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 INF = 0x3f3f3f3f;
using namespace std;
int N, a[1010];
int cur[10];
int dp[1010][(1 << 8)];
vector<int> cnt[9];
void checkmax(int &a, int b) { a = max(a, b); }
int check(int len) {
  int p;
  memset(cur, 0, sizeof(cur));
  for (int i = 0; i <= N; i++)
    for (int j = 0; j < (1 << 8); j++) dp[i][j] = -INF;
  dp[0][0] = 0;
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < (1 << 8); j++) {
      if (dp[i][j] == -INF) continue;
      for (int k = 0; k < 8; k++) {
        if (j & (1 << k)) continue;
        p = cur[k] + len - 1;
        if (p >= cnt[k].size()) continue;
        checkmax(dp[cnt[k][p] + 1][j | (1 << k)], dp[i][j]);
        p++;
        if (p >= cnt[k].size()) continue;
        checkmax(dp[cnt[k][p] + 1][j | (1 << k)], dp[i][j] + 1);
      }
    }
    cur[a[i]]++;
  }
  int ans = -INF;
  for (int i = 0; i <= N; i++) checkmax(ans, dp[i][(1 << 8) - 1]);
  if (ans == -INF) return -1;
  return ans * (len + 1) + (8 - ans) * len;
}
int main() {
  scanf("%d", &N);
  for (int i = 0; i <= 7; i++) cnt[i].clear();
  for (int i = 0; i < N; i++) {
    scanf("%d", &a[i]);
    a[i]--;
    cnt[a[i]].push_back(i);
  }
  int l, r;
  l = 1;
  r = N / 8 + 1;
  while (l + 1 < r) {
    int mid = (l + r) / 2;
    if (check(mid) != -1)
      l = mid;
    else
      r = mid;
  }
  int ans = check(l);
  if (ans == -1) {
    ans = 0;
    for (int i = 0; i < 8; i++)
      if (!cnt[i].empty()) ans++;
  }
  printf("%d\n", ans);
  return 0;
}
