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.
#include <bits/stdc++.h>
using namespace std;
int n, ans, b[1000001], a[2000000], p[2000000], h[10000001];
vector<int> t[10000001];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (p[a[i]] == 0) p[a[i]] = i;
    h[a[i]]++;
    ans = max(h[a[i]], ans);
  }
  for (int i = 1; i <= n; i++) {
    memset(b, 0, sizeof(b));
    int l = i;
    for (int j = i + 1; j <= n; j++)
      if (a[j] == a[i]) {
        for (int k = l + 1; k < j; k++)
          if (b[p[a[k]]] % 2 == 1) b[p[a[k]]]++;
        l = j;
      } else if (b[p[a[j]]] % 2 == 0)
        b[p[a[j]]]++;
    for (int j = 1; j <= n; j++) ans = max(ans, b[j] + 1);
  }
  cout << ans;
}
