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 MOD = 1e9 + 7;
const double PI = acos(-1.);
const double eps = 1e-10;
const int INF = 0x3f3f3f3f;
const long long INFll = 0x3f3f3f3f3f3f3f3f;
const int MAXN = 1e6 + 5;
const int MX = 123456;
const int inf = 0x3f3f3f3f;
const int MXE = 1e7;
struct MinCost_MaxFlow {
  struct Edge {
    int v, w, nxt;
    int cost;
  } E[MXE];
  int head[MX], tot, level[MX], pre[MX], d[MX];
  bool vis[MX];
  void init() {
    memset(head, -1, sizeof(head));
    tot = 0;
  }
  void add(int u, int v, int w, int cost) {
    E[tot].v = v;
    E[tot].w = w;
    E[tot].cost = cost;
    E[tot].nxt = head[u];
    head[u] = tot++;
    E[tot].v = u;
    E[tot].w = 0;
    E[tot].cost = -cost;
    E[tot].nxt = head[v];
    head[v] = tot++;
  }
  bool spfa(int s, int t) {
    memset(vis, 0, sizeof(vis));
    memset(d, 0x3f, sizeof(d));
    memset(pre, -1, sizeof(pre));
    queue<int> q;
    q.push(s);
    d[s] = 0;
    vis[s] = 1;
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      vis[u] = 0;
      for (int i = head[u]; ~i; i = E[i].nxt) {
        int w = E[i].w, v = E[i].v, cost = E[i].cost;
        if (w > 0 && d[v] > d[u] + cost) {
          d[v] = d[u] + cost;
          pre[v] = i;
          if (!vis[v]) {
            q.push(v);
            vis[v] = 1;
          }
        }
      }
    }
    return pre[t] != -1;
  }
  int solve(int s, int t, int &cost) {
    int flow = 0;
    cost = 0;
    while (spfa(s, t)) {
      int minFlow = inf;
      for (int i = pre[t]; ~i; i = pre[E[i ^ 1].v])
        minFlow = min(minFlow, E[i].w);
      for (int i = pre[t]; ~i; i = pre[E[i ^ 1].v]) {
        cost += minFlow * E[i].cost;
        E[i].w -= minFlow;
        E[i ^ 1].w += minFlow;
      }
      flow += minFlow;
    }
    return flow;
  }
} F;
int a[MAXN];
int main() {
  int n;
  scanf("%d", &n);
  F.init();
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int st = 0, en = 2 * n + 5;
  int add = en - 1;
  F.add(st, add, 4, 0);
  int magic = 3;
  for (int i = 1; i <= n; i++) {
    F.add(add, i, 1, 0);
    F.add(i + n, en, 1, 0);
    F.add(i, i + n, 1, -1);
    F.add(i, i + n, INF, 0);
    int cnt1 = 0, cnt2 = 0;
    for (int j = i + 1; j <= n; j++) {
      if (abs(a[i] - a[j]) == 1 && cnt1 <= magic) {
        F.add(i + n, j, 1, 0);
        cnt1++;
      }
      if (a[i] % 7 == a[j] % 7 && cnt2 <= magic) {
        F.add(i + n, j, 1, 0);
        cnt2++;
      }
    }
  }
  int flow, cost;
  flow = F.solve(st, en, cost);
  printf("%d\n", -cost);
  return 0;
}
