#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
long long power(long long a, long long n) {
  long long p = 1;
  while (n > 0) {
    if (n % 2) {
      p = p * a;
    }
    n >>= 1;
    a *= a;
  }
  return p;
}
long long power(long long a, long long n, long long mod) {
  long long p = 1;
  while (n > 0) {
    if (n % 2) {
      p = p * a;
      p %= mod;
    }
    n >>= 1;
    a *= a;
    a %= mod;
  }
  return p % mod;
}
long long add(long long a, long long b, long long mod = MOD) {
  return (a + b) % mod;
}
long long sub(long long a, long long b, long long mod = MOD) {
  return ((a - b) % mod + mod) % mod;
}
long long mul(long long a, long long b, long long mod = MOD) {
  return (a * b) % mod;
}
long long divide(long long a, long long b, long long mod = MOD) {
  return (a * power(b, mod - 2, mod)) % mod;
}
int seive[100005] = {0};
map<int, vector<int> > multiples;
void init() {
  for (long long i = 2; i < 100005; ++i) {
    if (!seive[i]) {
      seive[i] = i;
      for (long long j = i * i; j < 100005; j += i) {
        if (!seive[j]) {
          seive[j] = i;
        }
      }
    }
  }
}
set<int> factorize(int n) {
  set<int> fact;
  while (n > 1) {
    fact.insert(seive[n]);
    n /= seive[n];
  }
  return fact;
}
int main(int argc, char const *argv[]) {
  init();
  int n;
  cin >> n;
  map<int, int> dp;
  std::vector<pair<int, set<int> > > v(n);
  vector<int> temp(n);
  for (int i = 0; i < n; ++i) {
    cin >> v[i].first;
    temp[i] = v[i].first;
    v[i].second = factorize(v[i].first);
    for (std::set<int>::iterator it = v[i].second.begin();
         it != v[i].second.end(); ++it) {
      multiples[*it].push_back(v[i].first);
    }
  }
  dp[temp[0]] = 1;
  for (int i = 1; i < n; ++i) {
    for (std::set<int>::iterator it = v[i].second.begin();
         it != v[i].second.end(); ++it) {
      int currentDivisor = *it;
      int ans = *(--(lower_bound(multiples[currentDivisor].begin(),
                                 multiples[currentDivisor].end(), v[i].first)));
      dp[temp[i]] = max(dp[temp[i]], 1 + dp[ans]);
    }
  }
  int MAX = -INF;
  for (std::map<int, int>::iterator i = dp.begin(); i != dp.end(); ++i) {
    MAX = max(MAX, i->second);
  }
  cout << MAX;
  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;
const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
long long power(long long a, long long n) {
  long long p = 1;
  while (n > 0) {
    if (n % 2) {
      p = p * a;
    }
    n >>= 1;
    a *= a;
  }
  return p;
}
long long power(long long a, long long n, long long mod) {
  long long p = 1;
  while (n > 0) {
    if (n % 2) {
      p = p * a;
      p %= mod;
    }
    n >>= 1;
    a *= a;
    a %= mod;
  }
  return p % mod;
}
long long add(long long a, long long b, long long mod = MOD) {
  return (a + b) % mod;
}
long long sub(long long a, long long b, long long mod = MOD) {
  return ((a - b) % mod + mod) % mod;
}
long long mul(long long a, long long b, long long mod = MOD) {
  return (a * b) % mod;
}
long long divide(long long a, long long b, long long mod = MOD) {
  return (a * power(b, mod - 2, mod)) % mod;
}
int seive[100005] = {0};
map<int, vector<int> > multiples;
void init() {
  for (long long i = 2; i < 100005; ++i) {
    if (!seive[i]) {
      seive[i] = i;
      for (long long j = i * i; j < 100005; j += i) {
        if (!seive[j]) {
          seive[j] = i;
        }
      }
    }
  }
}
set<int> factorize(int n) {
  set<int> fact;
  while (n > 1) {
    fact.insert(seive[n]);
    n /= seive[n];
  }
  return fact;
}
int main(int argc, char const *argv[]) {
  init();
  int n;
  cin >> n;
  map<int, int> dp;
  std::vector<pair<int, set<int> > > v(n);
  vector<int> temp(n);
  for (int i = 0; i < n; ++i) {
    cin >> v[i].first;
    temp[i] = v[i].first;
    v[i].second = factorize(v[i].first);
    for (std::set<int>::iterator it = v[i].second.begin();
         it != v[i].second.end(); ++it) {
      multiples[*it].push_back(v[i].first);
    }
  }
  dp[temp[0]] = 1;
  for (int i = 1; i < n; ++i) {
    for (std::set<int>::iterator it = v[i].second.begin();
         it != v[i].second.end(); ++it) {
      int currentDivisor = *it;
      int ans = *(--(lower_bound(multiples[currentDivisor].begin(),
                                 multiples[currentDivisor].end(), v[i].first)));
      dp[temp[i]] = max(dp[temp[i]], 1 + dp[ans]);
    }
  }
  int MAX = -INF;
  for (std::map<int, int>::iterator i = dp.begin(); i != dp.end(); ++i) {
    MAX = max(MAX, i->second);
  }
  cout << MAX;
  return 0;
}

```