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, L = 1;
int arr[1000], wraya[1000];
vector<vector<int> > vec(9);
int dp[1000][1 << 8];
int solve(int ind, int msk) {
  if (ind == n) {
    if (msk == (1 << 8) - 1) return 0;
    return -1e9;
  }
  if (dp[ind][msk] != -1) return dp[ind][msk];
  int ret = solve(ind + 1, msk);
  if (!(msk & (1 << arr[ind]))) {
    int nmsk = msk | (1 << arr[ind]);
    if (L + wraya[ind] < vec[arr[ind]].size())
      ret = max(ret, solve(vec[arr[ind]][L + wraya[ind]] + 1, nmsk) + L + 1);
    if (L + wraya[ind] - 1 < vec[arr[ind]].size())
      ret = max(ret, solve(vec[arr[ind]][L + wraya[ind] - 1] + 1, nmsk) + L);
  }
  return dp[ind][msk] = ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    arr[i]--;
    vec[arr[i]].push_back(i);
    wraya[i] = vec[arr[i]].size() - 1;
  }
  int t = 8;
  for (int i = 0; i < 8; i++)
    if (!vec[i].size()) t--;
  if (t < 8) return cout << t, 0;
  int ans = t;
  for (int i = 1; i < 126; i++)
    memset(dp, -1, sizeof dp), L = i, ans = max(ans, solve(0, 0));
  cout << ans;
  return 0;
}
