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>).
"""
Code of Ayush Tiwari
Codeforces: servermonk
Codechef: ayush572000

"""
import sys
input = sys.stdin.buffer.readline
import math
def solution():
    n=int(input())
    l=list(map(int,input().split()))
    dp=[0]*(100005)
    cnt=[0]*(100005)
    ans=0
    for i in range(n):
        x=l[i]
        for j in range(1,int(x**0.5)+1):
            if x%j==0:
                cnt[i+1]=max(cnt[i+1],cnt[dp[x//j]]+1)
                if j>1:
                    cnt[i+1]=max(cnt[i+1],cnt[dp[j]]+1)
                ans=max(ans,cnt[i+1])
                dp[j]=i+1 
                dp[x//j]=i+1 
    print(ans)

solution()