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 int mx = 1e3 + 10;
const int oo = 1e9;
int a[mx];
int nxt[mx][8];
int f[1 << 8][mx];
vector<int> pos[8];
int idx[mx];
int main() {
  int n;
  n = 1e3;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    a[i] = rand() % 8 + 1;
    scanf("%d", a + i);
    --a[i];
    idx[i] = pos[a[i]].size();
    pos[a[i]].push_back(i);
  }
  int last[8];
  for (int i = 0; i < 8; i++) last[i] = -1;
  for (int i = n - 1; i >= 0; i--) {
    last[a[i]] = i;
    for (int j = 0; j < 8; j++) nxt[i][j] = last[j];
  }
  int ans = 0;
  for (int i = 0; i < 8; i++)
    if (last[i] != -1) ans++;
  int li = 1, ls = n / 8;
  while (li <= ls) {
    int k = (li + ls) >> 1;
    for (int b = 0; b < (1 << 8); b++) f[b][n] = -oo;
    f[0][n] = 0;
    for (int i = n - 1; i >= 0; i--) {
      f[0][i] = 0;
      for (int b = 1; b < (1 << 8); b++) {
        f[b][i] = -oo;
        for (int at = 0; at < 8; at++)
          if (b & (1 << at)) {
            int from = nxt[i][at];
            if (from == -1) break;
            from = idx[from];
            int to = from + k - 1;
            if (to < (int)pos[at].size()) {
              f[b][i] = max(f[b][i], f[b ^ (1 << at)][pos[at][to] + 1] + k);
            }
            to++;
            if (to < (int)pos[at].size()) {
              f[b][i] = max(f[b][i], f[b ^ (1 << at)][pos[at][to] + 1] + k + 1);
            }
            if (f[b][i] < 0) f[b][i] = -oo;
          }
      }
    }
    int bet = f[(1 << 8) - 1][0];
    if (bet > 0)
      li = k + 1;
    else
      ls = k - 1;
    ans = max(ans, bet);
  }
  printf("%d\n", ans);
  return 0;
}
