Squirrel Liss is interested in sequences. She also has preferences of integers. She thinks *n* integers *a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub> are good.

Now she is interested in good sequences. A sequence *x*<sub class="lower-index">1</sub>, *x*<sub class="lower-index">2</sub>, ..., *x*<sub class="lower-index">*k*</sub> is called good if it satisfies the following three conditions:
 -  The sequence is strictly increasing, i.e. *x*<sub class="lower-index">*i*</sub> &lt; *x*<sub class="lower-index">*i* + 1</sub> for each *i* (1 ≤ *i* ≤ *k* - 1). -  No two adjacent elements are coprime, i.e. *gcd*(*x*<sub class="lower-index">*i*</sub>, *x*<sub class="lower-index">*i* + 1</sub>) &gt; 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.
The input consists of two lines. The first line contains a single integer *n* (1 ≤ *n* ≤ 10<sup class="upper-index">5</sup>) — the number of good integers. The second line contains a single-space separated list of good integers *a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub> in strictly increasing order (1 ≤ *a*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">5</sup>; *a*<sub class="lower-index">*i*</sub> &lt; *a*<sub class="lower-index">*i* + 1</sub>).
import math 
m=int(input());
a=[int(x) for x in input().strip().split(" ")];
ans=[1 for x in a];
dp={};

for i in range(len(a)):
  
  n=a[i];
  for p in range(2,int(math.sqrt(n))+1):
    if(n%p==0):
      if(p in dp):
        ans[i]=max(ans[i],dp[p]+1);
    while(n%p==0):
      n=n/p;
  if(n>1):
    if(n in dp):
      ans[i]=max(ans[i],dp[n]+1);
      
  n=a[i];
  for p in range(2,int(math.sqrt(n))+1):
    if(n%p==0):
      if(p in dp):
        dp[p]=max(dp[p],ans[i]);
      else:
        dp[p]=ans[i];
    while(n%p==0):
      n=n/p;
  if(n>1):
    if(n in dp):
      dp[n]=max(dp[n],ans[i]);
    else:
      dp[n]=ans[i];
  #print(dp);
 
#print(ans);    
     
    

print(max(ans));
