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 Imx = 2147483647;
const int mod = 1000000007;
const long long Lbig = 2e18;
inline long long getnum() {
  register long long r = 0;
  register bool ng = 0;
  register char c;
  c = getchar();
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  if (c == '-') ng = 1, c = getchar();
  while (c >= '0' && c <= '9') r = r * 10 + c - '0', c = getchar();
  if (ng) r = -r;
  return r;
}
template <class T>
inline void putnum(T x) {
  if (x < 0) putchar('-'), x = -x;
  register short a[20] = {}, sz = 0;
  while (x > 0) a[sz++] = x % 10, x /= 10;
  if (sz == 0) putchar('0');
  for (int i = sz - 1; i >= 0; i--) putchar('0' + a[i]);
}
inline void putsp() { putchar(' '); }
inline void putendl() { putchar('\n'); }
inline char mygetchar() {
  register char c = getchar();
  while (c == ' ' || c == '\n') c = getchar();
  return c;
}
int n, a[5011];
int tabr[5011], tabc[5011][5011], dp[5011][5011], id0[5011], id1[5011],
    id2[5011];
int mxr[7], mxc[5011][7];
map<pair<int, int>, int> mp;
map<int, int> id;
int main() {
  n = getnum();
  for (int i = 1; i <= n; i++) a[i] = getnum(), id[a[i]] = i;
  for (int i = 1; i <= n; i++) {
    id0[i] = id[a[i]];
    id1[i] = id[a[i] + 1];
    id2[i] = id[a[i] - 1];
  }
  int mx = 0;
  for (int i = 0; i <= n; i++) {
    memset(mxr, 0, sizeof(mxr));
    memset(tabr, 0, sizeof(tabr));
    for (int j = 0; j <= n; j++) {
      if (i == j) continue;
      int &v = dp[i][j];
      v = (i > 0) + (j > 0);
      if (i > j) {
        v = max(v, mxc[j][a[i] % 7] + 1);
        if (id1[i]) v = max(v, tabc[j][id1[i]] + 1);
        if (id2[i]) v = max(v, tabc[j][id2[i]] + 1);
      } else {
        v = max(v, mxr[a[j] % 7] + 1);
        if (id1[j]) v = max(v, tabr[id1[j]] + 1);
        if (id2[j]) v = max(v, tabr[id2[j]] + 1);
      }
      if (j == 0)
        for (int t = 0; t < 7; t++) mxr[t] = max(mxr[t], v);
      else
        mxr[a[j] % 7] = max(mxr[a[j] % 7], v);
      if (i == 0)
        for (int t = 0; t < 7; t++) mxc[j][t] = max(mxc[j][t], v);
      else
        mxc[j][a[i] % 7] = max(mxc[j][a[i] % 7], v);
      tabr[id0[j]] = max(tabr[id0[j]], v);
      tabc[j][id0[i]] = max(tabc[j][id0[i]], v);
      mx = max(mx, v);
    }
  }
  cout << mx << endl;
  return 0;
}
