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 CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast", "unroll-loops", "omit-frame-pointer", "inline")
#pragma GCC option("arch=native", "tune=native", "no-zero-upper")
#pragma GCC target("avx2")
using namespace std;
const long long INF = 1e16;
const int maxn = 1e6 + 7;
const int mod = 1000000;
inline bool read(long long &num) {
  char in;
  bool IsN = false;
  in = getchar();
  if (in == EOF) return false;
  while (in != '-' && (in < '0' || in > '9')) in = getchar();
  if (in == '-') {
    IsN = true;
    num = 0;
  } else
    num = in - '0';
  while (in = getchar(), in >= '0' && in <= '9') {
    num *= 10, num += in - '0';
  }
  if (IsN) num = -num;
  return true;
}
long long n, m, p;
long long dp[5005][5005];
long long num[maxn];
long long mxp[100005], mxm[10];
int main() {
  long long ans = 0;
  read(n);
  for (int i = 1; i <= n; i++) read(num[i]);
  for (int i = 1; i <= n; i++) dp[0][i] = dp[i][0] = 1;
  for (int i = 0; i <= n; i++) {
    for (int k = 1; k <= i; k++) {
      mxp[num[k]] = max(mxp[num[k]], dp[i][k]);
      mxm[num[k] % 7] = max(mxm[num[k] % 7], dp[i][k]);
    }
    for (int j = i + 1; j <= n; j++) {
      dp[i][j] = max(dp[i][0] + 1, dp[i][j]);
      long long temp = max(mxp[num[j] - 1] + 1, mxp[num[j] + 1] + 1);
      dp[i][j] = max(mxm[num[j] % 7] + 1, dp[i][j]);
      dp[i][j] = max(temp, dp[i][j]);
      mxp[num[j]] = max(mxp[num[j]], dp[i][j]);
      mxm[num[j] % 7] = max(mxm[num[j] % 7], dp[i][j]);
      dp[j][i] = dp[i][j];
    }
    for (int k = 1; k <= n; k++) mxp[num[k]] = 0, mxm[num[k] % 7] = 0;
  }
  for (int i = 1; i <= n; i++) {
    for (int k = 1; k <= n; k++) ans = max(ans, dp[i][k]);
  }
  printf("%lld\n", ans);
  return 0;
}