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 maxn = 3005;
int n;
int a[maxn];
namespace Dinic {
const int maxv = 3005 * 2;
const int maxe = 1e7 + 5;
int tot, hd, tl;
int head[maxv];
int to[maxe], cap[maxe], cost[maxe], nxt[maxe];
int dis[maxv], h[maxv], vis[maxv], fv[maxv], fe[maxv];
int deg[maxv], que[maxv];
inline void init() { memset(head, -1, sizeof(head)); }
inline void add(int u, int v, int w, int c) {
  to[tot] = v, cap[tot] = w, cost[tot] = c, nxt[tot] = head[u], head[u] = tot++;
  return;
}
inline void addEdge(int u, int v, int w, int c) {
  add(u, v, w, c);
  add(v, u, 0, -c);
  return;
}
inline void topoSort() {
  for (int i = (0); i < (tot); ++i)
    if (cap[i]) ++deg[to[i]];
  for (int u = (0); u < (maxv); ++u)
    if (!deg[u]) que[tl++] = u;
  for (; hd < tl; ++hd) {
    int u = que[hd];
    for (int i = head[u]; ~i; i = nxt[i])
      if (cap[i]) {
        int v = to[i];
        if (!(--deg[v])) que[tl++] = v;
      }
  }
  return;
}
inline void dpOnDag(int s) {
  topoSort();
  memset(dis, 0x3f, sizeof(dis));
  dis[s] = 0;
  for (int i = (0); i < (tl); ++i) {
    int u = que[i];
    for (int j = head[u]; ~j; j = nxt[j])
      if (cap[j]) {
        int v = to[j], w = cost[j];
        dis[v] = min(dis[u] + w, dis[v]);
      }
  }
  return;
}
inline void dijkstra(int s) {
  memset(dis, 0x3f, sizeof(dis));
  memset(vis, 0, sizeof(vis));
  dis[s] = 0;
  for (int i = (0); i < (maxv); ++i) {
    int u = -1;
    for (int v = (0); v < (maxv); ++v)
      if (!vis[v] && (!~u || dis[v] < dis[u])) u = v;
    if (!~u) break;
    vis[u] = 1;
    for (int j = head[u]; ~j; j = nxt[j])
      if (cap[j]) {
        int v = to[j], w = cost[j];
        if (dis[v] > dis[u] + w + h[u] - h[v]) {
          dis[v] = dis[u] + w + h[u] - h[v];
          fv[v] = u;
          fe[v] = j;
        }
      }
  }
  return;
}
inline int minCostFlow(int s, int t) {
  dpOnDag(s);
  for (int u = (0); u < (maxv); ++u) h[u] += dis[u];
  int ret = 0;
  while (true) {
    dijkstra(s);
    if (dis[t] + h[t] > 0) break;
    int flow = 0x3f3f3f3f;
    for (int v = t; v != s; v = fv[v]) flow = min(flow, cap[fe[v]]);
    for (int v = t; v != s; v = fv[v]) {
      cap[fe[v]] -= flow;
      cap[fe[v] ^ 1] += flow;
    }
    ret += (dis[t] + h[t]) * flow;
    for (int u = (0); u < (maxv); ++u) h[u] += dis[u];
  }
  return ret;
}
}  // namespace Dinic
int main() {
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) scanf("%d", a + i);
  Dinic::init();
  for (int i = (0); i < (n); ++i) {
    Dinic::addEdge(n << 1 | 1, i << 1, 1, 0);
    Dinic::addEdge(i << 1 | 1, (n + 1) << 1, 1, -1);
    Dinic::addEdge(i << 1, i << 1 | 1, 1, 0);
    for (int j = (i + 1); j < (n); ++j)
      if (abs(a[i] - a[j]) == 1 || a[i] % 7 == a[j] % 7)
        Dinic::addEdge(i << 1 | 1, j << 1, 1, -1);
  }
  Dinic::addEdge(n + 1 << 1, (n + 1) << 1 | 1, 4, 0);
  printf("%d\n", -Dinic::minCostFlow(n << 1 | 1, (n + 1) << 1 | 1));
  return 0;
}
