#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
long long dx[] = {0, 0, 1, -1};
long long dy[] = {1, -1, 0, -1};
long long m, k, n, a[2000005], ans = 0, sm = 0, res = 0, dp[300005], d[300005];
multiset<long long> st1, st2;
string s1 = "one", s2 = "zero", s, ch = "";
vector<long long> v[300005], vnode, v1[50], v2[50], v3;
vector<pair<long long, long long> > vv;
vector<pair<long long, pair<long long, long long> > > vvv;
priority_queue<long long> pq;
map<pair<long long, long long>, long long> cost;
long long xa, xb, xc, ya, yb, yc, xres;
vector<long long> tree[8000005];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long i = 0, x, y, j = 0, z = 0, q, nb0 = 0, nb = 0, nbmax = 0, nbinf = 0,
            nb2 = 0, mm = 5e8, idx = 0, ok = 0, l, r, negatif = 0, positif = 0,
            l1, r1, t;
  multiset<long long>::iterator it, it1, it2;
  char cc;
  cin >> n;
  memset(dp, 0, sizeof dp);
  memset(d, 0, sizeof d);
  for (int i = 1; i <= n; i++) {
    cin >> x;
    for (j = 1; j * j <= x; j++) {
      if (x % j == 0) {
        dp[x] = max(dp[x], d[x / j]);
        if (j != 1) dp[x] = max(dp[x], d[j]);
      }
    }
    dp[x]++;
    for (j = 1; j * j <= x; j++) {
      if (x % j == 0) {
        d[j] = d[x / j] = dp[x];
      }
    }
    ans = max(ans, dp[x]);
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Create a solution in cpp for 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;
const long long mod = 1e9 + 7;
long long dx[] = {0, 0, 1, -1};
long long dy[] = {1, -1, 0, -1};
long long m, k, n, a[2000005], ans = 0, sm = 0, res = 0, dp[300005], d[300005];
multiset<long long> st1, st2;
string s1 = "one", s2 = "zero", s, ch = "";
vector<long long> v[300005], vnode, v1[50], v2[50], v3;
vector<pair<long long, long long> > vv;
vector<pair<long long, pair<long long, long long> > > vvv;
priority_queue<long long> pq;
map<pair<long long, long long>, long long> cost;
long long xa, xb, xc, ya, yb, yc, xres;
vector<long long> tree[8000005];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long i = 0, x, y, j = 0, z = 0, q, nb0 = 0, nb = 0, nbmax = 0, nbinf = 0,
            nb2 = 0, mm = 5e8, idx = 0, ok = 0, l, r, negatif = 0, positif = 0,
            l1, r1, t;
  multiset<long long>::iterator it, it1, it2;
  char cc;
  cin >> n;
  memset(dp, 0, sizeof dp);
  memset(d, 0, sizeof d);
  for (int i = 1; i <= n; i++) {
    cin >> x;
    for (j = 1; j * j <= x; j++) {
      if (x % j == 0) {
        dp[x] = max(dp[x], d[x / j]);
        if (j != 1) dp[x] = max(dp[x], d[j]);
      }
    }
    dp[x]++;
    for (j = 1; j * j <= x; j++) {
      if (x % j == 0) {
        d[j] = d[x / j] = dp[x];
      }
    }
    ans = max(ans, dp[x]);
  }
  cout << ans << endl;
  return 0;
}

```