#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
;
long long max(long long a, long long b) {
  if (a > b)
    return a;
  else
    return b;
}
long long min(long long a, long long b) {
  if (a > b)
    return b;
  else
    return a;
}
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long normalize(long long x, long long mod) {
  x %= mod;
  if (x < 0) x += mod;
  return x;
}
long long dx[4] = {0, 0, 1, -1};
long long dy[4] = {1, -1, 0, 0};
string abc = "abcdefghijklmnopqrstuvwxyz";
const double PI = acosl(-1);
const long long mod = 998244353;
const long long inf = 1e18 + 5;
inline long long add(long long a, long long b, long long p = mod) {
  long long c = a + b;
  if (c >= p) c -= p;
  return c;
}
inline long long sub(long long a, long long b, long long p = mod) {
  long long c = a - b;
  if (c < 0) c += p;
  return c;
}
inline long long mul(long long a, long long b, long long p = mod) {
  return (a * 1ll * b) % p;
}
long long floor1(long long n, long long k) {
  if (n % k == 0 || n >= 0) return n / k;
  return (n / k) - 1;
}
long long ceil1(long long n, long long k) { return floor1(n + k - 1, k); }
long long powm(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
const long long N = 1e5 + 5;
long long n;
long long a[N];
long long seive[N];
vector<long long> fac[N];
void solve() {
  cin >> n;
  for (long long i = (0); i < (n); ++i) cin >> a[i];
  for (long long i = (1); i < (N); ++i) seive[i] = 1;
  for (long long i = (2); i < (N); ++i) {
    if (seive[i]) {
      for (long long j = 1; j * i < N; ++j) {
        fac[i * j].push_back(i);
        seive[i * j] = 0;
      }
    }
  }
  long long ans = 1;
  long long dp[N] = {0};
  for (long long i = (0); i < (n); ++i) {
    long long len = 1;
    for (auto it : fac[a[i]]) {
      len = max(len, dp[it]);
    }
    ans = max(ans, len);
    for (auto it : fac[a[i]]) {
      dp[it] = max(len + 1, dp[it]);
    }
  }
  cout << ans;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(10);
  long long t = 1;
  while (t--) {
    solve();
  }
  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>
#pragma GCC optimize("Ofast")
using namespace std;
;
long long max(long long a, long long b) {
  if (a > b)
    return a;
  else
    return b;
}
long long min(long long a, long long b) {
  if (a > b)
    return b;
  else
    return a;
}
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long normalize(long long x, long long mod) {
  x %= mod;
  if (x < 0) x += mod;
  return x;
}
long long dx[4] = {0, 0, 1, -1};
long long dy[4] = {1, -1, 0, 0};
string abc = "abcdefghijklmnopqrstuvwxyz";
const double PI = acosl(-1);
const long long mod = 998244353;
const long long inf = 1e18 + 5;
inline long long add(long long a, long long b, long long p = mod) {
  long long c = a + b;
  if (c >= p) c -= p;
  return c;
}
inline long long sub(long long a, long long b, long long p = mod) {
  long long c = a - b;
  if (c < 0) c += p;
  return c;
}
inline long long mul(long long a, long long b, long long p = mod) {
  return (a * 1ll * b) % p;
}
long long floor1(long long n, long long k) {
  if (n % k == 0 || n >= 0) return n / k;
  return (n / k) - 1;
}
long long ceil1(long long n, long long k) { return floor1(n + k - 1, k); }
long long powm(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
const long long N = 1e5 + 5;
long long n;
long long a[N];
long long seive[N];
vector<long long> fac[N];
void solve() {
  cin >> n;
  for (long long i = (0); i < (n); ++i) cin >> a[i];
  for (long long i = (1); i < (N); ++i) seive[i] = 1;
  for (long long i = (2); i < (N); ++i) {
    if (seive[i]) {
      for (long long j = 1; j * i < N; ++j) {
        fac[i * j].push_back(i);
        seive[i * j] = 0;
      }
    }
  }
  long long ans = 1;
  long long dp[N] = {0};
  for (long long i = (0); i < (n); ++i) {
    long long len = 1;
    for (auto it : fac[a[i]]) {
      len = max(len, dp[it]);
    }
    ans = max(ans, len);
    for (auto it : fac[a[i]]) {
      dp[it] = max(len + 1, dp[it]);
    }
  }
  cout << ans;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(10);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```