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>
int main() {
  int n;
  scanf("%d", &n);
  std::vector<int> A(n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &A[i]);
    A[i]--;
  }
  std::vector<std::vector<int>> next(n + 1, std::vector<int>(n + 1, -1));
  std::vector<int> last(8, -1);
  for (int i = n - 1; i >= 0; --i) {
    if (last[A[i]] != -1) {
      for (int j = 0; j < n; ++j) {
        next[i][j + 1] = next[last[A[i]]][j];
      }
    }
    next[i][1] = i;
    last[A[i]] = i;
  }
  auto update = [](int &a, int b) {
    if (a < b) a = b;
  };
  auto calc = [&](int low) {
    std::vector<std::vector<int>> dp(n + 1, std::vector<int>(1 << 8, -1));
    dp[0][0] = 0;
    for (int i = 0; i < n; ++i) {
      for (int mask = 0; mask < 1 << 8; ++mask) {
        if (dp[i][mask] == -1) continue;
        update(dp[i + 1][mask], dp[i][mask]);
        if (~mask >> A[i] & 1) {
          for (int cnt = low; cnt <= low + 1 && next[i][cnt] != -1; ++cnt) {
            update(dp[next[i][cnt] + 1][mask | 1 << A[i]], dp[i][mask] + cnt);
          }
        }
      }
    }
    return dp[n][(1 << 8) - 1];
  };
  int tmp = 0;
  for (int x : A) tmp |= 1 << x;
  int result = __builtin_popcount(tmp);
  int l = 1, r = n / 8;
  while (l <= r) {
    int mid = l + r >> 1;
    tmp = calc(mid);
    if (tmp > result)
      result = tmp, l = mid + 1;
    else
      r = mid - 1;
  }
  printf("%d\n", result);
}
