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>
time_t st = clock();
using namespace std;
template <class stl>
void DBGSTL(stl a) {
  cout << "\tDEBUG:\n\t";
  for (typeof(a.begin()) i = a.begin(); i != a.end(); i++) {
    cerr << *i << " ";
  }
  cerr << "\n";
  return;
}
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
double mathround(double x) {
  int intx = (int)x;
  if (x < intx + 0.5) return intx;
  return intx + 1;
}
int n, mx = 1;
int p[4000];
int bf[4010][4010];
int dp[4010][4010];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> p[i];
  int ind = -1;
  for (int i = 0; i < n; i++) {
    ind = -1;
    for (int j = 0; j < n; j++) {
      if (p[j] == p[i]) ind = j;
      bf[i][j] = ind;
    }
  }
  int ext1, ext2;
  for (int i = 0; i < n; i++) {
    if (i == 0)
      dp[0][0] = 1;
    else {
      for (int j = i - 1; j >= 0; j--) {
        if (j == 0)
          dp[i][j] = 2;
        else {
          ext1 = bf[i][j - 1];
          if (ext1 == -1)
            dp[i][j] = 2;
          else if (ext1 == 0)
            dp[i][j] = 3;
          else {
            ext2 = bf[j][ext1 - 1];
            if (ext2 == -1)
              dp[i][j] = 3;
            else
              dp[i][j] = dp[ext1][ext2] + 2;
          }
        }
      }
    }
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      mx = max(mx, dp[i][j]);
    }
  cout << mx;
  return 0;
}
