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 long long mod = 1e9 + 7;
const long long MOD = 1e9 + 7;
inline long long add(long long x, long long y) {
  x += y;
  if (x >= mod) x -= mod;
  return x;
}
inline long long sub(long long x, long long y) {
  x -= y;
  if (x < 0) x += mod;
  return x;
}
inline long long mul(long long x, long long y) { return (x * 1ll * y) % mod; }
mt19937 rng32(chrono::steady_clock::now().time_since_epoch().count());
inline void io() {}
signed main() {
  long long n;
  cin >> n;
  long long arr[n];
  long long cnt = 1;
  map<long long, long long> mp;
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
    if (mp[arr[i]] == 0) mp[arr[i]] = cnt, cnt++;
  }
  for (long long i = 0; i < n; i++) arr[i] = mp[arr[i]];
  long long dp[n + 2][n + 2];
  for (long long i = 0; i < n + 2; i++)
    for (long long j = 0; j < n + 2; j++) dp[i][j] = 1;
  long long ans = 1;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < i; j++) {
      dp[i][arr[j]] = max(dp[i][arr[j]], 1 + dp[j][arr[i]]);
      ans = max(ans, dp[i][arr[j]]);
    }
  }
  cout << ans;
}
