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 MAXN = 1000, MAXB = 8;
int n, a[MAXN + 5];
vector<int> pos[MAXB];
int dp[MAXN + 5][1 << MAXB], cur[MAXN + 5];
int check(int k) {
  if (k == 0) {
    int Ans = 0;
    for (int i = 0; i < MAXB; i++)
      if (pos[i].size()) Ans++;
    return Ans;
  }
  memset(dp, -1, sizeof(dp));
  memset(cur, 0, sizeof(cur));
  dp[1][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int S = 0; S < 1 << MAXB; S++)
      if (~dp[i][S])
        for (int j = 0; j < MAXB; j++)
          if (!(S & (1 << j))) {
            int x = cur[j] + k - 1;
            if (x < pos[j].size())
              dp[pos[j][x]][S | (1 << j)] =
                  max(dp[pos[j][x]][S | (1 << j)], dp[i][S]);
            if (x + 1 < pos[j].size())
              dp[pos[j][x + 1]][S | (1 << j)] =
                  max(dp[pos[j][x + 1]][S | (1 << j)], dp[i][S] + 1);
          }
    cur[a[i]]++;
  }
  int Max = -1;
  for (int i = 1; i <= n; i++) Max = max(Max, dp[i][(1 << MAXB) - 1]);
  if (Max == -1) return 0;
  return k * 8 + Max;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), pos[--a[i]].push_back(i);
  int Ans;
  for (int l = 0, r = n / 8 + 1, Mid, tmp; l <= r;) {
    Mid = (l + r) >> 1;
    if (tmp = check(Mid))
      l = Mid + 1, Ans = tmp;
    else
      r = Mid - 1;
  }
  printf("%d\n", Ans);
  return 0;
}
