#include <bits/stdc++.h>
using namespace std;
long long Set(long long N, long long pos) { return N = N | (1 << pos); }
long long reset(long long N, long long pos) { return N = N & ~(1 << pos); }
bool check(long long N, long long pos) { return (bool)(N & (1 << pos)); }
void CI(long long &_x) { scanf("%I64d", &_x); }
void CO(long long &_x) { cout << _x; }
template <typename T>
void getarray(T a[], long long n) {
  for (long long i = 0; i < n; i++) cin >> a[i];
}
template <typename T>
void prLLIarray(T a[], long long n) {
  for (long long i = 0; i < n - 1; i++) cout << a[i] << " ";
  cout << a[n - 1] << endl;
}
const double EPS = 1e-9;
const long long INF = 0x7f7f7f7f;
long long dr8[8] = {1, -1, 0, 0, 1, -1, -1, 1};
long long dc8[8] = {0, 0, -1, 1, 1, 1, -1, -1};
long long dr4[4] = {0, 0, 1, -1};
long long dc4[4] = {-1, 1, 0, 0};
long long kn8r[8] = {1, 2, 2, 1, -1, -2, -2, -1};
long long kn8c[8] = {2, 1, -1, -2, -2, -1, 1, 2};
int V[100005];
int dp[100005];
int found[100005];
int main() {
  int n;
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> V[i];
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    vector<int> P;
    int ttt = V[i];
    for (int j = 2; j * j <= V[i]; j++)
      if (V[i] % j == 0) {
        P.push_back(j);
        while (V[i] % j == 0) {
          V[i] = V[i] / j;
        }
      }
    if (V[i] > 1) P.push_back(V[i]);
    int temp = 0;
    for (int j = 0; j < P.size(); j++) {
      temp = max(temp, 1 + dp[P[j]]);
    }
    for (int j = 0; j < P.size(); j++) {
      dp[P[j]] = temp;
    }
    dp[ttt] = temp;
    ans = max(ans, dp[ttt]);
  }
  cout << max(ans, 1) << "\n";
}

### Prompt

Your task is to create 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;
long long Set(long long N, long long pos) { return N = N | (1 << pos); }
long long reset(long long N, long long pos) { return N = N & ~(1 << pos); }
bool check(long long N, long long pos) { return (bool)(N & (1 << pos)); }
void CI(long long &_x) { scanf("%I64d", &_x); }
void CO(long long &_x) { cout << _x; }
template <typename T>
void getarray(T a[], long long n) {
  for (long long i = 0; i < n; i++) cin >> a[i];
}
template <typename T>
void prLLIarray(T a[], long long n) {
  for (long long i = 0; i < n - 1; i++) cout << a[i] << " ";
  cout << a[n - 1] << endl;
}
const double EPS = 1e-9;
const long long INF = 0x7f7f7f7f;
long long dr8[8] = {1, -1, 0, 0, 1, -1, -1, 1};
long long dc8[8] = {0, 0, -1, 1, 1, 1, -1, -1};
long long dr4[4] = {0, 0, 1, -1};
long long dc4[4] = {-1, 1, 0, 0};
long long kn8r[8] = {1, 2, 2, 1, -1, -2, -2, -1};
long long kn8c[8] = {2, 1, -1, -2, -2, -1, 1, 2};
int V[100005];
int dp[100005];
int found[100005];
int main() {
  int n;
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> V[i];
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    vector<int> P;
    int ttt = V[i];
    for (int j = 2; j * j <= V[i]; j++)
      if (V[i] % j == 0) {
        P.push_back(j);
        while (V[i] % j == 0) {
          V[i] = V[i] / j;
        }
      }
    if (V[i] > 1) P.push_back(V[i]);
    int temp = 0;
    for (int j = 0; j < P.size(); j++) {
      temp = max(temp, 1 + dp[P[j]]);
    }
    for (int j = 0; j < P.size(); j++) {
      dp[P[j]] = temp;
    }
    dp[ttt] = temp;
    ans = max(ans, dp[ttt]);
  }
  cout << max(ans, 1) << "\n";
}

```