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 INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1e9 + 7;
const int N = 4e3 + 7;
int n, a[N];
int dp[N][N];
void Init() {
  for (int i = (1); i < (n + 1); ++i) scanf("%d", a + i);
}
int Solve() {
  int ans = 1;
  for (int i = (1); i < (n + 1); ++i) dp[0][i] = 1;
  for (int i = (2); i < (n + 1); ++i) {
    int pre = 0;
    for (int j = (1); j < (i); ++j) {
      dp[j][i] = dp[pre][j] + 1;
      ans = max(ans, dp[j][i]);
      if (a[i] == a[j]) pre = j;
    }
  }
  return printf("%d\n", ans);
}
int main() {
  while (~scanf("%d", &n)) {
    Init();
    Solve();
  }
  return 0;
}
