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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
template <class T, class U>
inline void Max(T &a, U b) {
  if (a < b) a = b;
}
template <class T, class U>
inline void Min(T &a, U b) {
  if (a > b) a = b;
}
inline void add(int &a, int b) {
  a += b;
  while (a >= 1000000007) a -= 1000000007;
}
int pow(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = ans * (long long)a % 1000000007;
    a = (long long)a * a % 1000000007;
    b >>= 1;
  }
  return ans;
}
int dp[5005][5005], a[5005], last[100010], vis[100010], now, g[7];
int main() {
  int T, i = 0, j, k, ca = 0, m = 0, n;
  scanf("%d", &n);
  for (int i = 1; i < n + 1; i++) scanf("%d", &a[i]);
  for (int i = 1; i < n; i++) {
    for (int j = 0; j < i; j++) Max(dp[i][j], j == 0 ? 1 : 2);
    now++;
    for (int j = 0; j < 7; j++) g[j] = n + 1;
    for (int j = n + 1 - 1; j >= i + 1; j--) {
      last[a[j]] = j;
      g[a[j] % 7] = j;
      vis[a[j]] = now;
    }
    int first = a[i], second = n + 1;
    if (vis[first - 1] == now) {
      second = last[first - 1];
      for (int j = 0; j < i; j++) {
        Max(dp[second][j], dp[i][j] + 1);
      }
    }
    if (vis[first + 1] == now) {
      second = last[first + 1];
      for (int j = 0; j < i; j++) {
        Max(dp[second][j], dp[i][j] + 1);
      }
    }
    first %= 7;
    second = g[first];
    if (second <= n) {
      for (int j = 0; j < i; j++) {
        Max(dp[second][j], dp[i][j] + 1);
      }
    }
    for (int j = 1; j < i; j++) {
      first = a[j];
      second = n + 1;
      if (vis[first - 1] == now) {
        second = last[first - 1];
        Max(dp[second][i], dp[i][j] + 1);
      }
      if (vis[first + 1] == now) {
        second = last[first + 1];
        Max(dp[second][i], dp[i][j] + 1);
      }
      first %= 7;
      second = g[first];
      if (second <= n) {
        Max(dp[second][i], dp[i][j] + 1);
      }
    }
    for (int j = i + 1; j < n + 1; j++) {
      Max(dp[j][i], dp[i][0] + 1);
    }
  }
  int ans = 2;
  for (int i = 1; i < n + 1; i++) {
    for (int j = 0; j < i; j++) Max(ans, dp[i][j]);
  }
  printf("%d\n", ans);
  return 0;
}
