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;
vector<vector<int> > psi(10), dp;
vector<int> a;
int n;
int valMx(int i, int eleg, int m) {
  if (eleg == (1 << 8) - 1 || (i == n && m == 0)) return 0;
  if (i >= n) return -1e9;
  if (dp[i][eleg] > -2e9) return dp[i][eleg];
  int mx = valMx(i + 1, eleg, m), k = a[i];
  if (!(eleg & (1 << k - 1))) {
    int l = lower_bound(psi[k].begin(), psi[k].end(), i) - psi[k].begin();
    if (l + m - 1 < psi[k].size())
      mx = max(mx, valMx(psi[k][l + m - 1] + 1, (eleg | (1 << k - 1)), m) + m);
    if (l + m < psi[k].size())
      mx = max(mx, valMx(psi[k][l + m] + 1, (eleg | (1 << k - 1)), m) + m + 1);
  }
  return dp[i][eleg] = mx;
}
int obtRes() {
  int i = 0, j = 1000, m = (i + j) / 2;
  for (int l = 0; l < 12; l++) {
    dp = vector<vector<int> >(n + 2, vector<int>(1 << 8, -2e9));
    if (valMx(0, 0, m) >= 0)
      i = m;
    else
      j = m;
    m = (i + j) / 2;
    dp = vector<vector<int> >(n + 2, vector<int>(1 << 8, -2e9));
  }
  return valMx(0, 0, m);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  a = vector<int>(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    psi[a[i]].push_back(i);
  }
  cout << obtRes() << '\n';
  return 0;
}
