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 = 1005, inf = 1 << 30;
int f[N][256], n, a[N], sum[10], Num = 255;
vector<int> v[10];
int gao(int len) {
  memset(sum, 0, sizeof(sum));
  memset(f, -1, sizeof(f));
  f[1][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int s = 0; s <= Num; s++) {
      if (f[i][s] == -1) continue;
      for (int j = 0; j < 8; j++) {
        if (s & (1 << j)) continue;
        int t = len + sum[j] - 1;
        if (v[j].size() < t) continue;
        f[v[j][t - 1]][s | (1 << j)] =
            max(f[v[j][t - 1]][s | (1 << j)], f[i][s]);
        if (v[j].size() <= t) continue;
        f[v[j][t]][s | (1 << j)] = max(f[v[j][t]][s | (1 << j)], f[i][s] + 1);
      }
    }
    sum[a[i]]++;
  }
  int mx = -inf;
  for (int i = 1; i <= n; i++) mx = max(mx, f[i][Num]);
  if (mx == -1) return -1;
  return len * mx + (8 - mx) * (len - 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), a[i]--, v[a[i]].push_back(i);
  int ans = 0, l = 2, r = n;
  while (l <= r) {
    int mid = l + r >> 1;
    int t = gao(mid);
    if (t != -1)
      ans = max(t, ans), l = mid + 1;
    else
      r = mid - 1;
  }
  if (ans == 0) {
    for (int i = 0; i < 8; i++) {
      if (v[i].size()) ans++;
    }
  }
  printf("%d\n", ans);
  return 0;
}
