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;
int n, x, y;
bool f[1000010];
vector<int> v, b[1000010];
int ans = 0, res = 0;
int main(int argc, char **argv) {
  scanf("%d", &n);
  if (n == 1) {
    puts("1");
    return 0;
  }
  for (int i = 0; i < n; ++i) {
    scanf("%d", &x);
    b[x].push_back(i);
    if (!f[x]) {
      f[x] = 1;
      v.push_back(x);
    }
  }
  for (int i = 0; i < (int)v.size(); ++i) {
    for (int j = i; j < (int)v.size(); ++j) {
      x = v[i];
      y = v[j];
      if (i == j) {
        res = max(res, (int)b[x].size());
        continue;
      }
      bool ok = 0;
      int idx = -1;
      ans = 0;
      if (b[x][0] < b[y][0]) {
        while (idx <= (int)b[x].back() || idx <= (int)b[y].back()) {
          if (ok == 0) {
            if (idx == -1) {
              ++ans;
              idx = b[x][0];
            } else {
              int ii =
                  lower_bound(b[x].begin(), b[x].end(), idx) - b[x].begin();
              if (ii < (int)b[x].size()) {
                ++ans;
                idx = b[x][ii];
              } else
                idx = 1e9;
            }
          } else {
            if (idx == -1) {
              ++ans;
              idx = b[y][0];
            } else {
              int ii =
                  lower_bound(b[y].begin(), b[y].end(), idx) - b[y].begin();
              if (ii < (int)b[y].size()) {
                ++ans;
                idx = b[y][ii];
              } else
                idx = 1e9;
            }
          }
          ok ^= 1;
        }
      } else {
        swap(x, y);
        while (idx <= (int)b[x].back() || idx <= (int)b[y].back()) {
          if (ok == 0) {
            if (idx == -1) {
              ++ans;
              idx = b[x][0];
            } else {
              int ii =
                  lower_bound(b[x].begin(), b[x].end(), idx) - b[x].begin();
              if (ii < (int)b[x].size()) {
                ++ans;
                idx = b[x][ii];
              } else
                idx = 1e9;
            }
          } else {
            if (idx == -1) {
              ++ans;
              idx = b[y][0];
            } else {
              int ii =
                  lower_bound(b[y].begin(), b[y].end(), idx) - b[y].begin();
              if (ii < (int)b[y].size()) {
                ++ans;
                idx = b[y][ii];
              } else
                idx = 1e9;
            }
          }
          ok ^= 1;
        }
      }
      res = max(res, ans);
    }
  }
  cout << res << endl;
  return 0;
}
