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>
#pragma comment(linker, "/STACK:16777216")
using namespace std;
const int I = 11;
pair<int, int> a[5000];
int main() {
  int n;
  cin >> n;
  if (n == 1) {
    cout << 1;
    return 0;
  }
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = i + 1;
  }
  sort(a, a + n);
  int m1 = 1, m2 = 1;
  vector<vector<int>> v;
  for (int i = 0; i < n;) {
    int j = i;
    vector<int> t;
    while (j < n && a[j].first == a[i].first) t.push_back(a[j++].second);
    v.push_back(t);
    i = j;
  }
  int m = 0;
  for (int i = 0; i < v.size(); i++)
    for (int j = i; j < v.size(); j++) {
      int k = 1;
      if (v[i][0] < v[j][0]) {
        int jj = 0, ii = 0;
        while (ii < v[i].size() && jj < v[j].size()) {
          while (jj < v[j].size() && v[j][jj] < v[i][ii]) jj++;
          if (jj < v[j].size()) {
            k++;
            while (ii < v[i].size() && v[i][ii] < v[j][jj]) ii++;
            if (ii < v[i].size()) k++;
          }
        }
      } else {
        int jj = 0, ii = 0;
        while (ii < v[i].size() && jj < v[j].size()) {
          while (ii < v[i].size() && v[i][ii] <= v[j][jj]) ii++;
          if (ii < v[i].size()) {
            k++;
            while (jj < v[j].size() && v[j][jj] <= v[i][ii]) jj++;
            if (jj < v[j].size()) k++;
          }
        }
      }
      if (k > m) m = k;
    }
  cout << m;
  return 0;
}
