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>
#pragma GCC optimize("-Ofast", "-funroll-all-loops")
using namespace std;
const int inf = 0x3f3f3f3f;
const int N = 1e4 + 10, M = 1e6 + 10;
int n, a[N], s, s1, t, d[N], st[N], vis[N], cost;
int head[N], nex[M], to[M], w[M], flow[M], tot = 1;
inline void ade(int a, int b, int c, int d) {
  to[++tot] = b;
  nex[tot] = head[a];
  w[tot] = d;
  flow[tot] = c;
  head[a] = tot;
}
inline void add(int a, int b, int c, int d) {
  ade(a, b, c, d);
  ade(b, a, 0, -d);
}
inline int spfa() {
  queue<int> q;
  q.push(s);
  memset(st, 0, sizeof st);
  memset(d, -inf, sizeof d);
  d[s] = 0;
  while (q.size()) {
    int u = q.front();
    q.pop();
    vis[u] = 0;
    for (int i = head[u]; i; i = nex[i]) {
      if (flow[i] && d[to[i]] < d[u] + w[i]) {
        d[to[i]] = d[u] + w[i];
        if (!vis[to[i]]) q.push(to[i]), vis[to[i]] = 1;
      }
    }
  }
  return d[t] > 0;
}
int dfs(int x, int f) {
  if (x == t) return cost += d[t] * f, f;
  int fl = 0;
  st[x] = 1;
  for (int i = head[x]; i && f; i = nex[i]) {
    if (flow[i] && !st[to[i]] && d[to[i]] == d[x] + w[i]) {
      int mi = dfs(to[i], min(flow[i], f));
      flow[i] -= mi, flow[i ^ 1] += mi, fl += mi, f -= mi;
    }
  }
  return fl;
}
signed main() {
  cin >> n;
  t = n * 2 + 1;
  s1 = t + 1;
  add(s, s1, 4, 0);
  for (int i = 1; i <= n; i++)
    cin >> a[i], add(i, i + n, 1, 1), add(i, i + n, inf, 0);
  for (int i = 1; i <= n; i++) add(s1, i, 1, 0), add(i + n, t, 1, 0);
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++)
      if (a[j] == a[i] - 1) {
        add(i + n, j, inf, 0);
        break;
      }
    for (int j = i + 1; j <= n; j++)
      if (a[j] == a[i] + 1) {
        add(i + n, j, inf, 0);
        break;
      }
    for (int j = i + 1; j <= n; j++)
      if (a[j] % 7 == a[i] % 7) {
        add(i + n, j, inf, 0);
        break;
      }
  }
  while (spfa()) dfs(s, inf);
  cout << cost;
  return 0;
}
