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;
using ll = int64_t;
using P = pair<int, int>;
template <class T, class U>
ostream& operator<<(ostream& os, const pair<T, U>& p) {
  os << "p"
     << " = (" << p.first << ", " << p.second << ")";
  return os;
}
template <class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  cout << "v"
       << " = [";
  for (auto i : v) os << " " << i;
  os << " ]";
  return os;
}
const int inf = 1e8;
int n;
vector<int> a;
vector<int> in[8];
int dp[1025][1025];
int can(int len) {
  vector<int> cur(8);
  for (int(i) = (0); (i) < (n + 1); (i)++) {
    for (int(j) = (0); (j) < (1 << 9); (j)++) {
      dp[i][j] = -inf;
    }
  }
  dp[0][0] = 0;
  for (int(i) = (0); (i) < (n); (i)++) {
    for (int(j) = (0); (j) < (1 << 8); (j)++) {
      if (dp[i][j] == -inf) {
        continue;
      }
      for (int(k) = (0); (k) < (8); (k)++) {
        if (j & (1 << k)) {
          continue;
        }
        int it = cur[k] + len - 1;
        if (it >= in[k].size()) {
          continue;
        }
        dp[in[k][it] + 1][j | (1 << k)] =
            max(dp[in[k][it] + 1][j | (1 << k)], dp[i][j]);
        it++;
        if (it >= in[k].size()) {
          continue;
        }
        dp[in[k][it] + 1][j | (1 << k)] =
            max(dp[in[k][it] + 1][j | (1 << k)], dp[i][j] + 1);
      }
    }
    cur[a[i]]++;
  }
  int ret = -inf;
  for (int(i) = (0); (i) < (n + 1); (i)++) {
    ret = max(ret, dp[i][(1 << 8) - 1]);
  }
  if (ret == -inf) {
    return -1;
  }
  return ret * (len + 1) + (8 - ret) * len;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  a.resize(n);
  for (int(i) = (0); (i) < (n); (i)++) {
    cin >> a[i];
    a[i]--;
    in[a[i]].emplace_back(i);
  }
  int l = 0, r = n / 8;
  while (r - l > 1) {
    int m = (l + r) / 2;
    if (can(m) == -1) {
      r = m;
    } else {
      l = m;
    }
  }
  int ans = max(can(l), can(r));
  if (ans == -1) {
    ans = 0;
    for (int(i) = (0); (i) < (8); (i)++) {
      if (in[i].size() != 0) {
        ans++;
      }
    }
  }
  cout << ans << endl;
}
