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;
const int maxn = 1e3 + 10;
const int maxm = (1 << k) + 10;
int n, a[maxn], dp[maxn][maxm], ans, cur[k];
vector<int> p[8];
void init() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    a[i]--;
    p[a[i]].push_back(i);
  }
}
int get_ans(int len) {
  memset(cur, 0, sizeof(cur));
  memset(dp, -1, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 0; i < n; i++) {
    for (int mask = 0; mask < (1 << k); mask++) {
      if (dp[i][mask] == -1) continue;
      for (int v = 0; v < k; v++) {
        if (mask & (1 << v)) continue;
        int pos = cur[v] + len - 1;
        if (pos < p[v].size())
          dp[p[v][pos] + 1][mask | (1 << v)] =
              max(dp[p[v][pos] + 1][mask | (1 << v)], dp[i][mask]);
        pos++;
        if (pos < p[v].size()) {
          dp[p[v][pos] + 1][mask | (1 << v)] =
              max(dp[p[v][pos] + 1][mask | (1 << v)], dp[i][mask] + 1);
        }
      }
    }
    cur[a[i]]++;
  }
  int ret = -1;
  for (int i = 0; i <= n; i++) ret = max(ret, dp[i][(1 << k) - 1]);
  if (ret == -1) return ret;
  return ret * (len + 1) + (8 - ret) * len;
}
void solve() {
  int l = 1, r = n / k + 1;
  ans = -1;
  for (int times = 0; times <= 20; times++) {
    int mid = (l + r) / 2;
    ans = get_ans(mid);
    if (ans != -1) {
      l = mid;
    } else {
      r = mid;
    }
  }
  ans = max(get_ans(l), get_ans(r));
  if (ans == -1) {
    ans = 0;
    for (int i = 0; i < k; i++) {
      if (p[i].size() > 0) ans++;
    }
  }
  printf("%d\n", ans);
}
int main() {
  init();
  solve();
  return 0;
}
