#include <bits/stdc++.h>
using namespace std;
bool isprime[100001];
int T[100001], best[100001];
vector<long long> prime;
int main() {
  memset(isprime, true, sizeof(isprime));
  for (int i = 4; i < 100001; i += 2) isprime[i] = false;
  prime.push_back(2);
  for (long long i = 3; i < 100001; i += 2)
    if (isprime[i]) {
      prime.push_back(i);
      for (long long j = i * i; j < 100001; j += i * 2) isprime[j] = false;
    }
  int n, sol = 0;
  scanf("%d", &n);
  memset(best, 0, sizeof(best));
  for (int i = 0; i < n; i++) {
    int a, b;
    scanf("%d", &b);
    int T = 0;
    a = b;
    for (int j = 0; j < prime.size() && prime[j] * prime[j] <= a; j++)
      if (a % prime[j] == 0) {
        while (a % prime[j] == 0) a /= prime[j];
        if (best[prime[j]] > T) T = best[prime[j]];
      }
    if (a > 1 && best[a] > T) T = best[a];
    T++;
    a = b;
    for (int j = 0; j < prime.size() && prime[j] * prime[j] <= a; j++)
      if (a % prime[j] == 0) {
        while (a % prime[j] == 0) a /= prime[j];
        if (T > best[prime[j]]) best[prime[j]] = T;
      }
    if (a > 1 && T > best[a]) best[a] = T;
    if (T > sol) sol = T;
  }
  printf("%d\n", sol);
  return 0;
}

### Prompt

Develop a solution in CPP 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;
bool isprime[100001];
int T[100001], best[100001];
vector<long long> prime;
int main() {
  memset(isprime, true, sizeof(isprime));
  for (int i = 4; i < 100001; i += 2) isprime[i] = false;
  prime.push_back(2);
  for (long long i = 3; i < 100001; i += 2)
    if (isprime[i]) {
      prime.push_back(i);
      for (long long j = i * i; j < 100001; j += i * 2) isprime[j] = false;
    }
  int n, sol = 0;
  scanf("%d", &n);
  memset(best, 0, sizeof(best));
  for (int i = 0; i < n; i++) {
    int a, b;
    scanf("%d", &b);
    int T = 0;
    a = b;
    for (int j = 0; j < prime.size() && prime[j] * prime[j] <= a; j++)
      if (a % prime[j] == 0) {
        while (a % prime[j] == 0) a /= prime[j];
        if (best[prime[j]] > T) T = best[prime[j]];
      }
    if (a > 1 && best[a] > T) T = best[a];
    T++;
    a = b;
    for (int j = 0; j < prime.size() && prime[j] * prime[j] <= a; j++)
      if (a % prime[j] == 0) {
        while (a % prime[j] == 0) a /= prime[j];
        if (T > best[prime[j]]) best[prime[j]] = T;
      }
    if (a > 1 && T > best[a]) best[a] = T;
    if (T > sol) sol = T;
  }
  printf("%d\n", sol);
  return 0;
}

```