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 inf = 100001;
int cnt[9];
vector<int> pos[9];
int closest[9][1001];
void prec() {
  for (int i = 1; i <= 8; ++i)
    for (int j = 0; j <= 1000; ++j) closest[i][j] = inf;
  for (int i = 1; i <= 8; ++i) {
    for (int j = (int)pos[i].size() - 1; j >= 0; --j) {
      int fin = 0;
      if (j != 0) fin = pos[i][j - 1];
      for (int k = pos[i][j] - 1; k >= fin; --k) closest[i][k] = j;
    }
    if (pos[i].size() != 0) closest[i][0] = 0;
  }
}
bool fl = 0;
int check(int len) {
  int ans = 0;
  vector<int> perm(8);
  for (int i = 0; i < 8; ++i) perm[i] = i + 1;
  do {
    for (int mask = 0; mask < 1 << 8; ++mask) {
      int cur_ans = 0;
      int start = 0;
      for (int i = 0; i < 8; ++i) {
        fl = 0;
        int cur = perm[i];
        int next_pos = closest[cur][start];
        int need = len + ((mask >> i) & 1);
        if (need == 0) continue;
        if (next_pos == inf) {
          cur_ans = 0;
          i = 9;
          break;
        }
        if (next_pos + need - 1 >= pos[cur].size()) {
          cur_ans = 0;
          i = 9;
          break;
        }
        cur_ans += need;
        start = pos[cur][next_pos + need - 1];
      }
      ans = max(ans, cur_ans);
    }
  } while (next_permutation(perm.begin(), perm.end()));
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  vector<int> vc(n + 1);
  for (int i = 1; i <= n; ++i) {
    cin >> vc[i];
    cnt[vc[i]]++;
    pos[vc[i]].push_back(i);
  }
  prec();
  int max_len = n + 1;
  for (int i = 1; i <= 8; ++i) max_len = min(max_len, cnt[i]);
  int left = 0, right = max_len + 1;
  while (right - left > 1) {
    int m = (left + right) / 2;
    if (check(m) != 0)
      left = m;
    else
      right = m;
  }
  cout << check(left) << endl;
  return 0;
}
