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 = 6e3 + 5, M = 1e5 + 5;
const int INF = 0x3f3f3f3f;
int tot = 1, lnk[N], cur[N], ter[M], nxt[M], cap[M], cost[M], dis[N], ret;
bool vis[N];
void add(int u, int v, int w, int c) {
  ter[++tot] = v, nxt[tot] = lnk[u], lnk[u] = tot, cap[tot] = w, cost[tot] = c;
}
void addedge(int u, int v, int w, int c) { add(u, v, w, c), add(v, u, 0, -c); }
bool spfa(int s, int t) {
  memset(dis, 0x3f, sizeof(dis));
  memcpy(cur, lnk, sizeof(lnk));
  std::queue<int> q;
  q.push(s), dis[s] = 0, vis[s] = 1;
  while (!q.empty()) {
    int u = q.front();
    q.pop(), vis[u] = 0;
    for (int i = lnk[u]; i; i = nxt[i]) {
      int v = ter[i];
      if (cap[i] && dis[v] > dis[u] + cost[i]) {
        dis[v] = dis[u] + cost[i];
        if (!vis[v]) q.push(v), vis[v] = 1;
      }
    }
  }
  return dis[t] != INF;
}
int dfs(int u, int t, int flow) {
  if (u == t) return flow;
  vis[u] = 1;
  int ans = 0;
  for (int &i = cur[u]; i && ans < flow; i = nxt[i]) {
    int v = ter[i];
    if (!vis[v] && cap[i] && dis[v] == dis[u] + cost[i]) {
      int x = dfs(v, t, std::min(cap[i], flow - ans));
      if (x) ret += x * cost[i], cap[i] -= x, cap[i ^ 1] += x, ans += x;
    }
  }
  vis[u] = 0;
  return ans;
}
int mcmf(int s, int t) {
  int ans = 0;
  while (spfa(s, t)) {
    int x;
    while ((x = dfs(s, t, INF))) ans += x;
  }
  return ans;
}
int n, a[3005], nx[100005], nm[15];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  fill(nx + 1, nx + 100001, n + 1);
  fill(nm, nm + 7, n + 1);
  for (int i = n; i >= 1; i--) {
    addedge(n * 2 + 2, i * 2 - 1, 1, 0);
    addedge(i * 2 - 1, i * 2, 1, -1);
    addedge(i * 2, n * 2 + 1, 1, 0);
    addedge(i * 2 - 1, nx[a[i]] * 2 - 1, 1, 0);
    for (int j = i + 1; j <= n; j++)
      if (a[i] - 1 == a[j]) {
        addedge(i * 2, j * 2 - 1, 1, 0);
        break;
      }
    for (int j = i + 1; j <= n; j++)
      if (a[i] + 1 == a[j]) {
        addedge(i * 2, j * 2 - 1, 1, 0);
        break;
      }
    nx[a[i]] = i;
    addedge(i * 2 - 1, nm[a[i] % 7] * 2 - 1, 1, 0);
    for (int j = i + 1; j <= n; j++)
      if (a[i] % 7 == a[j] % 7) {
        addedge(i * 2, j * 2 - 1, 1, 0);
        break;
      }
    nm[a[i] % 7] = i;
  }
  addedge(0, n * 2 + 2, 4, 0);
  mcmf(0, n * 2 + 1);
  cout << -ret;
  return 0;
}