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;
typedef const int& ci;
int a[1000 + 1];
int pos[8 + 1][1000 + 1], cnt[8 + 1];
int pnt[1000 + 1];
bool dp[1000 + 1][1 << 8][8];
bool check(ci n, ci p) {
  const int num1 = p >> 3, num2 = p & 7, tmp = num1 - 1;
  if (!num1) {
    int cntt = 0;
    for (int i = 0; ++i <= 8; cntt += bool(cnt[i]))
      ;
    return p <= cntt;
  }
  memset(dp, 0, sizeof(dp));
  dp[0][0][0] = 1;
  for (int i = 0, nnum, nbit, npos, pr1; ++i <= n;) {
    memcpy(dp[i], dp[i - 1], sizeof(dp[i]));
    if ((npos = pnt[i]) < num1) continue;
    nnum = a[i];
    nbit = 1 << (nnum - 1), pr1 = pos[nnum][npos - tmp] - 1;
    if (npos == num1)
      for (int j = -1; ++j <= 255;) {
        (j & nbit) || (j |= nbit);
        for (int k = -1; ++k <= num2;) {
          if (dp[i][j][k]) continue;
          dp[i][j][k] = dp[pr1][j ^ nbit][k];
        }
      }
    else {
      int pr2 = pos[nnum][npos - num1] - 1;
      for (int j = -1; ++j <= 255;) {
        (j & nbit) || (j |= nbit);
        for (int k = -1; ++k <= num2;) {
          if (dp[i][j][k] || (dp[i][j][k] = dp[pr1][j ^ nbit][k])) continue;
          dp[i][j][k] = dp[pr2][j ^ nbit][k - 1];
        }
      }
    }
  }
  return dp[n][255][num2];
}
int main() {
  int n;
  scanf("%d", &n), memset(cnt, 0, sizeof(cnt));
  for (int i = 0; ++i <= n; pos[a[i]][pnt[i] = ++cnt[a[i]]] = i)
    scanf("%d", a + i);
  int l = 1, r = (1 + *min_element(cnt + 1, cnt + 9)) * 8;
  for (int mid; (mid = (l + r) >> 1) > l; (check(n, mid) ? l : r) = mid)
    ;
  printf("%d\n", l);
  return 0;
}
