#include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
template <typename T>
T GCD(T a, T b) {
  long long t;
  while (a) {
    t = a;
    a = b % a;
    b = t;
  }
  return b;
}
template <typename T>
string toString(T a) {
  return to_string(a);
}
template <typename T>
void toInt(string s, T& x) {
  stringstream str(s);
  str >> x;
}
inline int add(int x, int y) {
  x += y;
  if (x >= 1000000007) x -= 1000000007;
  return x;
}
inline int sub(int x, int y) {
  x -= y;
  if (x < 0) x += 1000000007;
  return x;
}
inline int mul(int x, int y) { return (x * 1ll * y) % 1000000007; }
inline int powr(int a, long long b) {
  int x = 1 % 1000000007;
  while (b) {
    if (b & 1) x = mul(x, a);
    a = mul(a, a);
    b >>= 1;
  }
  return x;
}
inline int inv(int a) { return powr(a, 1000000007 - 2); }
const int N = 1e5 + 1;
vector<int> vec;
int n;
int dp[N];
vector<int> primeFact(int n) {
  vector<int> ans;
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      ans.push_back(i);
      while (n % i == 0) {
        n /= i;
      }
    }
  }
  if (n > 1) ans.push_back(n);
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  int v;
  for (int i = 0; i < n; i++) {
    cin >> v;
    vec.push_back(v);
  }
  if (n == 1) {
    cout << 1 << '\n';
    return 0;
  }
  vector<int> fac = primeFact(vec[n - 1]);
  for (auto it : fac) {
    dp[it] = 1;
  }
  for (int i = n - 2; i >= 0; i--) {
    fac = primeFact(vec[i]);
    int mx = 0;
    for (auto it : fac) {
      dp[it] = dp[it] + 1;
      mx = max(mx, dp[it]);
    }
    for (auto it : fac) {
      dp[it] = mx;
    }
  }
  int ans = *max_element(dp, dp + N);
  cout << ans << '\n';
  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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
template <typename T>
T GCD(T a, T b) {
  long long t;
  while (a) {
    t = a;
    a = b % a;
    b = t;
  }
  return b;
}
template <typename T>
string toString(T a) {
  return to_string(a);
}
template <typename T>
void toInt(string s, T& x) {
  stringstream str(s);
  str >> x;
}
inline int add(int x, int y) {
  x += y;
  if (x >= 1000000007) x -= 1000000007;
  return x;
}
inline int sub(int x, int y) {
  x -= y;
  if (x < 0) x += 1000000007;
  return x;
}
inline int mul(int x, int y) { return (x * 1ll * y) % 1000000007; }
inline int powr(int a, long long b) {
  int x = 1 % 1000000007;
  while (b) {
    if (b & 1) x = mul(x, a);
    a = mul(a, a);
    b >>= 1;
  }
  return x;
}
inline int inv(int a) { return powr(a, 1000000007 - 2); }
const int N = 1e5 + 1;
vector<int> vec;
int n;
int dp[N];
vector<int> primeFact(int n) {
  vector<int> ans;
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      ans.push_back(i);
      while (n % i == 0) {
        n /= i;
      }
    }
  }
  if (n > 1) ans.push_back(n);
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  int v;
  for (int i = 0; i < n; i++) {
    cin >> v;
    vec.push_back(v);
  }
  if (n == 1) {
    cout << 1 << '\n';
    return 0;
  }
  vector<int> fac = primeFact(vec[n - 1]);
  for (auto it : fac) {
    dp[it] = 1;
  }
  for (int i = n - 2; i >= 0; i--) {
    fac = primeFact(vec[i]);
    int mx = 0;
    for (auto it : fac) {
      dp[it] = dp[it] + 1;
      mx = max(mx, dp[it]);
    }
    for (auto it : fac) {
      dp[it] = mx;
    }
  }
  int ans = *max_element(dp, dp + N);
  cout << ans << '\n';
  return 0;
}

```