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>).
#!/usr/bin/env python3
from sys import stdin

mem = [0 for i in range(int(1e5+1))]


def solve(tc):
    n = int(stdin.readline().strip())
    seq = list(map(int, stdin.readline().split()))

    ans = 0
    for i in range(n):
        num = seq[i]

        divisor = set()
        p = 2
        while p*p <= num:
            while num % p == 0:
                divisor.add(p)
                num //= p
            p += 1
        if num > 1:
            divisor.add(num)

        for dn in divisor:
            mem[seq[i]] = max(mem[seq[i]], mem[dn]+1)

        for dn in divisor:
            mem[dn] = mem[seq[i]]

        ans = max(ans, mem[seq[i]])

    print(ans)


tcs = 1
for tc in range(tcs):
    solve(tc)
