#include <bits/stdc++.h>
using namespace std;
const int sz = 1500000;
const long long MAX = 10000000000000000;
template <typename T>
T abs(T a) {
  return a < 0 ? -a : a;
}
vector<int> res;
string s;
void ans(int ind) {
  if (ind == s.length() - 1) {
    res.push_back(ind + 1);
    return;
  }
  if (s[ind] == 'l') {
    ans(ind + 1);
    res.push_back(ind + 1);
  } else {
    res.push_back(ind + 1);
    ans(ind + 1);
  }
  return;
}
int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<vector<int> > fact(n);
  for (long long i = 0; i < n; i++) {
    for (long long j = 2; j * j <= a[i]; j++) {
      if (a[i] % j == 0) {
        fact[i].push_back(j);
        if (a[i] / j != j) fact[i].push_back(a[i] / j);
      }
    }
    if (a[i] != 1) fact[i].push_back(a[i]);
  }
  vector<int> maxIndPrime(100001, -1);
  for (long long i = 0; i < fact[0].size(); i++) {
    maxIndPrime[fact[0][i]] = 0;
  }
  vector<int> ans(n, 1);
  for (int i = 1; i < n; i++) {
    for (long long j = 0; j < fact[i].size(); j++) {
      if (maxIndPrime[fact[i][j]] != -1) {
        ans[i] = ((ans[i]) > (ans[maxIndPrime[fact[i][j]]] + 1)
                      ? (ans[i])
                      : (ans[maxIndPrime[fact[i][j]]] + 1));
      }
      if (fact[i][j] != 1) maxIndPrime[fact[i][j]] = i;
    }
  }
  int maxAns = 0;
  for (long long i = 0; i < n; i++) {
    maxAns = ((maxAns) > (ans[i]) ? (maxAns) : (ans[i]));
  }
  cout << maxAns << endl;
  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;
const int sz = 1500000;
const long long MAX = 10000000000000000;
template <typename T>
T abs(T a) {
  return a < 0 ? -a : a;
}
vector<int> res;
string s;
void ans(int ind) {
  if (ind == s.length() - 1) {
    res.push_back(ind + 1);
    return;
  }
  if (s[ind] == 'l') {
    ans(ind + 1);
    res.push_back(ind + 1);
  } else {
    res.push_back(ind + 1);
    ans(ind + 1);
  }
  return;
}
int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<vector<int> > fact(n);
  for (long long i = 0; i < n; i++) {
    for (long long j = 2; j * j <= a[i]; j++) {
      if (a[i] % j == 0) {
        fact[i].push_back(j);
        if (a[i] / j != j) fact[i].push_back(a[i] / j);
      }
    }
    if (a[i] != 1) fact[i].push_back(a[i]);
  }
  vector<int> maxIndPrime(100001, -1);
  for (long long i = 0; i < fact[0].size(); i++) {
    maxIndPrime[fact[0][i]] = 0;
  }
  vector<int> ans(n, 1);
  for (int i = 1; i < n; i++) {
    for (long long j = 0; j < fact[i].size(); j++) {
      if (maxIndPrime[fact[i][j]] != -1) {
        ans[i] = ((ans[i]) > (ans[maxIndPrime[fact[i][j]]] + 1)
                      ? (ans[i])
                      : (ans[maxIndPrime[fact[i][j]]] + 1));
      }
      if (fact[i][j] != 1) maxIndPrime[fact[i][j]] = i;
    }
  }
  int maxAns = 0;
  for (long long i = 0; i < n; i++) {
    maxAns = ((maxAns) > (ans[i]) ? (maxAns) : (ans[i]));
  }
  cout << maxAns << endl;
  return 0;
}

```