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