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;
int n;
int a[1005];
int dp[1005][(1 << 8) + 1];
int nxt[9][1005][1005];
int X;
int solve(int idx, int mask) {
  if (mask == (1 << 8) - 1) return 0;
  if (dp[idx][mask] != -1) return dp[idx][mask];
  int &ans = dp[idx][mask];
  ans = -100000000;
  for (int i = 0; i < 8; ++i) {
    if (mask & (1 << i)) continue;
    int pos1 = nxt[i][idx][X];
    if (pos1 != -1) ans = max(ans, solve(pos1 + 1, mask | (1 << i)));
    int pos2 = nxt[i][idx][X + 1];
    if (pos2 != -1) ans = max(ans, 1 + solve(pos2 + 1, mask | (1 << i)));
  }
  return ans;
}
bool possible() {
  memset(dp, -1, sizeof(dp));
  if (solve(0, 0) < 0)
    return false;
  else
    return true;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  set<int> s;
  int i, j, k;
  cin >> n;
  for (i = 0; i < n; ++i) {
    cin >> a[i];
    a[i]--;
    s.insert(a[i]);
  }
  for (i = 0; i < 8; ++i) {
    for (k = 0; k < 1005; ++k) {
      for (j = 0; j < 1005; ++j) nxt[i][j][k] = -1;
    }
  }
  for (i = n - 1; i >= 0; --i) {
    for (j = 0; j < 8; ++j) {
      if (a[i] == j)
        nxt[j][i][1] = i;
      else
        nxt[j][i][1] = nxt[j][i + 1][1];
      for (k = 2; k < 1005; ++k) {
        if (a[i] == j)
          nxt[j][i][k] = nxt[j][i + 1][k - 1];
        else
          nxt[j][i][k] = nxt[j][i + 1][k];
      }
    }
  }
  int best = 0;
  int lo = 0, hi = n;
  while (lo <= hi) {
    int mid = (lo + hi) >> 1;
    X = mid;
    if (possible()) {
      best = mid;
      lo = mid + 1;
    } else {
      hi = mid - 1;
    }
  }
  int ans = (int)s.size();
  if (best > 0) {
    memset(dp, -1, sizeof(dp));
    X = best;
    ans = 8 * X + solve(0, 0);
  }
  cout << ans << '\n';
  return 0;
}
