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>
using namespace std;
const int MX = 5022;
int sv[MX][MX], a[MX];
short int minus1[MX][MX], plus1[MX][MX], mod7[MX][10], vs[100010];
int id;
inline int max(int a, int b) {
  if (a < b) return b;
  return a;
}
int go(int x, int y) {
  int &ret = sv[x][y];
  if (ret != -1) return ret;
  ret = 0;
  int mx = max(x, y);
  int xx = a[x] % 7;
  int yy = a[y] % 7;
  if (mod7[mx][xx] <= id) {
    ret = max(ret, 1 + go(mod7[mx][xx], y));
  }
  if (minus1[x][mx] <= id && minus1[x][mx] >= 8) {
    ret = max(ret, 1 + go(minus1[x][mx], y));
  }
  if (plus1[x][mx] <= id && plus1[x][mx] >= 8) {
    ret = max(ret, 1 + go(plus1[x][mx], y));
  }
  if (mod7[mx][yy] <= id) {
    ret = max(ret, 1 + go(x, mod7[mx][yy]));
  }
  if (minus1[y][mx] <= id && minus1[y][mx] >= 8) {
    ret = max(ret, 1 + go(x, minus1[y][mx]));
  }
  if (plus1[y][mx] <= id && plus1[y][mx] >= 8) {
    ret = max(ret, 1 + go(x, plus1[y][mx]));
  }
  return ret;
}
void pre() {
  int ta[10];
  for (int i = 0; i < 7; ++i) {
    ta[i] = id + 1;
  }
  for (int i = id; i >= 1; --i) {
    int t = a[i] % 7;
    for (int j = 0; j < 7; j++) {
      mod7[i][j] = ta[j];
    }
    ta[t] = i;
  }
  for (int i = 1; i <= id; ++i) {
    for (int j = 1; j <= id; ++j) {
      vs[a[j]] = id + 1;
    }
    for (int j = id; j > 7; --j) {
      if (j > i)
        vs[a[j]] = j;
      else {
        minus1[j][i] = vs[a[j] - 1];
        plus1[j][i] = vs[a[j] + 1];
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, x;
  for (int i = 1; i < 8; i++) {
    a[++id] = i;
  }
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    a[++id] = x;
  }
  pre();
  int Ans = 0;
  memset(sv, -1, sizeof sv);
  for (int i = 1; i <= 7; ++i) {
    for (int j = 1; j <= 7; ++j) {
      if (i != j) {
        Ans = max(Ans, go(i, j));
      }
    }
  }
  cout << Ans << endl;
  return 0;
}