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 = 1000 + 10;
const int inf = 1 << 30;
int n, a[N], ans = 0, dp[N][1 << 8], pos, mark, cnt;
vector<int> vec[8];
bool check(int len) {
  bool flag = 0;
  for (int i = 0; i <= n; i++)
    for (int j = 0; j < (1 << 8); j++) dp[i][j] = -inf;
  dp[0][0] = 0;
  for (int i = 0; i <= n; i++)
    for (int j = 0; j < (1 << 8); j++) {
      if (dp[i][j] == -inf) continue;
      for (int k = 0; k < 8; k++) {
        if (j & (1 << k)) continue;
        int down =
            upper_bound(vec[k].begin(), vec[k].end(), i) - vec[k].begin();
        int up = down + len - 1;
        if (up >= vec[k].size()) continue;
        pos = vec[k][up];
        dp[pos][j | (1 << k)] = max(dp[pos][j | (1 << k)], dp[i][j] + len);
        up++;
        if (up >= vec[k].size()) continue;
        pos = vec[k][up];
        dp[pos][j | (1 << k)] = max(dp[pos][j | (1 << k)], dp[i][j] + len + 1);
      }
      if (j == (1 << 8) - 1) flag = 1, ans = max(ans, dp[i][j]);
    }
  return flag;
}
void solve() {
  int l = 0, r = n / 8 + 1, mid;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (check(mid))
      l = mid + 1;
    else
      r = mid - 1;
  }
}
int main() {
  for (int i = 0; i < 8; i++) vec[i].clear();
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]), vec[a[i] - 1].push_back(i);
    if (mark & 1 << (a[i] - 1))
      ;
    else
      mark |= 1 << (a[i] - 1), cnt++;
  }
  if (mark == (1 << 8) - 1) {
    solve();
    printf("%d", ans);
  } else
    printf("%d\n", cnt);
}
