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).
import java.io.*;
public class Two_Melodies{
   public static void main(String[] args) throws IOException{
      BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
      int N = Integer.parseInt(br.readLine());
      String[] inp = br.readLine().split(" ");
      int[] a = new int[5001];
      int[] maxMod = new int[7];
      int[] maxNum = new int[100001];
      int[][] dp = new int[5001][5001];
      
      for(int i = 0; i < inp.length; i++){
         a[i+1] = Integer.parseInt(inp[i]);
      }
      
      int max = 0;
      for(int y = 0; y <= N; y++){
         maxMod = new int[7];
         maxNum = new int[100001];
         for(int x = 0; x < y; x++){
            if(dp[x][y] > maxMod[a[x]%7])
            	maxMod[a[x]%7] = dp[x][y];
            if(dp[x][y] > maxNum[a[x]])
            	maxNum[a[x]] = dp[x][y];
         }
         for(int x = y+1; x <= N; x++){
            int adj1 = 0;
            int adj2 = 0;
            if(a[x]==1){
               adj1 = 0;
               adj2 = maxNum[a[x]+1];
            }
            else if(a[x] == 100000){
               adj1 = maxNum[a[x]-1];
               adj2 = 0;   
            }
            else{
               adj1 = maxNum[a[x]-1];
               adj2 = maxNum[a[x]+1];
            }
            int max1 = Math.max(adj1+1, adj2+1);
            int max2 = Math.max(maxMod[a[x]%7]+1, dp[0][y]+1);
            int max3 = Math.max(max1,  max2);
            
            //System.out.print(max3+" "+x+" "+y+"\n");
            dp[x][y] = max3;
            dp[y][x] = max3;
            
            if(dp[x][y] > maxMod[a[x]%7])
            	maxMod[a[x]%7] = dp[x][y];
            if(dp[x][y] > maxNum[a[x]])
            	maxNum[a[x]] = dp[x][y];
            
            if(max3 > max)
               max = max3;
         }
      }
      
      System.out.println(max);
   }
}