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 a[1005];
int dp[1005][1 << 8];
bool vis[1005][1 << 8];
vector<int> pos[8];
int n;
int nextpos(int crt, int num, int k) {
  if (k == 0) return crt;
  const vector<int>& v = pos[num];
  int p = lower_bound(v.begin(), v.end(), crt) - v.begin();
  p = p + k - 1;
  if (p >= (int)v.size())
    return n;
  else
    return v[p];
}
int solve(int seglen) {
  memset(dp, 0, sizeof(dp));
  memset(vis, 0, sizeof(vis));
  for (int i = 0; i < n; ++i) vis[i][0] = true;
  int maxstate = (1 << 8) - 1;
  for (int i = 0; i < n; ++i) {
    for (int s = 0; s <= maxstate; ++s) {
      if (!vis[i][s]) continue;
      for (int num = 0; num < 8; ++num) {
        if (!(s & (1 << num))) {
          int p = nextpos(i, num, seglen);
          if (p < n) {
            dp[p][s | (1 << num)] = max(dp[p][s | (1 << num)], dp[i][s] + 1);
            vis[p][s | (1 << num)] = true;
          }
          p = nextpos(i, num, seglen - 1);
          if (p < n) {
            dp[p][s | (1 << num)] = max(dp[p][s | (1 << num)], dp[i][s]);
            vis[p][s | (1 << num)] = true;
          }
        }
      }
    }
  }
  int cnt = 0;
  for (int i = 0; i < n; ++i) cnt = max(cnt, dp[i][maxstate]);
  if (cnt == 0)
    return 0;
  else
    return (cnt * seglen) + (8 - cnt) * (seglen - 1);
}
int main(int argc, char* argv[]) {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    --a[i];
    pos[a[i]].push_back(i);
  }
  int maxlen = 0;
  for (int i = 0; i < 8; ++i) maxlen = max(maxlen, (int)pos[i].size());
  int ans = 0;
  for (int i = 0; i < 8; ++i) {
    if (pos[i].size() > 0) ++ans;
  }
  for (int len = 2; len <= maxlen; ++len) ans = max(ans, solve(len));
  printf("%d\n", ans);
  return 0;
}
