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, a[1010];
vector<int> g[9];
int dp[1010][256], offset[9];
void add(int &a, int b) { a = max(a, b); }
int check(int x) {
  for (int i = 0; i <= 8; i++) offset[i] = 0;
  memset(dp, -0x3f, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 256; j++)
      if (dp[i][j] >= 0) {
        for (int k = 0; k < 8; k++)
          if (!((1 << (k)) & j)) {
            int req_times = offset[k] + x - 1;
            if (req_times < g[k].size())
              add(dp[g[k][req_times]][j | (1 << (k))], dp[i][j] + x);
            if (++req_times < g[k].size())
              add(dp[g[k][req_times]][j | (1 << (k))], dp[i][j] + x + 1);
          }
      }
    offset[a[i]]++;
  }
  int ans = -212121212;
  for (int i = 1; i <= n; i++) ans = max(ans, dp[i][255]);
  if (ans < 0) return -1;
  return ans;
}
int main() {
  scanf("%d", &n);
  int ans = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    --a[i];
    g[a[i]].push_back(i);
  }
  for (int i = 0; i < 8; i++)
    if (g[i].size()) ans++;
  int l = 1, r = n + 1;
  for (int i = 0; i < 8; i++) r = min(r, (int)g[i].size());
  while (l <= r) {
    int mid = (l + r) >> 1;
    int p = check(mid);
    if (p != -1) {
      ans = p;
      l = mid + 1;
    } else
      r = mid - 1;
  }
  return printf("%d", ans), 0;
}
