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>).
from math import sqrt
def gcd(a, b):
    while(a!=0):
        tmp=a
        a=b%a
        b=tmp
    return b

n=int(input())
div=[[] for i in range(int(1e5)+1)]
for x in range(2,int(1e5)+1):
    for i in range(int(sqrt(x)), 1, -1):
        if (x % i == 0):
            div[x].append(i)
            div[x].append(x//i)
    div[x].append(x)

l=[int(i) for i in input().strip().split()]
dp=[0 for i in range(n)]
dpdiv=[0 for i in range(int(1e5)+1)]
dp[0]=1
for i in range(n):
    for x in range(len(div[l[i]])):
        dp[i]=max(dp[i], dpdiv[div[l[i]][x]]+1)
    dp[i]=max(dp[i], 1)
    for x in range(len(div[l[i]])):
        dpdiv[div[l[i]][x]]=dp[i]
mx=0
for i in range(n):
    mx=max(dp[i], mx)
print(mx)