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 N = 12005, M = 33005, Inf = 0x3f3f3f3f;
int n, a[N], lst[100005], ls[7], s, t, cost;
int eid = 1, he[N], nw[N], q[N], d[N], vs[N];
struct edge {
  int v, nx, c, w;
} e[M * 2];
void add(int u, int v, int c, int w) {
  e[++eid] = {v, he[u], c, w}, he[u] = eid;
  e[++eid] = {u, he[v], 0, -w}, he[v] = eid;
}
int spfa() {
  memset(d, 63, sizeof d), memcpy(nw, he, sizeof he);
  q[0] = s, d[s] = 0, vs[s] = 1;
  for (int l = 0, r = 0; l <= r; l++) {
    int u = q[l % N];
    vs[u] = 0;
    for (int i = he[u], v; (v = e[i].v); i = e[i].nx)
      if (e[i].c && d[u] + e[i].w < d[v]) {
        d[v] = d[u] + e[i].w;
        if (!vs[v]) q[++r % N] = v, vs[v] = 1;
      }
  }
  return d[t] < Inf;
}
int dfs(int u, int lim, int re = 0) {
  if (u == t) return lim;
  vs[u] = 1;
  for (int &i = nw[u], v; v = e[i].v; i = e[i].nx)
    if (!vs[v] && e[i].c && d[v] == d[u] + e[i].w) {
      int t = dfs(v, min(lim, e[i].c));
      e[i].c -= t, e[i ^ 1].c += t, re += t, lim -= t, cost += t * e[i].w;
      if (!lim) break;
    }
  if (lim) d[u] = Inf;
  vs[u] = 0;
  return re;
}
int main() {
  cin >> n;
  s = n * 4 + 1, t = s + 1;
  for (int i = (1); i <= (n); i++) scanf("%d", &a[i]);
  for (int i = (n); i >= (1); i--) {
    int i1 = i + n, i2 = i1 + n, i3 = i2 + n;
    int l2 = lst[a[i] - 1], L2 = lst[a[i] + 1], &l3 = ls[a[i] % 7];
    add(s, i, 1, 0), add(i, i1, 1, -1), add(i1, t, 1, 0);
    add(i2, i, 1, 0), add(i3, i, 1, 0);
    auto Add = [&](int i, int l) {
      if (l) add(i, l, 4, 0), add(i1, l, 1, 0);
    };
    Add(i2, l2), Add(i2, L2), Add(i3, l3), lst[a[i]] = i2, l3 = i3;
  }
  add(t, t + 1, 4, 0), t++;
  while (spfa()) dfs(s, Inf);
  cout << -cost;
  return 0;
}