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>
const long long mod = 1000000007ll;
const double pi = acos(-1.0);
using namespace std;
int n;
int nxt[10][1009][1009];
int ar[1009];
int dp[1009][1 << 10];
int DP(int len, int idx, int msk) {
  if (idx == -1) return -mod;
  if (msk == ((1 << 8) - 1)) return 0;
  if (idx >= n) return -mod;
  if (dp[idx][msk] != -1) return dp[idx][msk];
  int ans = -mod;
  for (int i = 0; i < 8; i++) {
    if (msk & 1 << i) continue;
    ans = max(ans, len + DP(len, nxt[i][idx][len], msk | (1 << i)));
    ans = max(ans, len + 1 + DP(len, nxt[i][idx][len + 1], msk | (1 << i)));
  }
  return dp[idx][msk] = ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  ;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> ar[i];
    ar[i]--;
  }
  memset(nxt, -1, sizeof(nxt));
  for (int k = 0; k < 8; k++) {
    for (int i = 0; i < n; i++) {
      int cur = 0;
      nxt[k][i][0] = i;
      for (int j = i; j < n; j++) {
        if (ar[j] == k) {
          ++cur;
          nxt[k][i][cur] = j + 1;
        }
      }
    }
  }
  int l = 0, h = 1000;
  int ans = 1;
  while (l <= h) {
    int m = (l + h) >> 1;
    memset(dp, -1, sizeof(dp));
    int val = DP(m, 0, 0);
    if (val >= 0)
      l = ++m;
    else
      h = --m;
    ans = max(ans, val);
  }
  memset(dp, -1, sizeof(dp));
  ans = max(ans, DP(h, 0, 0));
  cout << ans << endl;
  return 0;
}
