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 function(n , array):
    grid = [ {} for i in range(n)]

    if(n <= 2):
        print(n)
        return

    global_max = -10
    for i in range(n - 2, -1, -1):
        for j in range(i + 1, n):
            diff = array[i] - array[j]

            max_val = 1
            if((-diff) in grid[j].keys()):
                max_val = max(grid[j][(-diff)] + 1, max_val)

            if(diff in grid[i].keys()):
                max_val = max(max_val, grid[i][diff])
                grid[i][diff] = max_val
            else:
                grid[i][diff] = max_val

            global_max = max(global_max, max_val)

    print(global_max + 1)

n = int(input())
array = [ int(x) for x in input().split() ]
function(n, array)