#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
long long n;
vector<long long int> pre(100005), dp(100005, -1), minprime(100005);
vector<vector<long long int> > pf(100005);
long long lim = 1e5;
void fillpf(long long x) {
  long long st = x;
  set<long long> toa;
  while (x != 1) {
    toa.insert(minprime[x]);
    x /= minprime[x];
  }
  for (auto &&it : toa) pf[st].push_back(it);
}
void sieve() {
  for (long long int i = 0; i < lim + 1; i++) minprime[i] = i;
  for (long long i = 2; i * i <= lim; i++) {
    for (long long j = i; j <= lim; j += i) {
      if (i < minprime[j]) minprime[j] = i;
    }
  }
  for (long long int i = 1; i < lim + 1; i++) {
    if (pre[i] == 1) {
      fillpf(i);
    }
  }
}
long long solve(long long st) {
  if (dp[st] != -1) return dp[st];
  long long res = 1;
  long long mx = 0;
  for (auto &&it : pf[st]) {
    long long j = st + it;
    while (j <= 1e5 && pre[j] != 1) {
      j += it;
    }
    if (j <= 1e5 && pre[j] == 1) {
      mx = max(mx, solve(j));
    }
  }
  res += mx;
  dp[st] = res;
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    long long tmp;
    cin >> n;
    for (long long int i = 0; i < n; i++) {
      cin >> tmp;
      pre[tmp] = 1;
    }
    long long lim = 1e5;
    sieve();
    long long ans = 0;
    for (long long int i = 0; i < lim + 1; i++) {
      if (pre[i] == 1) ans = max(ans, solve(i));
    }
    cout << ans << "\n";
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 mod = 1e9 + 7;
long long n;
vector<long long int> pre(100005), dp(100005, -1), minprime(100005);
vector<vector<long long int> > pf(100005);
long long lim = 1e5;
void fillpf(long long x) {
  long long st = x;
  set<long long> toa;
  while (x != 1) {
    toa.insert(minprime[x]);
    x /= minprime[x];
  }
  for (auto &&it : toa) pf[st].push_back(it);
}
void sieve() {
  for (long long int i = 0; i < lim + 1; i++) minprime[i] = i;
  for (long long i = 2; i * i <= lim; i++) {
    for (long long j = i; j <= lim; j += i) {
      if (i < minprime[j]) minprime[j] = i;
    }
  }
  for (long long int i = 1; i < lim + 1; i++) {
    if (pre[i] == 1) {
      fillpf(i);
    }
  }
}
long long solve(long long st) {
  if (dp[st] != -1) return dp[st];
  long long res = 1;
  long long mx = 0;
  for (auto &&it : pf[st]) {
    long long j = st + it;
    while (j <= 1e5 && pre[j] != 1) {
      j += it;
    }
    if (j <= 1e5 && pre[j] == 1) {
      mx = max(mx, solve(j));
    }
  }
  res += mx;
  dp[st] = res;
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    long long tmp;
    cin >> n;
    for (long long int i = 0; i < n; i++) {
      cin >> tmp;
      pre[tmp] = 1;
    }
    long long lim = 1e5;
    sieve();
    long long ans = 0;
    for (long long int i = 0; i < lim + 1; i++) {
      if (pre[i] == 1) ans = max(ans, solve(i));
    }
    cout << ans << "\n";
  }
  return 0;
}

```