#include <bits/stdc++.h>
using namespace std;
using ll = long long;
ll max(ll a, ll b) { return (a > b ? a : b); }
ll min(ll a, ll b) { return (a > b ? b : a); }
ll a, b, c, d, e, n, m, k, q, x, y, z, w, t, cur, cnt, ma,
    mi = LONG_LONG_MAX, l, r, mid, sum, i, j, pro = 1, f;
double ans, slp;
ll arr[200005];
ll arr1[200005];
ll arr2[200005];
ll dp[200005];
ll arr3[1000][1000];
char carr[1000][1000];
vector<ll> adj[100001];
string s;
ll dfs(int i) {
  if (arr1[i]) a = 1;
  arr[i] = 1;
  int cnt = 1;
  for (auto c : adj[i]) {
    if (!arr[c]) cnt += dfs(c);
  }
  return cnt;
}
void solve() {
  cin >> n;
  for (ll i = 1; i <= 1e5; i++) {
    dp[i] = -1;
    arr1[i] = 1;
  }
  for (ll i = 1; i <= n; i++) scanf("%d", &arr[i]);
  for (ll i = 1; i <= n; i++) {
    for (int d = 2; d * d <= arr[i]; d++) {
      if (arr[i] % d == 0) {
        if (dp[d] != -1) arr1[i] = max(arr1[i], 1 + arr1[dp[d]]);
        if (dp[arr[i] / d] != -1)
          arr1[i] = max(arr1[i], 1 + arr1[dp[arr[i] / d]]);
        dp[d] = dp[arr[i] / d] = i;
      }
    }
    dp[arr[i]] = i;
    ma = max(ma, arr1[i]);
  }
  cout << ma;
}
int main() {
  cout.precision(9);
  ll t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
using ll = long long;
ll max(ll a, ll b) { return (a > b ? a : b); }
ll min(ll a, ll b) { return (a > b ? b : a); }
ll a, b, c, d, e, n, m, k, q, x, y, z, w, t, cur, cnt, ma,
    mi = LONG_LONG_MAX, l, r, mid, sum, i, j, pro = 1, f;
double ans, slp;
ll arr[200005];
ll arr1[200005];
ll arr2[200005];
ll dp[200005];
ll arr3[1000][1000];
char carr[1000][1000];
vector<ll> adj[100001];
string s;
ll dfs(int i) {
  if (arr1[i]) a = 1;
  arr[i] = 1;
  int cnt = 1;
  for (auto c : adj[i]) {
    if (!arr[c]) cnt += dfs(c);
  }
  return cnt;
}
void solve() {
  cin >> n;
  for (ll i = 1; i <= 1e5; i++) {
    dp[i] = -1;
    arr1[i] = 1;
  }
  for (ll i = 1; i <= n; i++) scanf("%d", &arr[i]);
  for (ll i = 1; i <= n; i++) {
    for (int d = 2; d * d <= arr[i]; d++) {
      if (arr[i] % d == 0) {
        if (dp[d] != -1) arr1[i] = max(arr1[i], 1 + arr1[dp[d]]);
        if (dp[arr[i] / d] != -1)
          arr1[i] = max(arr1[i], 1 + arr1[dp[arr[i] / d]]);
        dp[d] = dp[arr[i] / d] = i;
      }
    }
    dp[arr[i]] = i;
    ma = max(ma, arr1[i]);
  }
  cout << ma;
}
int main() {
  cout.precision(9);
  ll t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```