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.
n = int(input())
ls = list(map(int, input().split()))
dp = [[1 for i in range(n)] for j in range(n)]
laspos = [None] * (max(ls) + 1)
for i in range(n):
	for j in range(i):

		if laspos[ls[i]] is not None:
			dp[i][j] = 1 + dp[j][laspos[ls[i]]]
		else:
			dp[i][j] += 1
		laspos[ls[j]] = j

mx = -100000
for i in range(n):
	for j in range(n):
		mx = max(dp[i][j], mx)
print(mx)
