#include <bits/stdc++.h>
using namespace std;
long long prime[100005];
long long spf[100005];
void sieve() {
  memset(prime, 1, sizeof(prime));
  prime[0] = 0;
  prime[1] = 0;
  for (long long i = 1; i <= 100000; ++i) spf[i] = i;
  for (long long i = 2; i * i <= 100000; ++i) {
    if (prime[i]) {
      for (long long j = i * i; j <= 100000; j += i) {
        if (prime[j]) {
          prime[j] = 0;
          spf[j] = i;
        }
      }
    }
  }
}
vector<long long> solve(long long x) {
  vector<long long> v;
  v.push_back(spf[x]);
  while (x != 1) {
    long long sp = spf[x];
    x /= spf[x];
    if (x == 1) break;
    if (sp != spf[x]) v.push_back(spf[x]);
  }
  return v;
}
long long lol[100005];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  sieve();
  long long n;
  cin >> n;
  long long arr[n + 1];
  for (long long i = 1; i <= n; ++i) cin >> arr[i];
  long long dp[n + 1];
  for (long long i = 1; i <= n; ++i) dp[i] = 1;
  for (long long i = 1; i <= n; ++i) {
    vector<long long> v = solve(arr[i]);
    for (auto j : v) {
      dp[i] = max(dp[i], lol[j] + 1);
    }
    for (auto j : v) {
      if (lol[j] < dp[i]) {
        lol[j] = dp[i];
      }
    }
  }
  long long maxi = 0;
  for (long long i = 1; i <= n; ++i) maxi = max(maxi, dp[i]);
  cout << maxi << "\n";
  return 0;
}

### Prompt

Your task is to create 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;
long long prime[100005];
long long spf[100005];
void sieve() {
  memset(prime, 1, sizeof(prime));
  prime[0] = 0;
  prime[1] = 0;
  for (long long i = 1; i <= 100000; ++i) spf[i] = i;
  for (long long i = 2; i * i <= 100000; ++i) {
    if (prime[i]) {
      for (long long j = i * i; j <= 100000; j += i) {
        if (prime[j]) {
          prime[j] = 0;
          spf[j] = i;
        }
      }
    }
  }
}
vector<long long> solve(long long x) {
  vector<long long> v;
  v.push_back(spf[x]);
  while (x != 1) {
    long long sp = spf[x];
    x /= spf[x];
    if (x == 1) break;
    if (sp != spf[x]) v.push_back(spf[x]);
  }
  return v;
}
long long lol[100005];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  sieve();
  long long n;
  cin >> n;
  long long arr[n + 1];
  for (long long i = 1; i <= n; ++i) cin >> arr[i];
  long long dp[n + 1];
  for (long long i = 1; i <= n; ++i) dp[i] = 1;
  for (long long i = 1; i <= n; ++i) {
    vector<long long> v = solve(arr[i]);
    for (auto j : v) {
      dp[i] = max(dp[i], lol[j] + 1);
    }
    for (auto j : v) {
      if (lol[j] < dp[i]) {
        lol[j] = dp[i];
      }
    }
  }
  long long maxi = 0;
  for (long long i = 1; i <= n; ++i) maxi = max(maxi, dp[i]);
  cout << maxi << "\n";
  return 0;
}

```