#include <bits/stdc++.h>
using namespace std;
string convertTostr(unsigned long long x) {
  stringstream ss;
  ss << x;
  string st;
  st = ss.str();
  return st;
}
unsigned long long convertToint(string y) {
  unsigned long long num;
  stringstream ss(y);
  ss >> num;
  return num;
}
unsigned long long fpow(unsigned long long b, unsigned long long p,
                        unsigned long long m) {
  if (p == 0) return 1;
  if (p == 1) return b % m;
  unsigned long long res = fpow(b, p / 2, m) % m;
  res = (res * res) % m;
  if (p % 2 != 0) res = (res * (b % m)) % m;
  return res;
}
int gcd(int a, int b) {
  if (a % b == 0) {
    return b;
  } else {
    return gcd(b, a % b);
  }
}
int n;
vector<int> vec;
int arr[100005];
int fn(int index, int pre) {
  if (index == n) return 0;
  if (pre != -1 && arr[pre] != -1) return arr[pre];
  int x = 0, y = 0;
  if (pre == -1 || gcd(vec[index], pre) > 1) x = fn(index + 1, vec[index]) + 1;
  y = fn(index + 1, pre);
  return arr[pre] = max(x, y);
}
int main() {
  std::cin.tie(0);
  std::ios::sync_with_stdio(false);
  memset(arr, -1, sizeof arr);
  int a;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a;
    vec.push_back(a);
  }
  int dp[100005];
  memset(dp, 0, sizeof dp);
  for (int i = 0; i < n; i++) {
    int maxi = 1;
    for (int j = 2; j * j <= vec[i]; j++) {
      if (vec[i] % j == 0) {
        if (dp[j] + 1 > maxi) maxi = dp[j] + 1;
        if (vec[i] / j != 1) {
          if (dp[vec[i] / j] + 1 > maxi) maxi = dp[vec[i] / j] + 1;
        }
      }
    }
    dp[vec[i]] = maxi;
    for (int j = 2; j * j <= vec[i]; j++) {
      if (vec[i] % j == 0) {
        dp[j] = max(dp[j], maxi);
        dp[vec[i] / j] = max(maxi, dp[vec[i] / j]);
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < 100005; i++) {
    if (dp[i] > ans) ans = dp[i];
  }
  cout << ans;
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
string convertTostr(unsigned long long x) {
  stringstream ss;
  ss << x;
  string st;
  st = ss.str();
  return st;
}
unsigned long long convertToint(string y) {
  unsigned long long num;
  stringstream ss(y);
  ss >> num;
  return num;
}
unsigned long long fpow(unsigned long long b, unsigned long long p,
                        unsigned long long m) {
  if (p == 0) return 1;
  if (p == 1) return b % m;
  unsigned long long res = fpow(b, p / 2, m) % m;
  res = (res * res) % m;
  if (p % 2 != 0) res = (res * (b % m)) % m;
  return res;
}
int gcd(int a, int b) {
  if (a % b == 0) {
    return b;
  } else {
    return gcd(b, a % b);
  }
}
int n;
vector<int> vec;
int arr[100005];
int fn(int index, int pre) {
  if (index == n) return 0;
  if (pre != -1 && arr[pre] != -1) return arr[pre];
  int x = 0, y = 0;
  if (pre == -1 || gcd(vec[index], pre) > 1) x = fn(index + 1, vec[index]) + 1;
  y = fn(index + 1, pre);
  return arr[pre] = max(x, y);
}
int main() {
  std::cin.tie(0);
  std::ios::sync_with_stdio(false);
  memset(arr, -1, sizeof arr);
  int a;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a;
    vec.push_back(a);
  }
  int dp[100005];
  memset(dp, 0, sizeof dp);
  for (int i = 0; i < n; i++) {
    int maxi = 1;
    for (int j = 2; j * j <= vec[i]; j++) {
      if (vec[i] % j == 0) {
        if (dp[j] + 1 > maxi) maxi = dp[j] + 1;
        if (vec[i] / j != 1) {
          if (dp[vec[i] / j] + 1 > maxi) maxi = dp[vec[i] / j] + 1;
        }
      }
    }
    dp[vec[i]] = maxi;
    for (int j = 2; j * j <= vec[i]; j++) {
      if (vec[i] % j == 0) {
        dp[j] = max(dp[j], maxi);
        dp[vec[i] / j] = max(maxi, dp[vec[i] / j]);
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < 100005; i++) {
    if (dp[i] > ans) ans = dp[i];
  }
  cout << ans;
  return 0;
}

```