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>
inline void in(int &MAGNUS) { scanf("%d", &MAGNUS); }
inline void out(int MAGNUS) { printf("%d\n", MAGNUS); }
inline void in(int &MAGNUS, int &CLAR) { scanf("%d%d", &MAGNUS, &CLAR); }
inline void out(int MAGNUS, int CLAR) { printf("%d %d\n", MAGNUS, CLAR); }
inline void inl(long long &LIV) { scanf("%lld", &LIV); }
inline void inl(long long &LIV, long long &MART) {
  scanf("%lld%lld", &LIV, &MART);
}
inline void outl(long long LIV) { printf("%lld\n", LIV); }
inline void outl(long long LIV, long long MART) {
  printf("%lld %lld\n", LIV, MART);
}
using namespace std;
int N, x[1005], len, MAX = 0, memo[1005][(1 << 8) + 2], last[1005];
vector<int> idx[8];
int solve(int pos, int mask) {
  if (mask == (1 << 8) - 1) return 0;
  if (pos > N) return -1000000;
  if (memo[pos][mask] != -1) return memo[pos][mask];
  int ret = solve(pos + 1, mask), num = x[pos];
  for (int k = len; k <= len + 1; k++) {
    if (mask & (1 << num)) break;
    int next = last[pos] + k - 1;
    if (next < idx[num].size()) {
      ret = max(ret, k + solve(idx[num][next] + 1, mask | (1 << num)));
    }
  }
  return memo[pos][mask] = ret;
}
int main() {
  in(N);
  int ans = 0;
  for (int i = 1; i <= N; i++) {
    in(x[i]);
    x[i]--;
    int num = x[i];
    last[i] = idx[num].size();
    if (idx[num].size() == 0) ans++;
    idx[num].push_back(i);
  }
  int left = 1, right = (N / 8);
  while (left <= right) {
    len = (left + right) / 2;
    memset(memo, -1, sizeof(memo));
    int sol = solve(1, 0);
    if (sol > 0) {
      ans = max(ans, sol);
      left = len + 1;
    } else
      right = len - 1;
  }
  out(ans);
  return 0;
}
