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 long long mod = 1e9 + 7;
const int mn = 3e5 + 10;
vector<int> p[8];
int nxt[1010][8][140];
int b[8];
int perm[8];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    p[x - 1].push_back(i);
  }
  for (int i = 0; i <= n + 1; i++) {
    for (int j = 0; j < 8; j++) {
      int tar = lower_bound(p[j].begin(), p[j].end(), i) - p[j].begin() - 1;
      nxt[i][j][0] = i;
      for (int k = 1; k <= 128; k++) {
        if (tar + k < p[j].size())
          nxt[i][j][k] = p[j][tar + k] + 1;
        else
          nxt[i][j][k] = n + 1;
      }
    }
  }
  iota(perm, perm + 8, 0);
  int ans = 0;
  do {
    for (int i = 0; i < 256; i++) {
      for (int j = 0; j < 8; j++) b[j] = (i >> j) & 1;
      int cur = -1;
      for (int j = 64; j; j >>= 1) {
        int tri = cur + j;
        int v = 0;
        bool gud = 1;
        for (int x : perm) {
          if (v == n + 1) {
            gud = 0;
            break;
          }
          v = nxt[v][x][tri + b[x]];
        }
        if (v == n + 1) gud = 0;
        if (gud) cur += j;
      }
      ans = max(ans, 8 * cur + __builtin_popcount(i));
    }
  } while (next_permutation(perm, perm + 8));
  printf("%d", ans);
}
