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, A[1005], freq;
int dp[1005][260], nextpos[10][1005][1005];
int solve(int pos, int mask) {
  if (mask == 255) return 0;
  if (pos > N) return -1e6;
  if (dp[pos][mask] != -1) return dp[pos][mask];
  int i, ans = -1e7;
  if ((1 << (A[pos] - 1)) & mask)
    ans = max(ans, solve(pos + 1, mask));
  else {
    int maxi = solve(pos + 1, mask);
    if (nextpos[A[pos]][pos][freq]) {
      maxi = max(maxi, freq + solve(nextpos[A[pos]][pos][freq] + 1,
                                    (mask | (1 << (A[pos] - 1)))));
    }
    if (nextpos[A[pos]][pos][freq + 1]) {
      maxi = max(maxi, freq + 1 +
                           solve(nextpos[A[pos]][pos][freq + 1] + 1,
                                 (mask | (1 << (A[pos] - 1)))));
    }
    ans = max(ans, maxi);
  }
  return dp[pos][mask] = ans;
}
int main() {
  scanf("%d", &N);
  int i, j, k, ans = 0;
  set<int> S;
  for (i = 1; i <= N; i++) {
    scanf("%d", &A[i]);
    S.insert(A[i]);
  }
  for (i = 1; i <= 8; i++) {
    for (j = 1; j <= N; j++) {
      int f = 0;
      for (k = j; k <= N; k++) {
        if (A[k] == i) {
          f++;
          nextpos[i][j][f] = k;
        }
      }
    }
  }
  ans = S.size();
  for (i = 0; i <= 125; i++) {
    freq = i;
    memset(dp, -1, sizeof(dp));
    ans = max(ans, solve(1, 0));
  }
  printf("%d", ans);
}
