Vladik was bored on his way home and decided to play the following game. He took n cards and put them in a row in front of himself. Every card has a positive integer number not exceeding 8 written on it. He decided to find the longest subsequence of cards which satisfies the following conditions:

  * the number of occurrences of each number from 1 to 8 in the subsequence doesn't differ by more then 1 from the number of occurrences of any other number. Formally, if there are ck cards with number k on them in the subsequence, than for all pairs of integers <image> the condition |ci - cj| ≤ 1 must hold. 
  * if there is at least one card with number x on it in the subsequence, then all cards with number x in this subsequence must form a continuous segment in it (but not necessarily a continuous segment in the original sequence). For example, the subsequence [1, 1, 2, 2] satisfies this condition while the subsequence [1, 2, 2, 1] doesn't. Note that [1, 1, 2, 2] doesn't satisfy the first condition. 



Please help Vladik to find the length of the longest subsequence that satisfies both conditions.

Input

The first line contains single integer n (1 ≤ n ≤ 1000) — the number of cards in Vladik's sequence.

The second line contains the sequence of n positive integers not exceeding 8 — the description of Vladik's sequence.

Output

Print single integer — the length of the longest subsequence of Vladik's sequence that satisfies both conditions.

Examples

Input

3
1 1 1


Output

1

Input

8
8 7 6 5 4 3 2 1


Output

8

Input

24
1 8 1 2 8 2 3 8 3 4 8 4 5 8 5 6 8 6 7 8 7 8 8 8


Output

17

Note

In the first sample all the numbers written on the cards are equal, so you can't take more than one card, otherwise you'll violate the first condition.
#include <bits/stdc++.h>
static const int MAXN = 1004;
static const int RG = 8;
template <typename T>
inline void upd_max(T &var, const T val) {
  if (var < val) var = val;
}
int n, a[MAXN];
std::vector<int> occ[RG];
int occ_next[RG][MAXN];
int p[RG], xt_mask, base_len;
inline bool check() {
  int ptr = 0;
  for (int i = 0; i < RG; ++i) {
    int u = p[i];
    int this_len = base_len + ((xt_mask >> u) & 1);
    if (this_len <= 0) continue;
    ptr = occ_next[u][ptr];
    if ((ptr += (this_len - 1)) >= (int)occ[u].size()) return false;
    ptr = occ[u][ptr] + 1;
  }
  return true;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    --a[i];
    occ[a[i]].push_back(i);
  }
  for (int i = 0; i < RG; ++i) {
    int last = 0;
    for (int j = 0; j < (int)occ[i].size(); ++j)
      for (; last <= occ[i][j]; ++last) occ_next[i][last] = j;
    for (; last < MAXN; ++last) occ_next[i][last] = (int)occ[i].size();
  }
  for (int i = 0; i < RG; ++i) p[i] = i;
  int ans = 0;
  do {
    for (xt_mask = 0; xt_mask < (1 << RG); ++xt_mask) {
      int lo = -1, hi = MAXN / RG + 10;
      while (lo < hi - 1) {
        base_len = (lo + hi) >> 1;
        if (check())
          lo = base_len;
        else
          hi = base_len;
        if (ans >= __builtin_popcount(xt_mask) + hi * 8) break;
      }
      if (lo >= 0) upd_max(ans, __builtin_popcount(xt_mask) + lo * 8);
    }
  } while (std::next_permutation(p, p + RG));
  printf("%d\n", ans);
  return 0;
}
