#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
inline bool is_palindrome(const string& s) {
  return std::equal(s.begin(), s.end(), s.rbegin());
}
const long long MOD = 1000000007;
const long long INF = 1e9 + 5;
const double eps = 1e-7;
const double PI = acos(-1.0);
const int N = 1e5 + 5;
vector<int> factors[N];
int d[N];
const int MAXN = 100003;
int spf[MAXN];
void sieve() {
  spf[1] = 1;
  for (int i = 2; i < MAXN; i++) spf[i] = i;
  for (int i = 4; i < MAXN; i += 2) spf[i] = 2;
  for (int i = 3; i * i < MAXN; i++) {
    if (spf[i] == i) {
      for (int j = i * i; j < MAXN; j += i)
        if (spf[j] == j) spf[j] = i;
    }
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, ans = 1, currmax, x, tmp;
  cin >> n;
  memset(d, 0, sizeof(d));
  sieve();
  for (long long i = (long long)(0); i < (long long)(n); i++) {
    cin >> x;
    tmp = x;
    currmax = 0;
    while (x > 1) {
      currmax = max(currmax, d[spf[x]] + 1);
      x /= spf[x];
    }
    while (tmp > 1) {
      d[spf[tmp]] = currmax;
      tmp /= spf[tmp];
    }
  }
  for (long long i = (long long)(0); i < (long long)(N); i++) {
    ans = max(ans, d[i]);
  }
  cout << ans;
  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;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
inline bool is_palindrome(const string& s) {
  return std::equal(s.begin(), s.end(), s.rbegin());
}
const long long MOD = 1000000007;
const long long INF = 1e9 + 5;
const double eps = 1e-7;
const double PI = acos(-1.0);
const int N = 1e5 + 5;
vector<int> factors[N];
int d[N];
const int MAXN = 100003;
int spf[MAXN];
void sieve() {
  spf[1] = 1;
  for (int i = 2; i < MAXN; i++) spf[i] = i;
  for (int i = 4; i < MAXN; i += 2) spf[i] = 2;
  for (int i = 3; i * i < MAXN; i++) {
    if (spf[i] == i) {
      for (int j = i * i; j < MAXN; j += i)
        if (spf[j] == j) spf[j] = i;
    }
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, ans = 1, currmax, x, tmp;
  cin >> n;
  memset(d, 0, sizeof(d));
  sieve();
  for (long long i = (long long)(0); i < (long long)(n); i++) {
    cin >> x;
    tmp = x;
    currmax = 0;
    while (x > 1) {
      currmax = max(currmax, d[spf[x]] + 1);
      x /= spf[x];
    }
    while (tmp > 1) {
      d[spf[tmp]] = currmax;
      tmp /= spf[tmp];
    }
  }
  for (long long i = (long long)(0); i < (long long)(N); i++) {
    ans = max(ans, d[i]);
  }
  cout << ans;
  return 0;
}

```