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
a=int(input())
z=list(map(int,input().split()))
from collections import *
from bisect import *
dp=[[0 for i in range(len(z))] for i in range(len(z))]
for i in range(len(z)):
    dp[0][i]=1
al=defaultdict(list)
maxa=0

        
for i in range(len(z)):
    al[z[i]].append(i)
    maxa=max(maxa,len(al[z[i]]))
for i in range(len(z)):
    index=-1
    for j in range(i):
        if(z[i]!=z[j]):
            if(index==-1):
                dp[i][j]=max(2,dp[i][j])
            else:
                dp[i][j]=max(dp[i][j],dp[j][index]+1)
        else:
            index=j
        maxa=max(dp[i][j],maxa)
        
       
    


print(maxa)
