B. Good Sequences

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Sample test(s)

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.

The code is self explanatory

#include<iostream>

#include<bits/stdc++.h>

using namespace std;

int arr[100010];

int dp[100010];

int last[100110];

int main()

{

int n;

cin>>n;

for(int i=1;i<=n;i++)

{

 cin>>arr[i];

}

for(int i=1;i<=100010;i++)

last[i]=-1;

memset(dp,0,sizeof(dp));

for(int i=1;i<=n;i++)

{

dp[i]=1;

   int val=arr[i];

   int range=sqrt(val);

 //  cout<<"her for i="<<" "<<i<<endl;

   int j;

for( j=2;j<=range;j++)

   {

    bool f=0;

   

      while(val%j==0 && val!=0)

      {

           f=1;

val=val/j;

 

  }

  if(f)

  {

//   cout<<"inisde "<<endl;

      if(last[j]!=-1)

      {

//  cout<<"last  "<<last[j]<<endl;

        dp[i]=max(dp[i],1+dp[last[j]]);

       }

//        cout<<"setting last of "<<j <<" as "<<i<<endl;

last[j]=i;  

  }

}

// cout<<"setting last of cal as "<<val<<" "<<i<<endl;

if(val>1)

dp[i]=max(dp[i],1+dp[last[val]]);

last[val]=i;

        }

        int ans=0;

        for(int i=1;i<=n;i++)

        { 

 ans=max(ans,dp[i]);

//  cout<<dp[i]<<" "<<endl;

}

cout<<ans<<endl;

return 0;

}