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 dat[1000];
int dp[1000][256];
int visit[1000][256];
vector<int> indices[8];
int vtidx[1000];
int n, mv;
int D(int idx, int st) {
  if (idx == n) return st == 255 ? 0 : -10000;
  if (visit[idx][st] != mv) {
    visit[idx][st] = mv;
    dp[idx][st] = D(idx + 1, st);
    if (!(st & (1 << dat[idx]))) {
      if (vtidx[idx] + mv - 1 < indices[dat[idx]].size()) {
        dp[idx][st] =
            max(dp[idx][st], D(indices[dat[idx]][vtidx[idx] + mv - 1] + 1,
                               st | (1 << dat[idx])) +
                                 mv);
        if (vtidx[idx] + mv < indices[dat[idx]].size())
          dp[idx][st] = max(
              dp[idx][st],
              D(indices[dat[idx]][vtidx[idx] + mv] + 1, st | (1 << dat[idx])) +
                  mv + 1);
      }
    }
  }
  return dp[idx][st];
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &dat[i]);
    dat[i]--;
    vtidx[i] = indices[dat[i]].size();
    indices[dat[i]].push_back(i);
  }
  int ans = 0;
  for (int i = 0; i < 8; i++)
    if (indices[i].empty() == false) ans++;
  for (int i = 1; i <= 125; i++) {
    mv = i;
    ans = max(ans, D(0, 0));
  }
  printf("%d", ans);
}
