Problem: 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).

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class CF813D {
	static class A {
		int a, i, c;
		A(int a, int i) {
			this.a = a; this.i = i;
		}
	}
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		StringTokenizer st = new StringTokenizer(br.readLine());
		A[] aa = new A[n];
		for (int i = 0; i < n; i++) {
			int a = Integer.parseInt(st.nextToken());
			aa[i] = new A(a, i);
		}
		Arrays.sort(aa, (p, q) -> p.a - q.a);
		int[] bb = new int[n];
		int k = 0;
		for (int i = 0, b = 0; i < n; i++) {
			if (aa[i].a > b)
				bb[k++] = b = aa[i].a;
			aa[i].c = k - 1;
		}
		Arrays.sort(aa, (p, q) -> p.i - q.i);
		int[][] dp = new int[n + 1][n + 1];
		for (int y = 0; y <= n; y++) {
			int[] maxmod = new int[7];
			int[] maxnum = new int[k];
			for (int x = 0; x <= n; x++) {
				if (x == y)
					continue;
				if (x > y) {
					int a = aa[x - 1].a;
					int r = aa[x - 1].a % 7;
					int c = aa[x - 1].c;
					dp[x][y] = dp[0][y] + 1;
					dp[x][y] = Math.max(dp[x][y], maxmod[r] + 1);
					if (c - 1 >= 0 && bb[c - 1] == a - 1)
						dp[x][y] = Math.max(dp[x][y], maxnum[c - 1] + 1);
					if (c + 1 <  k && bb[c + 1] == a + 1)
						dp[x][y] = Math.max(dp[x][y], maxnum[c + 1] + 1);
				} else
					dp[x][y] = dp[y][x];
				if (x > 0) {
					int r = aa[x - 1].a % 7;
					int c = aa[x - 1].c;
					maxmod[r] = Math.max(maxmod[r], dp[x][y]);
					maxnum[c] = Math.max(maxnum[c], dp[x][y]);
				}
			}
		}
		int max = 0;
		for (int x = 2; x <= n; x++)
			for (int y = 1; y < x; y++)
				max = Math.max(max, dp[x][y]);
		System.out.println(max);
	}
}