#include <bits/stdc++.h>
using namespace std;
const long long N = (int)1e5 + 74, OO = 0x3f3f3f3f, MOD = (int)1e9 + 7;
int a[N], dp[N];
vector<int> divisors[N];
void generateDivisors() {
  bitset<N> primes;
  primes.set();
  primes[0] = 0;
  primes[1] = 0;
  for (int i = 2; i < N; ++i)
    if (primes[i])
      for (int j = i; j < N; j += i) primes[j] = 0, divisors[j].push_back(i);
}
class DGoodSequences {
 public:
  void solve(istream& cin, ostream& cout) {
    int n;
    cin >> n;
    for (int i = 0; i < n; ++i) cin >> a[i];
    generateDivisors();
    int ans = 1;
    for (int i = 0; i < n; ++i) {
      int mx = 1;
      for (int d : divisors[a[i]]) {
        mx = max(mx, dp[d] + 1);
        ans = max(ans, mx);
      }
      for (int d : divisors[a[i]]) {
        dp[d] = mx;
      }
    }
    cout << ans << '\n';
    memset(dp, 0, sizeof(dp));
  }
};
int main() {
  ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
  DGoodSequences solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}

### Prompt

Your challenge is to write 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;
const long long N = (int)1e5 + 74, OO = 0x3f3f3f3f, MOD = (int)1e9 + 7;
int a[N], dp[N];
vector<int> divisors[N];
void generateDivisors() {
  bitset<N> primes;
  primes.set();
  primes[0] = 0;
  primes[1] = 0;
  for (int i = 2; i < N; ++i)
    if (primes[i])
      for (int j = i; j < N; j += i) primes[j] = 0, divisors[j].push_back(i);
}
class DGoodSequences {
 public:
  void solve(istream& cin, ostream& cout) {
    int n;
    cin >> n;
    for (int i = 0; i < n; ++i) cin >> a[i];
    generateDivisors();
    int ans = 1;
    for (int i = 0; i < n; ++i) {
      int mx = 1;
      for (int d : divisors[a[i]]) {
        mx = max(mx, dp[d] + 1);
        ans = max(ans, mx);
      }
      for (int d : divisors[a[i]]) {
        dp[d] = mx;
      }
    }
    cout << ans << '\n';
    memset(dp, 0, sizeof(dp));
  }
};
int main() {
  ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
  DGoodSequences solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}

```