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, a, f[10][2010], g[10], h[10], l[10], d[10][1010];
int check(int x) {
  int mx = -1;
  for (int k = 0; k < (1 << 8); k++) {
    for (int i = 0; i < 8; i++) {
      h[i] = i + 1;
      if (k & (1 << i))
        l[i + 1] = x - 1;
      else
        l[i + 1] = x;
    }
    int cnt = 0;
    for (int i = 1; i <= 8; i++) cnt += l[i];
    if (cnt <= mx) continue;
    do {
      int now = 0;
      for (int i = 0; i < 8 && now != 0x3f3f3f3f; i++) {
        int t = h[i];
        if (l[t] > 0) now = f[t][d[t][now] + l[t] - 1];
      }
      if (now != 0x3f3f3f3f) {
        mx = cnt;
        break;
      }
    } while (next_permutation(h, h + 8));
  }
  return mx;
}
int Sol() {
  int l = 1, r = 0x3f3f3f3f, mid, ans = 0;
  for (int i = 1; i <= 8; i++) r = min(r, g[i] + 1);
  while (l <= r) {
    mid = (l + r) >> 1;
    int x = check(mid);
    ans = max(ans, x);
    if (x != -1)
      l = mid + 1;
    else
      r = mid - 1;
  }
  return ans;
}
int main() {
  memset(f, 0x3f3f3f3f, sizeof(f));
  memset(g, 0, sizeof(g));
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a);
    f[a][g[a]++] = i;
  }
  for (int i = 1; i <= 8; i++) {
    int cur = 0;
    for (int j = 0; j <= n; j++) {
      if (j >= f[i][cur]) cur++;
      d[i][j] = cur;
    }
  }
  printf("%d\n", Sol());
  return 0;
}
