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>
using namespace std;
inline int read() {
  int x = 0;
  char c = getchar();
  while (!isdigit(c)) c = getchar();
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  return x;
}
int n, ans;
int x[1010], num[10];
int f[1010][(1 << 8) + 10];
vector<int> at[10];
bool check(int tim) {
  memset(num, 0, sizeof(num));
  memset(f, 250, sizeof(f));
  f[0][0] = 0;
  int tot = -0x3f3f3f3f;
  for (int i = 0; i < n; ++i) {
    ++num[x[i]];
    for (int j = 0; j < (1 << 8); ++j)
      if (f[i][j] >= 0)
        for (int k = 0; k < 8; ++k) {
          if (j & (1 << k)) continue;
          int now = num[k] + tim - 1;
          if (now < at[k].size())
            f[at[k][now]][j | (1 << k)] =
                max(f[at[k][now]][j | (1 << k)], f[i][j]);
          if (++now < at[k].size())
            f[at[k][now]][j | (1 << k)] =
                max(f[at[k][now]][j | (1 << k)], f[i][j] + 1);
        }
  }
  for (int i = 0; i <= n; ++i) tot = max(tot, f[i][(1 << 8) - 1]);
  if (tot < 0) return false;
  ans = max(ans, 8 * tim + tot);
  return true;
}
int main() {
  n = read();
  x[0] = 9;
  for (int i = 1; i <= n; ++i) x[i] = read() - 1, at[x[i]].push_back(i);
  int l = 0, r = n / 8 + 1;
  for (int i = 0; i < 8; ++i) r = min(r, (int)at[i].size());
  if (!r) {
    for (int i = 0; i < 8; ++i)
      if (at[i].size()) ++ans;
    printf("%d\n", ans);
    return 0;
  }
  while (l <= r) {
    int mid = l + r >> 1;
    if (check(mid))
      l = mid + 1;
    else
      r = mid - 1;
  }
  printf("%d\n", ans);
  return 0;
}
