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 MAXN = 3e3 + 10;
const int inf = 1e9;
struct Edge {
  int to, nxt, w, c;
};
int n, S, S0, T, cnt_e = 1;
int a[MAXN];
int head[MAXN * 10], curh[MAXN * 10];
int dis[MAXN * 10];
bool inq[MAXN * 10];
Edge e[MAXN * 20];
inline void addedge(int u, int v, int w, int c, bool fg) {
  e[++cnt_e].to = v;
  e[cnt_e].nxt = head[u];
  e[cnt_e].w = w;
  e[cnt_e].c = c;
  head[u] = cnt_e;
  if (fg) addedge(v, u, 0, -c, false);
}
inline bool spfa() {
  for (int i = 1; i <= 4 * n + 3; i++) curh[i] = head[i];
  for (int i = 1; i <= 4 * n + 3; i++) dis[i] = -inf, inq[i] = false;
  queue<int> q;
  q.push(S);
  dis[S] = 0;
  inq[S] = true;
  while (!q.empty()) {
    int u = q.front();
    inq[u] = false;
    q.pop();
    for (int i = head[u]; i; i = e[i].nxt) {
      int v = e[i].to, w = e[i].w, c = e[i].c;
      if (dis[v] >= dis[u] + c || !w) continue;
      dis[v] = dis[u] + c;
      if (!inq[v]) q.push(v), inq[v] = true;
    }
  }
  return dis[T] != -inf;
}
int cost;
inline int dfs(int u, int flow) {
  if (u == T || !flow) return flow;
  inq[u] = true;
  int res = 0;
  for (int &i = curh[u]; i; i = e[i].nxt) {
    int v = e[i].to, w = e[i].w, c = e[i].c;
    if (dis[v] != dis[u] + c || !w || inq[v]) continue;
    int k = dfs(v, min(flow, w));
    e[i].w -= k;
    e[i ^ 1].w += k;
    flow -= k;
    res += k;
    cost += k * c;
    if (!flow) break;
  }
  inq[u] = false;
  return res;
}
inline void dinic() {
  while (spfa()) dfs(S, inf);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  S = 4 * n + 1, S0 = 4 * n + 2, T = 4 * n + 3;
  addedge(S, S0, 4, 0, true);
  for (int i = 1; i <= n; i++) {
    addedge(S0, i, 1, 0, true);
    addedge(i, i + n, 1, 1, true);
    addedge(i + n, T, 1, 0, true);
    addedge(i + 2 * n, i, 1, 0, true);
    addedge(i + 3 * n, i, 1, 0, true);
    for (int j = i + 1; j <= n; j++)
      if (a[i] - a[j] == 1) {
        addedge(i + n, j + 3 * n, inf, 0, true);
        break;
      }
    for (int j = i + 1; j <= n; j++)
      if (a[j] - a[i] == 1) {
        addedge(i + n, j + 3 * n, inf, 0, true);
        break;
      }
    for (int j = i + 1; j <= n; j++)
      if (a[i] == a[j]) {
        addedge(i + 3 * n, j + 3 * n, inf, 0, true);
        break;
      }
    for (int j = i + 1; j <= n; j++)
      if (a[i] % 7 == a[j] % 7) {
        addedge(i + n, j + 2 * n, inf, 0, true);
        break;
      }
    for (int j = i + 1; j <= n; j++)
      if (a[i] % 7 == a[j] % 7) {
        addedge(i + 2 * n, j + 2 * n, inf, 0, true);
        break;
      }
  }
  dinic();
  printf("%d\n", cost);
  return 0;
}