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;
const int dx4[5] = {0, -1, 0, 1, 0};
const int dy4[5] = {0, 0, 1, 0, -1};
const int dx8[9] = {0, -1, 0, 1, 0, 1, 1, -1, -1};
const int dy8[9] = {0, 0, 1, 0, -1, 1, -1, 1, -1};
const int maxn = 1e5 + 5;
const double PI = acos(-1.0);
const long long mod = 1e9 + 7;
long long po(long long a, long long b, long long mod) {
  long long res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (a == 0) {
    return b;
  } else {
    return gcd(b % a, a);
  }
}
void YES() {
  puts("YES");
  exit(0);
}
void Yes() { puts("Yes"); }
void NO() {
  puts("NO");
  exit(0);
}
void No() { puts("No"); }
void one() {
  puts("-1");
  exit(0);
}
int n;
int val[5005];
int cnt[maxn], monum[15];
int maxmod[5005], maxnum[maxn];
int dp[5005][5005];
void solve() {
  scanf("%d", &(n));
  for (int i = 1; i <= n; i++) {
    scanf("%d", &(val[i]));
  }
  int ans = 0;
  for (int i = 0; i <= n; i++) {
    memset(maxmod, 0, sizeof(maxmod));
    memset(maxnum, 0, sizeof(maxnum));
    for (int j = 1; j < i; j++) {
      maxmod[val[j] % 7] = max(maxmod[val[j] % 7], dp[i][j]);
      maxnum[val[j]] = max(maxnum[val[j]], dp[i][j]);
    }
    for (int j = i + 1; j <= n; j++) {
      dp[i][j] = max(dp[i][0] + 1, dp[i][j]);
      dp[i][j] = max(maxmod[val[j] % 7] + 1, dp[i][j]);
      dp[i][j] = max(maxnum[val[j] + 1] + 1, dp[i][j]);
      dp[i][j] = max(maxnum[val[j] - 1] + 1, dp[i][j]);
      maxmod[val[j] % 7] = max(maxmod[val[j] % 7], dp[i][j]);
      maxnum[val[j]] = max(maxnum[val[j]], dp[i][j]);
      dp[j][i] = dp[i][j];
      ans = max(ans, dp[i][j]);
    }
  }
  printf("%d\n", ans);
}
int main() {
  solve();
  return 0;
}
