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 min1() { return INT_MAX; }
template <typename... Args>
int min1(int a, Args... args) {
  return min(a, min1(args...));
}
const long double PI = 3.141592653589793238462643383279;
const long double EPS = 1e-7;
const long long INF = 1e9 + 7;
const long long MOD = INF;
const long long maxn = 1e6 + 2;
int dp[4001][4001];
int szh[maxn];
int main() {
  int n;
  cin >> n;
  for (int i = (1); i < (n + 1); ++i) {
    int a;
    cin >> a;
    if (!szh[a]) szh[a] = i;
    for (int j = (1); j < (i); ++j) {
      if (dp[szh[a]][j]) dp[j][szh[a]] = dp[szh[a]][j] + 1;
      if (!dp[j][szh[a]]) dp[j][szh[a]] = 2;
    }
  }
  if (n == 1) {
    cout << 1;
    return 0;
  }
  int ans = 0;
  for (int i = (1); i < (n + 1); ++i)
    for (int j = (1); j < (n + 1); ++j) ans = max(ans, dp[i][j]);
  cout << ans;
}
