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 IN() {
  int c, f, x;
  while (!isdigit(c = getchar()) && c != '-')
    ;
  c == '-' ? (f = 1, x = 0) : (f = 0, x = c - '0');
  while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + c - '0';
  return !f ? x : -x;
}
const int N = 1000 + 19;
int A[N], f[1 << 8][N], sum[8][N];
int n, mx;
void Work(int s) {
  memset(f, 200, sizeof(f));
  for (int i = 0; i < n + 1; i++) f[0][i] = 0;
  for (int t = 0; t < 1 << 8; t++) {
    for (int j = 0; j < 8; j++)
      if (t >> j & 1) {
        int k1 = 0, k2 = 0;
        for (int i = 1; i < n + 1; i++) {
          while (k1 < i && sum[j][i] - sum[j][k1 + 1] >= s) k1++;
          while (k2 < i && sum[j][i] - sum[j][k2 + 1] >= s + 1) k2++;
          if (sum[j][i] - sum[j][k1] >= s) {
            f[t][i] = max(f[t][i], f[t ^ (1 << j)][k1] + s);
          }
          if (sum[j][i] - sum[j][k2] >= s + 1) {
            f[t][i] = max(f[t][i], f[t ^ (1 << j)][k2] + s + 1);
          }
        }
      }
  }
  if (s == 0) {
    for (int t = 0; t < 1 << 8; t++) mx = max(mx, f[t][n]);
  } else {
    mx = max(mx, f[(1 << 8) - 1][n]);
  }
}
int main() {
  n = IN();
  for (int i = 1; i < n + 1; i++) A[i] = IN() - 1;
  for (int i = 0; i < 8; i++) {
    for (int j = 1; j < n + 1; j++) sum[i][j] = sum[i][j - 1] + (A[j] == i);
  }
  for (int t = 0; 8 * t <= n; t++) Work(t);
  printf("%d\n", mx);
}
