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;
vector<int> v[1000010];
struct ln {
  int belong, i;
} ss[4010];
int s[4010], n;
bool cmp(ln a, ln b) { return a.i < b.i; }
int judge(int a, int b) {
  int cnt = 0;
  for (int i = 0; i < v[a].size(); i++) {
    ss[cnt].i = v[a][i];
    ss[cnt++].belong = 1;
  }
  for (int i = 0; i < v[b].size(); i++) {
    ss[cnt].i = v[b][i];
    ss[cnt++].belong = 2;
  }
  sort(ss, ss + cnt, cmp);
  int p = 1;
  int t = ss[0].belong;
  for (int i = 1; i < cnt; i++) {
    if (ss[i].belong != t) {
      t = ss[i].belong;
      p++;
    }
  }
  return p;
}
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; i++) {
      int x;
      scanf("%d", &x);
      v[x].push_back(i);
    }
    int cnt = 0, maxx = 0;
    for (int i = 1; i < 1000010; i++) {
      if (v[i].size()) {
        s[cnt++] = i;
        int tt = v[i].size();
        maxx = max(maxx, tt);
      }
    }
    for (int i = 0; i < cnt; i++) {
      for (int j = 0; j < cnt; j++) {
        if (i != j) {
          int tmp = judge(s[i], s[j]);
          maxx = max(maxx, tmp);
        }
      }
    }
    cout << maxx << endl;
    for (int i = 0; i < cnt; i++) v[s[i]].clear();
  }
  return 0;
}
