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 M = 1001;
int n, m, x, y, mid, a[M], nxt[M][M], dp[M][(1 << 8) + 2];
int calc(int i, int mask) {
  if (i == n) {
    if (mask == ((1 << 8) - 1) || mid == 1) return 0;
    return -1e9;
  }
  int &k = dp[i][mask];
  if (k != -1) return k;
  k = calc(i + 1, mask);
  if ((mask & (1 << a[i])) == 0) {
    if (nxt[i][mid] != -1)
      k = max(k, mid + calc(nxt[i][mid] + 1, mask | (1 << a[i])));
    if (mid > 1 && nxt[i][mid - 1] != -1)
      k = max(k, mid - 1 + calc(nxt[i][mid - 1] + 1, mask | (1 << a[i])));
  }
  return k;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
  }
  memset(nxt, -1, sizeof nxt);
  for (int i = 0; i < n; i++) {
    int c = 0;
    for (int j = i; j < n; j++)
      if (a[j] == a[i]) nxt[i][++c] = j;
  }
  int lo = 1, hi = n, ans = 1;
  while (lo <= hi) {
    mid = lo + hi >> 1;
    memset(dp, -1, sizeof dp);
    int k = calc(0, 0);
    ans = max(ans, k);
    if (k > 0)
      lo = mid + 1;
    else
      hi = mid - 1;
  }
  cout << ans;
  return 0;
}
