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;
int a[1050];
int nxt[10][1050][1050];
int cnt[10];
int ans;
int dp[1 << 8][10];
int pre[10][1 << 8];
int pres[10];
void umin(int &a, int b) {
  if (a == -1)
    a = b;
  else if (a > b)
    a = b;
}
bool work(int t) {
  memset(dp, -1, sizeof(dp));
  if (t == 0) {
    set<int> s;
    for (int i = 1; i <= n; i++) s.insert(a[i]);
    ans = s.size();
    return true;
  }
  dp[0][0] = 0;
  for (int i = 1; i <= 8; i++) {
    for (int j = 1; j <= pres[i]; j++) {
      int sta = pre[i][j];
      for (int k = 0; k <= i; k++) {
        for (int wei = 0; wei < 8; wei++) {
          if ((1 << wei) & sta) {
            int yuan = sta ^ (1 << wei);
            if (k == 0) {
              if (dp[yuan][0] != -1 && nxt[wei + 1][dp[yuan][0]][t] != -1)
                umin(dp[sta][0], nxt[wei + 1][dp[yuan][0]][t]);
            } else {
              if (dp[yuan][k] != -1 && nxt[wei + 1][dp[yuan][k]][t] != -1) {
                umin(dp[sta][k], nxt[wei + 1][dp[yuan][k]][t]);
              }
              if (dp[yuan][k - 1] != -1 &&
                  nxt[wei + 1][dp[yuan][k - 1]][t + 1] != -1) {
                umin(dp[sta][k], nxt[wei + 1][dp[yuan][k - 1]][t + 1]);
              }
            }
          }
        }
      }
    }
  }
  for (int i = 8; i >= 0; i--) {
    if (dp[(1 << 8) - 1][i] != -1) {
      ans = t * 8 + i;
      return true;
    }
  }
  return false;
}
int main() {
  memset(pres, 0, sizeof(pres));
  for (unsigned int i = 1; i < 1 << 8; i++) {
    int k = __builtin_popcount(i);
    pre[k][++pres[k]] = i;
  }
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  memset(nxt, -1, sizeof(nxt));
  for (int i = n; i >= 0; i--) {
    memset(cnt, 0, sizeof(cnt));
    for (int j = i + 1; j <= n; j++) {
      nxt[a[j]][i][++cnt[a[j]]] = j;
    }
  }
  ans = -1;
  int l = 0;
  int r = n / 8;
  while (l <= r) {
    if (r - l <= 1) {
      if (work(r)) break;
      work(l);
      break;
    }
    int mid = (l + r) >> 1;
    if (work(mid))
      l = mid;
    else
      r = mid;
  }
  printf("%d\n", ans);
}
