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, mx;
int arr[N], memo[N][1 << 8];
vector<int> adj[9];
int sh[N];
int solve(int i, int mask) {
  if (i == n) return mask == (1 << 8) - 1 || mx == 1 ? 0 : -1e9;
  int &ret = memo[i][mask];
  if (~ret) return ret;
  ret = solve(i + 1, mask);
  if (!(mask & (1 << arr[i])) && adj[arr[i]].size() - sh[i] >= mx)
    ret = max(
        ret, mx + solve(adj[arr[i]][sh[i] + mx - 1] + 1, mask | (1 << arr[i])));
  if (mx != 1 && !(mask & (1 << arr[i])) &&
      adj[arr[i]].size() - sh[i] >= mx - 1)
    ret = max(
        ret,
        mx - 1 + solve(adj[arr[i]][sh[i] + mx - 2] + 1, mask | (1 << arr[i])));
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", arr + i);
    arr[i]--;
    sh[i] = adj[arr[i]].size();
    adj[arr[i]].push_back(i);
  }
  for (mx = 130; mx >= 1; mx--) {
    memset(memo, -1, sizeof memo);
    int ans = solve(0, 0);
    if (ans >= 0) {
      printf("%d", ans);
      return 0;
    }
  }
}
