;
      }
    }
  for (int i = 3; i <= (n + 3); i += 2)
    if (isprime[i]) prime[++prime[0]] = i;
}
void solve() {
  memset(pos, -1, 4 * (n + 2));
  int ans = 0;
  for (int i = 1; i <= n; ++i) {
    dp[i] = 1;
    for (int j = 1; j <= prime[0] && prime[j] * prime[j] <= num[i]; ++j)
      if (num[i] % prime[j] == 0) {
        if (pos[prime[j]] != -1) {
          dp[i] = max(dp[i], dp[pos[prime[j]]] + 1);
        }
        pos[prime[j]] = i;
        while (num[i] % prime[j] == 0) num[i] /= prime[j];
      }
    if (num[i] > 1) {
      if (pos[num[i]] != -1) dp[i] = max(dp[i], dp[pos[num[i]]] + 1);
      pos[num[i]] = i;
    }
    if (ans < dp[i]) ans = dp[i];
  }
  printf("%d\n", ans);
}
int main(int argc, char** argv) {
  scanf("%d", &n);
  getPrime();
  for (int i = 1; i <= n; ++i) scanf("%d", num + i);
  solve();
}

```