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.
import sys
input = sys.stdin.readline
#for _ in range(int(input())):
n=int(input())
ans=1
arr=[int(x) for x in input().split()]
for i in range(n):
    temp=set()
    d={}
    #ans=0
    for j in range(i+1,n):
        if arr[j]==arr[i]:
            temp=set()
        else:
            if arr[j] not in temp:
                if arr[j] in d:
                    d[arr[j]]+=1
                else:
                    d[arr[j]]=1
                temp.add(arr[j])
    number=-1
    maxi=-1
    #print(d,temp)
    for i in d:
        if maxi<d[i]:
            number=i
            maxi=d[i]
    #print(number,maxi)
    if number in temp:
        ans=max(ans,2*maxi)
    else:
        ans=max(ans,2*maxi+1)
d={}
for i in arr:
    if i in d:
        d[i]+=1
    else:
        d[i]=1
for i in d:
    ans=max(ans,d[i])
print(ans)