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;
const int maxn = 4001;
int a[maxn], dp[maxn][maxn], n, e;
map<int, int> m;
int ID(int v) {
  if (m[v] == 0) m[v] = e++;
  return m[v];
}
void solve() {
  memset(dp, 0, sizeof dp);
  for (int i = 0; i < n; i++)
    for (int j = 1; j <= e; j++) dp[j][a[i]] = dp[a[i]][j] + 1;
}
int main() {
  int v;
  while (scanf("%d", &n) == 1) {
    m.clear();
    e = 1;
    for (int i = 0; i < n; i++) {
      scanf("%d", &v);
      a[i] = ID(v);
    }
    solve();
    int m = -1;
    for (int i = 1; i <= e; i++)
      for (int j = 1; j <= e; j++) m = max(dp[i][j], m);
    printf("%d\n", m);
  }
  return 0;
}
