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 N = 4000;
int n, v[N], dp[N][N], nxt[N][N], at[N], freq[N];
vector<int> hs;
int calc(int a, int b) {
  if (b == n) return 0;
  int &res = dp[a][b];
  if (res != -1) return res;
  res = 0;
  int x = nxt[b][v[a]];
  if (x < n && nxt[x][v[b]] < n)
    res = calc(x, nxt[x][v[b]]) + 2;
  else if (x < n)
    res = 1;
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &v[i]);
    hs.push_back(v[i]);
    for (int j = i; j < n; ++j) dp[i][j] = dp[j][i] = -1;
  }
  sort(hs.begin(), hs.end());
  int res = 0;
  for (int i = 0; i < n; ++i) {
    v[i] = lower_bound(hs.begin(), hs.end(), v[i]) - hs.begin();
    ++freq[v[i]];
    res = max(res, freq[v[i]]);
  }
  for (size_t i = 0; i < hs.size(); ++i) at[i] = n;
  for (int i = n - 1; i >= 0; --i) {
    for (size_t j = 0; j < hs.size(); ++j) nxt[i][j] = at[j];
    at[v[i]] = i;
  }
  for (int i = 0; i < n; ++i)
    for (int j = i + 1; j < n; ++j) res = max(res, calc(i, j) + 2);
  printf("%d\n", res);
  return 0;
}
