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>).
# list(map(int,input().split()))
import math
from collections import Counter

n = int(input())
a = list(map(int, input().split()))
dp = [-1] * n

primes=[[] for _ in range(a[-1]+1)]
for i in range(2,a[-1]+1):
    if not primes[i]:
        for j in range(i,a[-1]+1,i):
            primes[j].append(i)
def recur(a, i, prev):
    if i >= len(a):
            return 0
    elif dp[i] != -1:
        return dp[i]
    else:
        if list(filter(lambda x:x in primes[prev], primes[a[i]])):
            return 1 + recur(a, i + 1, a[i])
        else:
            return max(recur(a, i + 1, prev), recur(a, i + 1, a[i]))


print(1+recur(a, 1, a[0]))
