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;
template <class T, class U>
void maximize(T &x, U y) {
  if (x < y) x = y;
}
template <class T, class U>
void minimize(T &x, U y) {
  if (x > y) x = y;
}
template <class T>
T Abs(T x) {
  return (x < (T)0 ? -x : x);
}
template <class T, class U, class V>
T addmod(T x, U y, V mod) {
  return ((x + y) % mod + mod) % mod;
}
template <class T, class U, class V>
T submod(T x, U y, V mod) {
  return ((x - y) % mod + mod) % mod;
}
template <class T, class U, class V>
T mulmod(T x, U y, V mod) {
  return (long long)x * y % mod;
}
namespace task {
const int inf = 1e9;
const int N = 1e3 + 3;
int dp[N][((1LL) << (9))], a[N], next[N];
int n, ans = 1;
int f(int i, int mask, int lim) {
  if (i >= n) {
    int nb = __builtin_popcount(mask);
    if (nb == 8) return 0;
    if (lim <= 1) return 0;
    return -inf;
  }
  if (~dp[i][mask]) return dp[i][mask];
  int ret = -inf;
  for (int p = i + 1, cnt = 1; p <= n; p = next[p], ++cnt) {
    if (!(((mask) >> (a[p])) & 1) and (cnt == lim or cnt == lim - 1))
      maximize(ret, cnt + f(p, mask | ((1LL) << (a[p])), lim));
    maximize(ret, f(p, mask, lim));
  }
  return dp[i][mask] = ret;
}
void solve() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i], --a[i];
  for (int i = 0; i < 8; ++i) {
    int p = n + 1;
    for (int j = n; j >= 1; --j)
      if (a[j] == i) next[j] = p, p = j;
  }
  int lo = 0, hi = n;
  for (int mid = lo + (hi - lo) / 2; lo <= hi; mid = lo + (hi - lo) / 2) {
    memset(dp, -1, sizeof(dp));
    int len = f(0, 0, mid);
    if (len >= 0)
      lo = mid + 1, maximize(ans, len);
    else
      hi = mid - 1;
  }
  cout << ans;
}
}  // namespace task
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  task::solve();
}
