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 mod = 1000000000 + 7;
int addm(int& a, int b) { return (a += b) < mod ? a : a -= mod; }
template <class T, class U>
bool smin(T& a, U b) {
  return a > b ? (a = b, 1) : 0;
}
template <class T, class U>
bool smax(T& a, U b) {
  return a < b ? (a = b, 1) : 0;
}
int n, a[1000];
pair<int, int> dp[2][1 << 8][8][130];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
  }
  fill(dp[0][0][0], dp[2][0][0], pair<int, int>{INT_MIN, INT_MIN});
  for (int i = 0; i < 8; i++) dp[0][0][i][0] = {INT_MAX, 0};
  int p = 0;
  for (int i = 0; i < n; i++, p = 1 - p)
    for (int mask = 0; mask < (1 << 8); mask++)
      for (int j = 0; j < 8; j++)
        for (int k = 0; k <= 125; k++) {
          if ((1 << j) & mask) continue;
          if (a[i] == j) {
            smax(dp[1 - p][mask][j][k + 1], dp[p][mask][j][k]);
          } else {
            pair<int, int>& cur = dp[p][mask][j][k];
            pair<int, int> nw;
            if (cur.first < k) {
              nw.first = cur.first;
              nw.second = cur.second + 1;
            } else if (cur.first == k) {
              nw.first = cur.first;
              nw.second = cur.second;
            } else {
              nw.first = k;
              nw.second = __builtin_popcount(mask);
            }
            smax(dp[1 - p][mask | (1 << j)][a[i]][1], nw);
          }
          smax(dp[1 - p][mask][j][k], dp[p][mask][j][k]);
        }
  for (int mask = 0; mask < (1 << 8); mask++)
    for (int j = 0; j < 8; j++)
      for (int k = 0; k <= 125; k++) {
        if ((1 << j) & mask) continue;
        pair<int, int>& cur = dp[p][mask][j][k];
        pair<int, int> nw;
        if (cur.first < k) {
          nw.first = cur.first;
          nw.second = cur.second + 1;
        } else if (cur.first == k) {
          nw.first = cur.first;
          nw.second = cur.second;
        } else {
          nw.first = k;
          nw.second = __builtin_popcount(mask);
        }
        smax(dp[p][mask | (1 << j)][0][0], nw);
      }
  pair<int, int> ans = dp[p][(1 << 8) - 1][0][0];
  long long M = ans.first * 8 + ans.second;
  for (int mask = 0; mask < (1 << 8); mask++) smax(M, dp[p][mask][0][0].second);
  cout << M << endl;
}
