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 K = 8, N = 1e3 + 7;
int v[N];
int urm[N][N][K];
int ok(int val) {
  int perm[K], cate(1);
  for (int i = 0; i < K; ++i) perm[i] = i, cate *= (i + 1);
  int ans(-1);
  for (int _ = 0; _ < cate; ++_) {
    for (int msk = 0; msk < (1 << K); ++msk) {
      int i(1);
      bool ok(0);
      for (int poz = 0; poz < K; ++poz) {
        if ((msk & (1 << perm[poz])) == 0 && val == 0) continue;
        i = urm[i][((msk & (1 << perm[poz])) != 0) + val][perm[poz]];
        if (i == 0) {
          ok = 1;
          break;
        }
      }
      if (!ok) {
        int potans(val * K);
        for (int poz = 0; poz < K; ++poz) potans += ((msk & (1 << poz)) != 0);
        ans = max(ans, potans);
      }
    }
    if (_ < cate - 1) next_permutation(perm, perm + K);
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> v[i];
    v[i]--;
    for (int j = i; j >= 1 && (v[j] != v[i] || j == i); --j)
      urm[j][1][v[i]] = i;
  }
  for (int k = 2; k < n; ++k)
    for (int i = 1; i <= n; ++i)
      for (int val = 0; val < K; ++val)
        if (urm[i][k - 1][val] != 0)
          urm[i][k][val] = urm[urm[i][k - 1][val] + 1][1][val];
  int pas = 1 << 6, r(0);
  while (pas) {
    if (r + pas <= n / 8 && ok(r + pas) != -1) r += pas;
    pas >>= 1;
  }
  cout << ok(r);
  return 0;
}
