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.
def main():
    n = int(raw_input())
    a = map(int, raw_input().split())

    if n <= 2:
        print(n)
        return
    maxi = 2
    res = [{} for i in range(1000001)]
    for i in range(n):
        if a[i] in res[a[i]]: continue
        if a.count(a[i]) == 1: continue

        for j in range(i+1,n):
            if not a[j] in res[a[i]] and a[j] != a[i]:
                res[a[i]][a[j]] = 2
            if a[j] != a[i] and res[a[i]][a[j]] % 2 == 1:
                res[a[i]][a[j]] += 1
            if a[j] == a[i]:
                for k in res[a[i]].keys():
                    if res[a[i]][k] % 2 == 0:
                        res[a[i]][k] += 1
        res[a[i]][a[i]] = -1
        maxi = max(maxi,max(res[a[i]].values()),a.count(a[i]))
    print(maxi)
main()