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 dp[1010][(1 << 8)], ct[9];
bool vis[1010][(1 << 8)], ok;
int arr[1010], n, run, quan;
vector<int> numbers[9];
set<int> masks;
int solve(int ind, int mask) {
  if (mask == (1 << 8) - 1) return !(ok = true);
  if (ind == n) return -(1 << 30);
  masks.insert(mask);
  if (vis[ind][mask]) return dp[ind][mask];
  vis[ind][mask] = true;
  if ((mask >> arr[ind]) & 1) {
    return dp[ind][mask] = solve(ind + 1, mask);
  } else {
    int pos =
        lower_bound(numbers[arr[ind]].begin(), numbers[arr[ind]].end(), ind) -
        numbers[arr[ind]].begin();
    int next = solve(ind + 1, mask);
    int withr = -(1 << 30);
    int withr1 = -(1 << 30);
    assert(numbers[arr[ind]][pos] == ind);
    if (pos + run - 1 < numbers[arr[ind]].size())
      withr = run + solve(numbers[arr[ind]][pos + run - 1] + 1,
                          (mask | (1 << arr[ind])));
    if (pos + run < numbers[arr[ind]].size())
      withr1 =
          run + 1 +
          solve(numbers[arr[ind]][pos + run] + 1, (mask | (1 << arr[ind])));
    return dp[ind][mask] = max(next, max(withr, withr1));
  }
}
int main() {
  scanf("%d", &n);
  for (int e = 0; e < n; e++) scanf("%d", arr + e), arr[e]--;
  for (int e = 0; e < n; e++) numbers[arr[e]].push_back(e);
  int ans = 0;
  for (int e = 0; e < n; e++) ct[arr[e]] = 1;
  for (int e = 0; e < 9; e++) ans += ct[e];
  int lo = 1, hi = n / 8 + 4;
  while (lo <= hi) {
    run = (lo + hi) >> 1;
    memset(vis, 0, sizeof vis);
    int cur = solve(0, 0);
    if (cur > 0)
      ans = cur, lo = run + 1;
    else
      hi = run - 1;
  }
  printf("%d\n", ans);
  return 0;
}
