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 MAX = 1001;
int A[MAX], n;
int dp[MAX][1 << 8], len;
int vis[MAX][1 << 8];
vector<int> idx[9];
bool get_next(int pos, int len, int& x, vector<int>& V) {
  if (!len) {
    x = pos;
    return 1;
  } else {
    auto p = lower_bound(V.begin(), V.end(), pos) - V.begin();
    if (p + len <= V.size()) {
      x = V[p + len - 1];
      return 1;
    } else
      return 0;
  }
}
int solve(int pos, int msk) {
  if (!msk) return 0;
  if (vis[pos][msk]) return dp[pos][msk];
  vis[pos][msk] = 1;
  int &ret = dp[pos][msk], x;
  ret = !len ? 0 : -(len + 1100);
  for (int d = 0; d < (8); d++)
    if (msk & (1 << d)) {
      if (get_next(pos, len, x, idx[d]))
        ret = max(ret, solve(x + 1, msk ^ (1 << d)) + len);
      if (get_next(pos, len + 1, x, idx[d]))
        ret = max(ret, solve(x + 1, msk ^ (1 << d)) + len + 1);
    }
  return ret;
}
int f(int x) {
  len = x;
  memset(vis, 0, sizeof vis);
  return solve(0, (1 << 8) - 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < (n); i++) scanf("%d", A + i), idx[--A[i]].push_back(i);
  int s = 0, e = n;
  while (s < e) {
    int m = s + (e - s + 1) / 2;
    if (f(m) > 0)
      s = m;
    else
      e = m - 1;
  }
  s = 0;
  while (e - s > 1) {
    int m1 = s + (e - s + 1) / 3, m2 = m1 + (e - s + 1) / 3;
    if (f(m1) < f(m2))
      s = m1 + 1;
    else
      e = m2 - 1;
  }
  cout << max(f(s), f(e)) << endl;
  return 0;
}
