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>
const int N = 1010;
int n, i, j, k, x, a[N], g[N][N][8], v[8], f[256][10], ans;
inline void umin(int& a, int b) {
  if (a > b) a = b;
}
inline void umax(int& a, int b) {
  if (a < b) a = b;
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]), a[i]--;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n + 5; j++)
      for (k = 0; k < 8; k++) g[i][j][k] = n + 10;
    for (j = 0; j < 8; j++) v[j] = 0;
    for (j = i; j <= n; j++) g[i][++v[a[j]]][a[j]] = j;
  }
  for (x = 0; x * 8 <= n; x++) {
    for (i = 0; i < 256; i++)
      for (j = 0; j <= 8; j++) f[i][j] = n + 10;
    f[0][0] = 1;
    for (i = 0; i < 256; i++)
      for (j = 0; j <= 8; j++)
        if (f[i][j] <= n)
          for (k = 0; k < 8; k++)
            if (!(i >> k & 1)) {
              umin(f[i | (1 << k)][j], g[f[i][j]][x][k] + 1);
              umin(f[i | (1 << k)][j + 1], g[f[i][j]][x + 1][k] + 1);
            }
    for (j = 0; j <= 8; j++)
      if (f[255][j] <= n + 1) umax(ans, (8 - j) * x + j * (x + 1));
  }
  printf("%d", ans);
}
