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;
const int Len = 1e3 + 5;
vector<int> pos[15];
int dp[Len][1 << 10], num[15], n, color[Len], Ans = -1;
int check(int mid) {
  memset(num, 0, sizeof num);
  memset(dp, -1, sizeof dp);
  dp[1][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < (1 << 8); j++) {
      if (dp[i][j] == -1) continue;
      for (int k = 0; k < 8; k++) {
        if (j & (1 << k)) continue;
        int idx = num[k + 1] + mid - 1;
        if (idx >= pos[k + 1].size()) continue;
        dp[pos[k + 1][idx]][j ^ (1 << k)] =
            max(dp[i][j], dp[pos[k + 1][idx]][j ^ (1 << k)]);
        if (idx + 1 >= pos[k + 1].size()) continue;
        dp[pos[k + 1][idx + 1]][j ^ (1 << k)] =
            max(dp[pos[k + 1][idx + 1]][j ^ (1 << k)], dp[i][j] + 1);
      }
    }
    num[color[i]]++;
  }
  int ans = -1;
  for (int i = 1; i <= n; i++) ans = max(ans, dp[i][(1 << 8) - 1]);
  if (ans == -1) return -1;
  return ans * (mid + 1) + (8 - ans) * mid;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &color[i]);
    pos[color[i]].push_back(i);
  }
  int l = 1, r = n / 8;
  while (l + 1 < r) {
    int mid = (l + r) >> 1;
    if (check(mid) != -1)
      l = mid;
    else
      r = mid;
  }
  Ans = max(check(l), check(r));
  if (Ans == -1) {
    Ans = 0;
    for (int i = 1; i <= 8; i++)
      if (pos[i].size()) Ans++;
  }
  printf("%d\n", Ans);
  return 0;
}
