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 n_max = 1010;
int F[n_max][1 << 8];
int N[n_max][8][n_max];
int A[n_max];
int n, m;
int n_test;
int get_bit(int x, int i) { return x & (1 << i); }
int add_bit(int x, int i) { return x | (1 << i); }
int cal(int x, int s) {
  if (s + 1 == (1 << 8)) return 0;
  if (F[x][s] >= 0) return F[x][s];
  int f = INT_MIN;
  for (int i = 0; i < 8; ++i)
    if (get_bit(s, i) == 0) {
      if (N[x][i][m] <= n) f = max(f, cal(N[x][i][m], add_bit(s, i)) + m);
      if (N[x][i][m + 1] <= n)
        f = max(f, cal(N[x][i][m + 1], add_bit(s, i)) + m + 1);
    }
  return F[x][s] = f;
}
int cal(int x) {
  m = x;
  fill(F[0], F[n + 1], -1);
  return cal(0, 0);
}
void enter() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> A[i], --A[i];
}
void solve() {
  for (int i = 0; i <= n; ++i)
    for (int j = 0; j < 8; ++j) {
      N[i][j][0] = i;
      for (int k = 1; k <= n + 1; ++k) {
        int iter = N[i][j][k - 1] + 1;
        while (iter <= n && A[iter] != j) ++iter;
        N[i][j][k] = iter;
      }
    }
  int l = 0, r = n;
  while (l < r) {
    int mid = l + (r - l) / 2;
    if (cal(mid + 1) >= 0)
      l = mid + 1;
    else
      r = mid;
  }
  cout << cal(l) << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  if (ifstream("../test.inp"))
    cin.rdbuf((new ifstream("../test.inp"))->rdbuf());
  n_test = 1;
  while (n_test--) {
    enter();
    solve();
  }
}
