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;
int n, lis[1111], dp[1111][1 << 8];
int solve(int x) {
  for (int i = (0); i <= ((int)(1010) - 1); i++)
    for (int j = (1); j <= ((1 << 8) - 1); j++) dp[i][j] = -1000000007;
  if (!x) {
    int cntt[8];
    memset(cntt, 0, sizeof(cntt));
    for (int i = (1); i <= (n); i++) cntt[lis[i]] = 1;
    int ans = 0;
    for (int i = (0); i <= ((int)(8) - 1); i++) ans += cntt[i];
    return ans;
  }
  for (int i = (1); i <= (n); i++)
    for (int mask = (0); mask <= ((int)((1 << 8)) - 1); mask++)
      for (int las = (0); las <= ((int)(8) - 1); las++)
        if ((1 << las) & mask) {
          int p = i;
          int cnt = 0;
          while (p) {
            if (lis[p] == las) cnt++;
            if (cnt == x) break;
            p--;
          }
          if (p) {
            dp[i][mask] = max(dp[i][mask], x + dp[p - 1][mask ^ (1 << las)]);
            p--;
            while (p) {
              if (lis[p] == las) break;
              p--;
            }
            if (p)
              dp[i][mask] =
                  max(dp[i][mask], x + 1 + dp[p - 1][mask ^ (1 << las)]);
          }
          dp[i][mask] = max(dp[i][mask], dp[i - 1][mask]);
        }
  return max(0, dp[n][(1 << 8) - 1]);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = (1); i <= (n); i++) {
    cin >> lis[i];
    lis[i]--;
  }
  int high = n / 8 + 3;
  int low = 0;
  int ans = 0;
  while (low < high) {
    int chs = (high + low) / 2;
    int tmp = solve(chs);
    ans = max(ans, tmp);
    if (tmp)
      low = chs + 1;
    else
      high = chs;
  }
  cout << ans;
}
