Author note: I think some of you might remember the problem "Two Melodies" from Eductational Codeforces Round 22. Now it's time to make it a bit more difficult!

Alice is a composer, and recently she had recorded two tracks that became very popular. Now she has got a lot of fans who are waiting for new tracks. 

This time Alice wants to form four melodies for her tracks.

Alice has a sheet with n notes written on it. She wants to take four such non-empty non-intersecting subsequences that all of them form a melody and sum of their lengths is maximal.

Subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.

Subsequence forms a melody when each two adjacent notes either differ by 1 or are congruent modulo 7.

You should write a program which will calculate maximum sum of lengths of such four non-empty non-intersecting subsequences that all of them form a melody.

Input

The first line contains one integer number n (4 ≤ n ≤ 3000).

The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 105) — notes written on a sheet.

Output

Print maximum sum of lengths of such four non-empty non-intersecting subsequences that all of them form a melody.

Examples

Input

5
1 3 5 7 9


Output

4


Input

5
1 3 5 7 2


Output

5

Note

In the first example it is possible to compose 4 melodies by choosing any 4 notes (and each melody will consist of only one note).

In the second example it is possible to compose one melody with 2 notes — {1, 2}. Remaining notes are used in other three melodies (one note per each melody).
#include <bits/stdc++.h>
using namespace std;
int dp[5002][7];
int dp2[5002][5002];
int dp3[5002];
int a[5002];
int n;
vector<int> v;
int mp[100002];
int main() {
  memset(mp, -1, sizeof(mp));
  cin >> n;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    v.push_back(a[i]);
  }
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  for (int i = 0; i < v.size(); i++) {
    mp[v[i]] = i;
  }
  for (int i = 0; i < n; i++) {
    int f = 1;
    for (int k = 0; k < i; k++) {
      int ne = dp[k][a[i] % 7];
      if (mp[a[i] - 1] != -1) {
        ne = max(ne, dp2[k][mp[a[i] - 1]]);
      }
      if (mp[a[i] + 1] != -1) {
        ne = max(ne, dp2[k][mp[a[i] + 1]]);
      }
      ne++;
      dp[k][a[i] % 7] = max(dp[k][a[i] % 7], ne);
      dp2[k][mp[a[i]]] = max(dp2[k][mp[a[i]]], ne);
      dp[i][a[k] % 7] = max(dp[i][a[k] % 7], ne);
      dp2[i][mp[a[k]]] = max(dp2[i][mp[a[k]]], ne);
      f = max(f, ne);
    }
    dp3[i] = 1;
    for (int k = 0; k < i; k++) {
      if (a[k] % 7 == a[i] % 7 || abs(a[i] - a[k]) == 1) {
        dp3[i] = max(dp3[i], dp3[k] + 1);
      }
    }
    for (int k = 0; k < i; k++) {
      dp[k][a[i] % 7] = max(dp[k][a[i] % 7], dp3[k] + 1);
      dp2[k][mp[a[i]]] = max(dp2[k][mp[a[i]]], dp3[k] + 1);
      dp[i][a[k] % 7] = max(dp[i][a[k] % 7], dp3[k] + 1);
      dp2[i][mp[a[k]]] = max(dp2[i][mp[a[k]]], dp3[k] + 1);
    }
  }
  int ans = 2;
  for (int i = 0; i < 5002; i++) {
    for (int j = 0; j < 5002; j++) {
      ans = max(ans, dp2[i][j]);
    }
  }
  cout << ans << endl;
  return 0;
}
