Gena loves sequences of numbers. Recently, he has discovered a new type of sequences which he called an almost arithmetical progression. A sequence is an almost arithmetical progression, if its elements can be represented as:

  * a1 = p, where p is some integer; 
  * ai = ai - 1 + ( - 1)i + 1·q (i > 1), where q is some integer. 



Right now Gena has a piece of paper with sequence b, consisting of n integers. Help Gena, find there the longest subsequence of integers that is an almost arithmetical progression.

Sequence s1, s2, ..., sk is a subsequence of sequence b1, b2, ..., bn, if there is such increasing sequence of indexes i1, i2, ..., ik (1 ≤ i1 < i2 < ... < ik ≤ n), that bij = sj. In other words, sequence s can be obtained from b by crossing out some elements.

Input

The first line contains integer n (1 ≤ n ≤ 4000). The next line contains n integers b1, b2, ..., bn (1 ≤ bi ≤ 106).

Output

Print a single integer — the length of the required longest subsequence.

Examples

Input

2
3 5


Output

2


Input

4
10 20 10 30


Output

3

Note

In the first test the sequence actually is the suitable subsequence. 

In the second test the following subsequence fits: 10, 20, 10.
d, n, t = 0, int(input()), list(map(int, input().split()))

p = {a: 0 for a in set(t)}

for i in range(n):

    a = t[i]

    if not a in p: continue

    p.pop(a)

    s = t.count(a) - 1

    if 2 * s < d: continue

    if s > d: d = s

    k = i + 1

    for j in range(k, n):

        if t[j] == a:

            for b in set(t[k: j]):

                if b in p: p[b] += 2

            k = j + 1

    for b in set(t[k: n]):

        if b in p: p[b] += 1

    for b in p:

        if p[b] > d: d = p[b]

        p[b] = 0

print(d + 1)



# Made By Mostafa_Khaled