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 N = 1e3 + 5;
int n, l, r, mid, ans;
int a[N], sum[8], f[N][1 << 8];
vector<int> v[8];
inline int jay(int mid) {
  memset(f, 0, sizeof(f));
  for (register int i = 0; i < n; ++i)
    for (register int p = 0; p <= 255; ++p) {
      for (register int j = 0; j < 8; ++j)
        if (!((1 << j) & p)) {
          if (v[j][v[j].size() - mid] > i) {
            int pos = upper_bound(v[j].begin(), v[j].end(), i) - v[j].begin();
            f[v[j][pos + mid - 1]][p | (1 << j)] =
                max(f[v[j][pos + mid - 1]][p | (1 << j)], f[i][p] + mid);
            if (pos + mid < v[j].size())
              f[v[j][pos + mid]][p | (1 << j)] =
                  max(f[v[j][pos + mid]][p | (1 << j)], f[i][p] + mid + 1);
          }
        }
    }
  int maxn = 0;
  for (register int i = 1; i <= n; ++i)
    for (register int p = 0; p <= 255; ++p) maxn = max(maxn, f[i][p]);
  return maxn;
}
int main() {
  scanf("%d", &n);
  for (register int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]), a[i]--, sum[a[i]]++;
    v[a[i]].push_back(i);
  }
  sort(sum, sum + 8);
  l = 1;
  r = sum[0];
  if (!r) {
    for (register int i = 0; i < 8; ++i)
      if (sum[i]) ans++;
    printf("%d\n", ans);
    return 0;
  }
  while (l <= r) {
    mid = l + r >> 1;
    int now = jay(mid);
    if (now >= mid * 8)
      ans = now, l = mid + 1;
    else
      r = mid - 1;
  }
  printf("%d\n", ans);
  return 0;
}
