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;
int a[1001], last[10];
int f[1001][1 << 9];
vector<int> g[10];
int check(int len) {
  memset(last, 0, sizeof(last));
  memset(f, -1, sizeof(f));
  f[0][0] = 0;
  for (int i = 0; i < n; i++) {
    for (int s = 0; s < (1 << 8); s++)
      if (f[i][s] >= 0) {
        for (int j = 0; j < 8; j++)
          if (!((1 << j) & s)) {
            int now = last[j] + len - 1;
            if (now < g[j].size())
              f[g[j][now]][s | (1 << j)] =
                  max(f[g[j][now]][s | (1 << j)], f[i][s]);
            if (now + 1 < g[j].size())
              f[g[j][now + 1]][s | (1 << j)] =
                  max(f[g[j][now + 1]][s | (1 << j)], f[i][s] + 1);
          }
      }
    last[a[i] - 1]++;
  }
  int tot = -1e9;
  for (int i = 0; i <= n; i++) tot = max(tot, f[i][(1 << 8) - 1]);
  if (tot < 0) return 0;
  ans = max(ans, 8 * len + tot);
  return 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    g[a[i] - 1].push_back(i);
  }
  int l = 0, r = n / 8 + 1;
  for (int i = 0; i < 8; i++) r = min(r, (int)g[i].size());
  if (!r) {
    for (int i = 0; i < 8; i++)
      if (g[i].size()) ans++;
    printf("%d\n", ans);
    return 0;
  }
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (check(mid))
      l = mid + 1;
    else
      r = mid - 1;
  }
  printf("%d\n", ans);
  return 0;
}
