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, mem[256][1010];
vector<int> occs[8];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int a;
    scanf("%d", &a);
    occs[a - 1].push_back(i);
  }
  int mx = 0;
  for (int i = 0; i < 8; i++) {
    mx += occs[i].size() > 0;
  }
  for (int q = 1; q < 125; q++) {
    for (int i = 0; i < 256; i++)
      for (int j = 0; j <= n; j++) mem[i][j] = INT_MIN;
    mem[0][0] = 0;
    for (int i = 0; i <= n; i++) {
      int lens[8];
      for (int k = 0; k < 8; k++) {
        auto next = lower_bound(occs[k].begin(), occs[k].end(), i);
        lens[k] = distance(occs[k].begin(), next);
      }
      for (int j = 0; j < (1 << 8); j++) {
        for (int k = 0; k < 8; k++) {
          if (!((j >> k) & 1)) {
            int len = lens[k];
            if (len + q - 1 < occs[k].size()) {
              mem[j | (1 << k)][occs[k][len + q - 1] + 1] = max(
                  mem[j | (1 << k)][occs[k][len + q - 1] + 1], mem[j][i] + q);
            }
            if (len + q < occs[k].size()) {
              mem[j | (1 << k)][occs[k][len + q] + 1] = max(
                  mem[j | (1 << k)][occs[k][len + q] + 1], mem[j][i] + q + 1);
            }
          }
        }
      }
    }
    for (int j = 0; j <= n; j++) {
      mx = max(mx, mem[0xFF][j]);
    }
  }
  printf("%d\n", mx);
}
