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, ans = -0x3f3f3f3f, a[1005], cnt[9], f[1005][1 << 8];
vector<int> v[9];
int read() {
  int x = 0, f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) x = (x << 3) + (x << 1) + ch - 48;
  return x * f;
}
bool check(int mid) {
  memset(cnt, 0, sizeof(cnt));
  memset(f, -1, sizeof(f));
  f[1][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int S = 0; S < 1 << 8; S++)
      if (~f[i][S]) {
        for (int j = 1; j <= 8; j++)
          if (!(S & (1 << (j - 1)))) {
            int x = cnt[j] + mid;
            if (x > v[j].size()) continue;
            f[v[j][x - 1] + 1][S | (1 << (j - 1))] =
                max(f[v[j][x - 1] + 1][S | (1 << (j - 1))], f[i][S]);
            if (++x > v[j].size()) continue;
            f[v[j][x - 1] + 1][S | (1 << (j - 1))] =
                max(f[v[j][x - 1] + 1][S | (1 << (j - 1))], f[i][S] + 1);
          }
      }
    cnt[a[i]]++;
  }
  int tt = -1;
  for (int i = 1; i <= n + 1; i++) tt = max(tt, f[i][(1 << 8) - 1]);
  if (tt == -1) return 0;
  ans = max(ans, tt * (mid + 1) + (8 - tt) * mid);
  return 1;
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) a[i] = read(), v[a[i]].push_back(i);
  int l = 1, r = n, mid;
  while (l <= r) {
    mid = l + r >> 1;
    check(mid) ? l = mid + 1 : r = mid - 1;
  }
  if (ans == -0x3f3f3f3f) {
    ans = 0;
    for (int i = 1; i <= 8; i++)
      if (v[i].size()) ans++;
  }
  printf("%d\n", ans);
  return 0;
}
