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.
from sys import stdin,stdout
nmbr = lambda: int(stdin.readline())
lst = lambda: list(map(int,input().split()))
def fn(a):
    mp={}
    b=[]
    cnt=0
    for v in a:
        if v in mp:b+=[mp[v]]
        else:
            b+=[cnt]
            mp[v]=cnt
            cnt+=1
    return b
for i in range(1):#nmbr()):
    n=nmbr()
    a=fn(lst())
    N=max(a)+1;ans=1
    dp=[[1 for _ in range(N)]for j in range(n)]
    for i in range(1,n):
        for j in range(i):
            dp[i][a[j]]=1+dp[j][a[i]]
            ans=max(ans,dp[i][a[j]])
    # print(*dp,sep='\n')
    print(ans)